In this article, we shall study to solve problems of probability based on the concept of the binomial distribution.

Example – 01:

An unbiased coin is tossed 5 times. Find the probability of getting a) three heads b) at least 4 heads

Solution:

In this case number of trials = n = 5, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 3 heads (X = 3):

∴ P(X = 3) = ⁵C₃ (1/2)³ (1/2)^{5 – 3}

∴ P(X = 3) = ⁵C₃ (1/2)³ (1/2)²

∴ P(X = 3) = 10 x (1/2)⁵ = 10 x (1/32) = 5/16 = 0.3125

The probability of getting at least 4 heads (X ≥ 4):

∴ P(X ≥ 4) = P(X = 4) + P(X = 5)

∴ P(X ≥ 4) = ⁵C₄ (1/2)⁴ (1/2)^{5 – 4} + ⁵C₅ (1/2)⁵ (1/2)^{5 – 5}

∴ P(X ≥ 4) = 5 x (1/2)⁴ (1/2)¹ + 1 x  (1/2)⁵ (1/2)⁰

∴ P(X ≥ 4) = 5 x (1/2)⁵ + 1 x  (1/2)⁵

∴ P(X ≥ 4) = 5 x (1/32) + 1 x  (1/32) = 6/32 = 3/16 = 0.1875

Ans: The probability of getting exactly 3 heads is 5/16 or 0.3125 and the probability of getting at least 4 heads is 3/16 or 0.1875

Example – 02:

An unbiased coin is tossed 8 times. Find the probability of getting head a) exactly 5 times, b) a larger number of times than the tail, and c) at least once.

Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = ⁸C₅ (1/2)⁵ (1/2)^{8 – 5}

∴ P(X = 5) =  56 x (1/2)⁵ (1/2)³

∴ P(X = 5) =  56 x (1/2)⁸ = 56 x (1/256) = 56/256 = 7/32 = 0.2188

The probability of getting more heads than tail (X ≥ 5):

∴ P(X ≥ 5) = P(X = 5) + P(X = 6)  + P(X = 7) + P(X = 8)

∴ P(X ≥ 5) =  ⁸C₅ (1/2)⁵ (1/2)^{8 – 5} +  ⁸C₆ (1/2)⁶ (1/2)^{8 – 6} +  ⁸C₇ (1/2)⁷ (1/2)^{8 – 7}  + ⁸C₈ (1/2)⁸ (1/2)^{8 – 8}

∴ P(X ≥ 5) =  56 x (1/2)⁵ (1/2)³ + 28 x (1/2)⁶ (1/2)² +  8 x (1/2)⁷ (1/2)¹  + 1 x  (1/2)⁸ (1/2)⁰

∴ P(X ≥ 5) =  56 x (1/2)⁸  + 28 x (1/2)⁸ +  8 x (1/2)⁸  + 1 x  (1/2)⁸

∴ P(X ≥ 5) = (56 + 28 + 8 + 1)x  (1/256) = 93/256 = 0.3633 

The probability of getting atleast one head (X ≥ 1):

∴ P(X ≥ 1) =  1 – P(X = 0)

∴ P(X ≥ 1) = 1 –   ⁸C₀ (1/2)⁰ (1/2)^{8 – 0}

∴ P(X ≥ 1) =  1 x (1/2)⁰ (1/2)⁸

∴ P(X ≥ 1) =  1 – 1/256 = 255/256 = 0.9961

Ans: The probability of getting exactly 5 heads is 7/32 or 0.2188

The probability of getting a head a larger number of times than the tail is 93/256 or 0.3633

The probability of getting atleast one head is 255/256 or 0.9961

Example – 03:

An unbiased coin is tossed 9 times. Find the probability of getting head a) exactly 5 times, b) in the first four tosses, and tails in the last five tosses.

Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = ⁹C₅ (1/2)⁵ (1/2)^{9 – 5}

∴ P(X = 5) =  126 x (1/2)⁵ (1/2)⁴

∴ P(X = 5) =  126 x (1/2)⁹ = 126 x (1/512) = 63/256 = 0.2461

The probability of gettingin head in first four tosses and tails in last five tosses :

∴ P(X) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/512 = 0.001953

Ans: The probability of getting exactly 5 heads is 63/256 or 0.2461 and the probability of getting head in the first four tosses and tails in the last five tosses is 1/512 or 0.001953

Example – 04:

If the chance that out of 10 telephone lines one of the line is busy at any instant is 0.2. What is the chance that 5 of the lines are busy?

Solution:

In this case number of lines = n = 10, Probability of getting line busy (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting 5 lines busy (X = 5):

∴ P(X = 5) = ¹⁰C₅ (0.2)⁵ (0.8)^{10 – 5}

∴ P(X = 5) = ¹⁰C₅ (0.2)⁵ (0.8)⁵

∴ P(X = 5) = 252 x  (0.2 x 0.8)⁵ = 252 x (0.16)⁵ = 0.0264

Example – 05:

Each of five questions on a multiple-choice examination has four choices, only one of which is correct. The student is attempting to guess the answers. The random variable X is the number of questions answer correctly. What is the probability that the student will get a) exactly three correct answers? b) atmost three correct answers? c) at least one correct answer.

Solution:

In this case number of trials = n = 5, Probability of getting correct answer (success) = 1/4

∴ p = 1/4 and q = 1 – p = 1 – 1/4 = 3/4

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of getting exactly 3 answers correct (X = 3):

∴ P(X = 3) = ⁵C₃ (1/4)³ (3/4)^{5 – 3}

∴ P(X = 3) = 10 x (1/4)³ (3/4)²

∴ P(X = 3) = 10 x  (1/64) (9/16) = 90/1024 = 45/512 = 0.0879

The probability of getting atmost 3 correct answers (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1)  + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = ⁵C₀ (1/4)⁰ (3/4)^{5 – 0} + ⁵C₁ (1/4)¹ (3/4)^{5 – 1}+ ⁵C₂ (1/4)² (3/4)^{5 – 2} +  ⁵C₃ (1/4)³ (3/4)^{5 – 3}

∴ P(X ≤ 3) = 1 x 1 x  (3/4)⁵+ 5 x  (1/4)¹ (3/4)⁴  + 10 x (1/4)² (3/4)³ + 10 x (1/4)³ (3/4)²

∴ P(X ≤ 3) = (243/1024) + 5 x  (1/4) x (81/256) + 10 x (1/16) (27/64) + 10 x (1/64) (9/16)

∴ P(X ≤ 3) = (243/1024) + (405/1024) + (270/1024) + (90/1024)

∴ P(X ≤ 3) = 1008/2024 = 63/64 = 0.9844

The probability of getting atleast 1 correct answers (X ≥ 1):

∴ P(X ≥ 1) = 1 – P(X = 0)

∴ P(X ≥ 1) = 1 –  ⁵C₀ (1/4)⁰ (3/4)^{5 – 0 3}

∴ P(X ≥ 1) = 1 –  1 x 1 x  (3/4)⁵

∴ P(X ≥ 1) = 1-  (243/1024) = 781/1024 = 0.7627

Ans: The probability of getting exactly 3 answers correct is 45/512 or 0.0879, the probability of getting atmost 3 correct answers is 63/64 or 0.9844, the probability of getting atleast 1 correct answer is 781/1024 or 0.7627

Example – 06:

The probability of hitting a target in any shot is 0.2. If 10 shots are fired, find the probability that the target will be heat atleast twice

Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability of hitting the target atleast twice (X ≥ 2):

∴ P(X ≥ 2) = 1 – { P(X = 0) + P(X = 1)}

∴ P(X ≥ 2) = 1 – { ¹⁰C₀ (0.2)⁰ (0.8)^{10 – 0} + ¹⁰C₁ (0.2)¹ (0.8)^{10 – 1}}

∴ P(X ≥ 2) = 1 – { 1 x  1 x  (0.8)¹⁰ + 10 x  (0.2) (0.8)⁹}

∴ P(X ≥ 2) = 1 – (0.8 + 2) (0.8)⁹ = 1 – (2.8) (0.8)⁹

∴ P(X ≥ 2) = 1 – 0.3758= 0.6242

Ans: The probability of hitting the target atleast twice is 0.6242

Example – 07:

The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly two will miss the target.

Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

Exactly two miss the target implies 8 bombs hit the target

The probability exactly two bombs miss the target  (X = 2):

∴ P(X = 8) = ¹⁰C₈ (0.8)⁸ (0.2)^{10 – 8}

∴ P(X = 8) = 45 x (0.8)⁸ (0.2)²

∴ P(X = 8) = 0.3020

Ans: The probability exactly two bombs miss the target is 0.3020

Example – 08:

In a town, 80% of all the families own a television set. If 10 families are interviewed at random, find the probability that a) seven families own a television set b) atmost three families own a television set.

Solution:

In this case number of trials = n = 10, Probability of family having atelevision st is 80% = 80/100 = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that 7 families have television (X = 7):

∴ P(X = 7) = ¹⁰C₇ (0.8)⁷ (0.2)^{10 – 7}

∴ P(X = 7) = 120 x (0.8)⁷ (0.2)³ = 0.2013

The probability that atmost 3 families have television (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = ¹⁰C₀ (0.8)⁰ (0.2)^{10 – 0} + ¹⁰C₁ (0.8)¹ (0.2)^{10 – 1} + ¹⁰C₂ (0.8)² (0.2)^{10 – 2} + ¹⁰C₃ (0.8)³ (0.2)^{10 – 3}

∴ P(X ≤ 3) = 1 x  1 x (0.2)¹⁰ + 10 x  (0.8) (0.2)⁹ + 45 x (0.8)² (0.2)⁸ + 120 x (0.8)³ (0.2)⁷

∴ P(X ≤ 3) = 0.0008644

Ans: The probability that 7 families have television is 0.2013 and the probability that atmost 3 families have television is 0.0008644

Example – 09:

The probability that a person who undergoes a kidney operation will recover is 0.7, Find the probability that of the six patients who undergo similar operations: a) none will recover, b) all will recover, c) half of them will recover and iv) aleast half will recover.

Solution:

In this case number of trials = n = 6, Probability of recovery after operation (success) = 0.7

∴ p = 0.7 and q = 1 – p = 1 – 0.7 = 0.3

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that none will recover (X = 0):

∴ P(X = 0) = ⁶C₀ (0.7)⁰ (0.3)^{6 – 0}

∴∴ P(X = 0) = 1 x 1 x (0.3)⁶  = 0.000729

The probability that all will recover (X = 6):

∴ P(X = 6) = ⁶C₆ (0.7)⁶ (0.3)^{6 – 6}

∴∴ P(X = 6) = 1 x (0.7)⁶ x 1 = 0.1176

The probability that halff of them will recover (X = 3):

∴ P(X = 3) = ⁶C₃ (0.7)³ (0.3)^{6 – 3}

∴ P(X = 3) = 20 x (0.7)³ (0.3)³ = 0.1852

The probability that atleast half of them will recover (X ≥ 3):

∴ P(X ≥ 3) = P(X = 3) + P(X = 4) + P(x = 5) + P(X = 6)

∴ P(X ≥ 3) = ⁶C₃ (0.7)³ (0.3)^{6 – 3}  +  ⁶C₄ (0.7)⁴ (0.3)^{6 – 4}  +  ⁶C₅ (0.7)⁵ (0.3)^{6 – 5}  +  ⁶C₆ (0.7)⁶ (0.3)^{6 – 6}

∴ P(X ≥ 3) = 20 x (0.7)³ (0.3)³  + 10 x (0.7)⁴ (0.3)²  +  6 x (0.7)⁵ (0.3)¹  + 1 x (0.7)⁶ (0.3)⁰

∴ P(X ≥ 3) = 0.9294

Ans: The probability that none will recover is 0.000729. The probability that all will recover is 0.00086441176. The probability that half of them will recover is 0.1852. The probability that atleast half of them will recover is 0.9294

Example – 10:

Centres for disease control have determined that when a person is given a vaccine, the probability that the person will develop immunity to a virus is 0.8. If eight people are given the vaccine, find the probability that a) none will develop immunity, b) exactly one will develop immunity, and c) all will develop immunity

Solution:

In this case number of trials = n = 8, Probability taht person develops immunity (success) = 0.78

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

The probability that none will develop immunity (X = 0):

∴ P(X = 0) = ⁸C₀ (0.8)⁰ (0.2)^{8 – 0}

∴ P(X = 0) = 1 x 1 x (0.2)⁸  = 0.00000256

The probability that exactly 4 will develop immunity (X = 4):

∴ P(X = 4) = ⁸C₄ (0.8)⁴ (0.2)^{8 – 4}

∴ P(X = 4) = 70 x (0.8)⁴ (0.2)⁴  = 0.04587

The probability that all will develop immunity (X = 8):

∴ P(X = 8) = ⁸C₈ (0.8)⁸ (0.2)^{8 – 8}

∴ P(X = 8) = 1 x (0.8)⁸ x 1 = 0.1678

Example – 11:

A machine has fourteen identical components that function independently. It will stop working if three or more components fail. If the probability that the component fails is 0.1. Find the probability that the machine will be working.

Solution:

In this case number of trials = n = 14, Probability that component fails (success) = 0.1

∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

Machine will stop working if three or more components fail.

Hence machine will be working if less than three components fail

The probability that machine is working (X < 3):

∴ P(X < 3) = P(X = 0) + P(X = 1) + P(x = 2)

∴ P(X < 3) = ¹⁴C₀ (0.1)⁰ (0.9)^{14 – 0} + ¹⁴C₁ (0.1)¹ (0.9)^{14 – 1} + ¹⁴C₂ (0.1)² (0.9)^{14 – 2}

∴ P(X < 3) = 1x 1 x  (0.9)¹⁴ + 14 x  (0.1)¹ (0.9)¹³ + 91 x  (0.1)² (0.9)¹²

∴ P(X < 3) = (1 x  (0.9)² + 14 x  0.1 x (0.9) + 91 x  (0.1)² )(0.9)¹²

∴ P(X < 3) = (0.81 + 1.26 + 0.91 )(0.9)¹²

∴ P(X < 3) =0.8416

Example – 12:

The probability that a person picked at random will support a constitutional amendment requiring an annual balanced budget is 0.8. If nine individuals are interviewed and they respond independently. What is the probability that at least two-thirds of them will support the amendment?

Solution:

In this case number of trials = n = 9, Probability that support the ammendment (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = ⁿC_r p^r q^{n – r}

atleast two third of nine i.e. atleast 6 supports the ammendment

The probability that two third support ammendment (X ≥ 6):

∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(x = 8) + P(x = 9)

∴ P(X ≥ 6) = ⁹C₆ (0.8)⁶ (0.2)^{9 – 6} + ⁹C₇ (0.8)⁷ (0.2)^{9 – 7} + ⁹C₈ (0.8)⁸ (0.2)^{9 – 8} + ⁹C₉ (0.8)⁹ (0.2)^{9 – 9}

∴ P(X ≥ 6) = 84 x (0.8)⁶ (0.2)³ + 36 x (0.8)⁷ (0.2)² + 9 x (0.8)⁸ (0.2)¹ + 1x (0.8)⁹ (0.2)⁰

∴ P(X ≥ 6) = 84 x (0.8)⁶ (0.2)³ + 36 x (0.8)⁷ (0.2)² + 9 x (0.8)⁸ (0.2)¹ + 1x (0.8)⁹ x 1

∴ P(X ≥ 6) = 0.9143

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Binomial Distribution appeared first on The Fact Factor.

In this article, we shall study to find the probability of an event when data normally distributed is given.

Area Under Normal Curve (0 < x< z)

Example – 01:

A sample of 100 dry battery cells tested to find the length of life produced the following results. Mean = μ = 12 hours, standard deviation = σ = 3 hours. Assuming that the data are normally distributed, what percentage of battery cells are expressed to have the life a) more than 15 hours, b) less than 6 hours, and c) between 10 hours and 14 hours. Given:

Solution:

We have Mean = μ = 12 hours, standard deviation = σ = 3 hours, Total number of objects = N = 100

P(life of battery more than 15 hours)

The standardized value of x = 15 is z = (x – μ)/σ

= (15 – 12)/3 = 3/3 = 1

P(z >1) = area under the standard normal curve to the right of z = 1

P(z >1) = (Area to the right of z = 0) – (Area between z = 0 and z = 1)

P(z >1) = 0.5 – 0.3413 = 0.1587

P(life of battery less than 6 hours)

The standardized value of x = 6 is z = (x – μ)/σ

= (6 – 12)/3 = – 6/3 = – 2

P(z < – 2) = area under the standard normal curve to the left of z = -2

P(z < -2) = (Area to the left of z = 0) – (Area between z = 0 and z = -2)

P(z >1) = 0.5 – 0.4772 = 0.0228

P(life of battery between 10 hours and 14 hours)

The standardized value of x = 10 is z = (x – μ)/σ

= (10 – 12)/3 = – 2/3 = – 0.67

The standardized value of x = 14is z = (x – μ)/σ

= (14 – 12)/3 =  2/3 =  0.67

P(- 0.67 < z < 0.67) = area under the standard normal curve between

z = 0.67 and z = – 0.67

P(- 0.67 < z < 0.67) = (Area between z = 0 and z = – 0.67) + (Area between z = 0 and z = 0.67)

P(- 0.67 < z < 0.67) = 2 x (Area between z = 0 and z = – 0.67)

= 2 x 0.2486 = 0.4972

Ans: 15.87% batteries have life more than 15 hours.

2.28% batteries have life less than 6 hours.

49.72% batteries have life between 10 hours and 14 hours.

Example – 02:

In a certain examination, 500 students appeared. Means score is 68 and SD 8. Assuming that the data are normally distributed find the number of students scoring a) less than 50 and b) more than 60.

Solution:

We have Mean = μ = 68, standard deviation = σ = 8, Total number of students = N = 500

P(marks less than 50)

The standardized value of x = 50 is z = (x – μ)/σ

= (50 – 68)/8 = – 18/8 = – 2.25

P(z  < – 2.25) = area under the standard normal curve to the left of z = – 2.25

P(z < – 2.25) = (Area to the right of z = 0) – (Area between z = 0 and z = -2.25)

P(z < – 2.5) = 0.5 – 0.4878 = 0.0122

Number of students got less than 50 marks = N x P(z < – 2.25)

= 500 x 0.0122 = 6 (Appox.)

P(marks more than 60)

The standardized value of x = 50 is z = (x – μ)/σ

= (60 – 68)/8 = – 8/8 = – 1

P(z  > – 1) = area under the standard normal curve to the right of z = – 1

P(z > – 1) = (Area to the right of z = 0) + (Area between z = 0 and z = -1)

P(z > – 1) = 0.5 + 0.3413 = 0.8413

Number of students got more than 60 marks = N x P(z > – 1)

= 500 x 0.8413 = 421 (Appox.)

Ans: Number of students got less than 50 marks are 6

Number of students got more than 60 marks are 421

Example – 03:

Sacks of sugar-packed by an automatic loader having an average weight of 100 kg and with a standard deviation of 0.250 kg. Assuming that the data are normally distributed, find the chance of sack weighing less than 99.5 kg.

Solution:

We have Mean = μ = 100 kg, standard deviation = σ = 0.250 kg

P(weight less than 99.5 kg)

The standardized value of x = 99.5 is z = (x – μ)/σ

= (99.5 – 100)/0.250 = – 0.5/0.250 = – 2

P(z  < – 2) = area under the standard normal curve to the left of z = – 2

P(z < – 2) = (Area to the right of z = 0) – (Area between z = 0 and z = -2)

P(z < – 2) = 0.5 – 0.4772 = 0.0228

Ans: The chance of sack weighing less than 99.5 kg is 0.0228 or 2.28%

Example – 04:

In a test 0f 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for a) between 1920 hours and 2160 hours and b) more than 2150 hours. Given A(2) = 0.4772, A(1.83) = 0.4664

Solution:

We have Mean = μ = 2040 hours, standard deviation = σ = 60 hours, Total number of bulbs = N = 2000

P(Working life between 1920 hours and 2160 hours)

The standardized value of x = 1920 is z = (x – μ)/σ

= (1920 – 2040)/60 = – 120/60 = – 2

The standardized value of x = 2160 is z = (x – μ)/σ

= (2160 – 2040)/60 = 120/60 = 2

P(- 2 < z < 2) = area under the standard normal curve between z = -2 and z = 2

P(- 2 < z < 2) = (Area between z = 0 and z = -2) + (Area between z = 0 and z = 2)

P(- 2 < z < 2)  = 2 x (Area between z = 0 and z = 2) = 2 x 0.4772 = 0.9544

Number of bulbs having life between 1920 hours and 2160 hours

= N x P(- 2 < z < 2)

= 2000 x 0.9544 =1909 (Appox.)

P(Working life morethan 2150 hours)

The standardized value of x = 2150 is z = (x – μ)/σ

= (2150 – 2040)/60= 110/60 = 1.83

P(z  > 1.83) = area under the standard normal curve to the right of z = 1.83

P(z > 1.83) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.83)

P(z > 1.83) = 0.5 – 0.4664 = 0.0336

Number of bulbs having life more than 2150 hours

= N x P(z > 1.83)

= 2000 x 0. 0336 = 67 (Appox.)

Ans: The number of bulbs having life between 1920 hours and 2160 hours is 1909 and the number of bulbs having life more than 2150 hours is 67

Example – 05:

In a test of 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for a) between 1920 hours and 2160 hours and b) more than 2150 hours. Given A(2) = 0.4772, A(1.83) = 0.4664

Solution:

We have Mean = μ = 2040 hours, standard deviation = σ = 60 hours, Total number of bulbs = N = 2000

P(Working life between 1920 hours and 2160 hours)

The standardized value of x = 1920 is z = (x – μ)/σ

= (1920 – 2040)/60 = – 120/60 = – 2

The standardized value of x = 2160 is z = (x – μ)/σ

= (2160 – 2040)/60 = 120/60 = 2

P(- 2 < z < 2) = area under the standard normal curve between z = -2 and z = 2

P(- 2 < z < 2) = (Area betweenz = 0 and z = -2) + (Area between z = 0 and z = 2)

P(- 2 < z < 2)  = 2 x (Area between z = 0 and z = 2) = 2 x 0.4772 = 0.9544

Number of bulbs having life between 1920 hours and 2160 hours

= N x P(- 2 < z < 2)

= 2000 x 0.9544 =1909 (Appox.)

P(Working life morethan 2150 hours)

The standardized value of x = 2150 is z = (x – μ)/σ

= (2150 – 2040)/60= 110/60 = 1.83

P(z  > 1.83) = area under the standard normal curve to the right of z = 1.83

P(z > 1.83) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.83)

P(z > 1.83) = 0.5 – 0.4664 = 0.0336

Number of bulbs having life more than 2150 hours

= N x P(z > 1.83)

= 2000 x 0. 0336 = 67 (Appox.)

Ans: The number of bulbs having life between 1920 hours and 2160 hours is 1909 and the number of bulbs having life more than 2150 hours is 67

Example – 06:

The scores of 1000 students have a mean 14 and standard deviation 2.5. Assuming the data to be normally distributed, find a) how many students secured marks between 12 and 15? and b) How many student score more than 18. Given A(0.8) 0.2882, A(0.4) 0.1554, A(1.6) = 0.4452.

Solution:

We have Mean = μ = 14, standard deviation = σ = 2.5, Total number of students = N = 1000

P(score between 12 and 15)

The standardized value of x = 12 is z = (x – μ)/σ

= (12 – 14)/2.5 = – 2/2.5 = – 0.8

The standardized value of x = 15 is z = (x – μ)/σ

= (15 – 14)/2.5 = 1/2.5 = 0.4

P(- 0.8 < z < 0.4) = area under the standard normal curve between z = – 0.8 and z = 0.4

P(- 0.8 < z < 0.4) = (Area betweenz = 0 and z = – 0.8) + (Area between z = 0 and z = 0.4)

P(- 0.8 < z < 0.4)  = 0.2881 + 0.1554 = 0.4435

Number of students who score between 12 and 14 

= N x P(- 0.8 < z < 0.4)

= 1000 x 0.4435 = 444 (Appox.)

P(Score more than 18)

The standardized value of x = 18 is z = (x – μ)/σ

= (18 – 14)/2.5= 4/2.5 = 1.6

P(z  > 1.6) = area under the standard normal curve to the right of z = 1.6

P(z > 1.6) = (Area to the right of z = 0) – (Area between z = 0 and z = 1.6)

P(z > 1.6) = 0.5 – 0.4452 = 0.0548

Number of students who score more than 18 

= N x P(z > 1.6)

= 1000 x 0. 0548 = 55 (Appox.)

Ans: The number of students who score between 12 and 14 is 444 and the number of students who score more than 18 is 55

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Probability: Normal Distribution 02 appeared first on The Fact Factor.

The normal distribution refers to a family of continuous probability distributions described by the normal equation.

on the domain x ∈ (- ∞, ∞)

where x is a normal random variable, μ is the mean, σ is the standard deviation,

Thus the normal distribution can be completely specified by two parameter mean (μ) and standard deviation (σ) and is represented as N(μ, σ).

Mathematicians called this distribution a normal distribution, a physicist called it a Gaussian distribution, and scientists called it a bell curve due to its bell-like shape.

The normal distribution with mean μ = 0 and standard deviation, σ = 1 is called the standard normal distribution. It is denoted by N(0, 1).

Characteristics of a Normal Distribution

• The normal curve is symmetrical about the mean μ. It is perfectly symmetrical around its center. That is, the right side of the center is a mirror image of the left side.
• The mean is at the middle and divides the area into halves. The center of a normal distribution is located at its peak, and 50% of the data lies above the mean, while 50% lies below. It means that the mean, median, and mode are all equal in a normal distribution.
• There is also only one mode, or peak, in a normal distribution.
• Normal distributions are continuous and have tails that are asymptotic.
• The total area under the curve is equal to 1;
• It is completely determined by its mean and standard deviation (SD) σ (or variance σ²)
• Approximately 68 % of the data lies within 1 SD of the mean. Approximately 95 % of the data lies within 2 SD of the mean.  Approximately 99.7 % of the data lies within 3 SD of the mean.

The Z – score:

The number of standard deviations from the mean is called the standard score or z – score.

An arbitrary normal distribution can be converted to a standard normal distribution by changing variables to z.

Empirical Rules for z – Scores:

• Approximately 68 % of the data lies within 1 SD of the mean. 
• Approximately 95 % of the data lies within 2 SD of the mean. 
• Approximately 99.7 % of the data lies within 3 SD of the mean.

Importance of z – Score:

Z-Scores tell us whether a particular score is equal to the mean, below the mean or above the mean of a bunch of scores. They can also tell us how far a particular score is away from the mean.

We can use Z-scores to standardize scores from different groups of data. Then we can compare raw scores from different groups of data.

Example – 01:

95 % of students at the college are between 1.1 m and 1.7 m tall. Find mean and the standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (1.1 + 1.7)/2 = 2.8/2 = 1.4 m

From empirical rule states that approximately 95 % of the data lies within 2 SD of the mean on either side.

Therefore the total deviation is 4 S.D.

4 S.D. = 1.7 – 1.1

4 S.D. = 0.6

1 S.D. = σ = 0.15 m

Ans: Mean = 1.4 m and the standard deviation is 0.15 m

Example – 02:

95 % of students in a class of 100 weigh between 62 kg and 90 kg. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (62 + 90)/2 = 152/2 = 76 kg

From empirical rule states that approximately 95 % of the data lies within 2 SD of the mean on either side.

Therefore the total deviation is 4 S.D.

4 S.D. = 90 – 62

4 S.D. = 28

1 S.D. = σ = 7 kg

Ans: Mean = 76 kg and standard deviation is 7 kg

Example – 03:

68 % of marks of students in a certain test are between 51 and 64. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (51 + 64)/2 = 115/2 = 57.5 kg

From empirical rule states that approximately 68 % of the data lies within 1 SD of the mean on either side.

Therefore the total deviation is 2 S.D.

2 S.D. = 64 – 51

2 S.D. = 13

1 S.D. = σ = 6.5 kg

Ans: Mean marks = 57.5 and standard deviation in marks is 6.5

Example – 04:

99.7 % of electrical components produced by a machine have lengths between 1.176 cm and 1.224 cm. Find mean and standard deviation assuming a normal distribution.

Solution:

Mean = μ = average of the two values given = (1.176 + 1.224)/2 = 2.4/2 = 1. 2 cm

From empirical rule states that approximately 99.7 % of the data lies within 3 SD of the mean on either side.

Therefore the total deviation is 6 S.D.

6 S.D. = 1.224 – 1.176

6 S.D. = 0.048

1 S.D. = σ = 0.008 cm

Ans: Mean length = 1.2 cm and standard deviation in length is 0.008 cm

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Probability: Normal Distribution appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of a single ball from a collection of identical but different coloured balls. e.g. An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. If one ball is drawn at random from the urn. Find the probability that of getting a red ball.

Problems based on the Draw of a Single Ball:

Example – 01:

An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. If one ball is drawn at random from the urn. Find the probability that

Solution:

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

one ball out of 20 can be drawn by ²⁰C₁ ways

Hence n(S) = ²⁰C₁ = 20

a) a red ball

Let A be the event of getting a red ball

There are 9 red balls in the urn

1 red ball out of 9 red balls can be drawn by ⁹C₁ ways

∴ n(A) = ⁹C₁  = 9

By the definition P(A) = n(A)/n(S) =9/20

Therefore the probability of getting a red ball is 9/20.

b) a white ball

Let B be the event of getting a white ball

There are 7 white balls in the urn

1 white ball out of 7 white balls can be drawn by ⁷C₁ ways

∴ n(B) = ⁷C₁  = 7

By the definition P(B) = n(B)/n(S) =7/20

Therefore the probability of getting a white ball is 7/20.

c) a black ball

Let C be the event of getting a black ball

There are 4 black balls in the urn

1 black ball out of 4 black balls can be drawn by ⁴C₁ ways

∴ n(C) = ⁴C₁  = 4

By the definition P(C) = n(C)/n(S) = 4/20 = 1/5

Therefore the probability of getting a black ball is 1/5.

d) not a red ball

Let D be the event of getting not a red ball

There are 7 + 4 = 11 non-red balls

1 non-red ball out of 11 non-red balls can be drawn by ¹¹C₁ ways

∴ n(D) = ¹¹C₁  = 11

By the definition P(D) = n(D)/n(S) = 11/20

Therefore the probability of getting not red ball is 11/20

e) not a white ball

Let E be the event of getting not a white ball

There are 9 + 4 = 13  non-white balls

1 non-white ball out of 13 non-white balls can be drawn by ¹³C₁ ways

∴ n(E) = ¹³C₁  = 13

By the definition P(E) = n(E)/n(S) = 13/20

Therefore the probability of getting not white ball is 13/20

f) not a black ball

Let F be the event of getting not a black ball

There are 9 + 7 = 16 non-black balls

1 non-black ball out of 16 non-black balls can be drawn by ¹⁶C₁ ways

∴ n(F) = ¹⁶C₁  = 16

By the definition P(F) = n(F)/n(S) = 16/20 = 4/5

Therefore the probability of getting not a black ball is 4/5.

g) a red ball or a black ball

Let G be the event of getting not a red ball

There are 9 + 4 = 13 red or black balls

1 red or a  black ball out of 13 can be drawn by ¹³C₁ ways

∴ n(G) = ¹³C₁  = 13

By the definition P(G) = n(G)/n(S) = 13/20

Therefore the probability of getting a red ball or a black ball is 13/20.

h) a red ball or a white ball

Let H be the event of getting a red ball or a white ball

There are 9 + 7 = 16 red or white balls

1 red or a  white ball out of 16 can be drawn by ¹⁶C₁ ways

∴ n(H) = ¹⁶C₁  = 16

By the definition P(H) = n(H)/n(S) = 16/20 = 4/5

Therefore the probability of getting a red ball or a white ball is 4/5.

i) a black ball or a white ball

Let J be the event of getting a black ball or a white ball

There are 4 + 7 = 11 black or white balls

1 black or a  white ball out of 11 can be drawn by ¹¹C₁ ways

∴ n(J) = ¹¹C₁  = 11

By the definition P(J) = n(J)/n(S) = 11/20

Therefore the probability of getting a black ball or a white ball is 11/20.

Example – 02:

An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. Two balls are drawn at random from the urn. Find the probability that

Solution:

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

Two balls out of 20 can be drawn by ²⁰C₂ ways

Hence n(S) = ²⁰C₂ = 10 x 19

a) both red balls

Let A be the event of getting both red balls

There are 9 red balls in the urn

2 red balls out of 9 red balls can be drawn by ⁹C₂ ways

∴ n(A) = ⁹C₂  = 9 x 4

By the definition P(A) = n(A)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting both red balls is 18/95

b) no red ball

Let B be the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(B) = ¹¹C₂  = 11 x 5

By the definition P(B) = n(B)/n(S) = (11 x 5)/(10 x 19) = 11/38

Therefore the probability of getting no red ball is 11/76

c) atleast one red ball

Let C be the event of getting atleast one red ball

Hence C is the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 non-red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(C) = ¹¹C₂  = 11 x 5

By the definition P(C) = n(C)/n(S) = (11 x 5)/(10 x 19) = 11/38

Now, P(C) = 1 – P(C) = 1 – 11/38 = 27/38

Therefore the probability of getting atleast one red ball is 27/38

d) exactly one red ball

Let D be the event of getting exactly one red ball i.e. one red and 1 non red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(D) = ⁹C₁  x  ¹¹C₁  = 9  x 11

By the definition P(D) = n(D)/n(S) = (9 x 11)/(10 x 19) = 99/190

Therefore the probability of getting exactly one red ball is 99/190.

e) at most one red ball

Let E be the event of getting at most one red ball There are two possibilities

Case – 1: Getting no red ball and two non-red balls or

Case – 2: Getting 1 red ball and 1 non-red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

∴ n(E) = ⁹C₀  x  ¹¹C₂ +  ⁹C₁  x  ¹¹C₁ = 1 x 11 x 5 + 9 x 11 = 55 + 99 = 154

By the definition P(E) = n(E)/n(S) = 154/(10 x 19) = 77/95

Therefore the probability of getting at most one red ball is 77/95.

f) one is red and other is white

Let F be the event of getting one red and one white ball

There are 9 red and 7 white balls in the urn

∴ n(F) = ⁹C₁  x  ⁷C₁ = 9 x 7

By the definition P(F) = n(F)/n(S) = (9 x 7)/(10 x 19) = 63/190

Therefore the probability of getting one red and other white ball is 63/190

g) one is red and other is black

Let G be the event of getting one red and one black ball

There are 9 red and 4 black balls in the urn

∴ n(G) = ⁹C₁  x  ⁴C₁ = 9 x 4

By the definition P(G) = n(G)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting one red and other black ball is 18/95

h) one is white and the other is black

Let H be the event of getting one white and one black ball

There are 7 white and 4 black balls in the urn

∴ n(H) = ⁷C₁  x  ⁴C₁ = 7 x 4

By the definition P(H) = n(H)/n(S) = (7 x 4)/(10 x 19) = 14/95

Therefore the probability of getting one white and other black ball is 14/95.

i) both are of same colour

Let J be the event of getting balls of same colour.

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(J) = ⁹C₂  +  ⁷C₂ +  ⁴C₂ = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(J) = n(J)/n(S) = 63/(10 x 19) = 63/190

Therefore the probability of getting both balls of same colour is 63/190

j) both are of not of the same colour

Let K be the event of getting balls not of the same colour.

Hence K is the event of getting the balls of same colour

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(K) = ⁹C₂  +  ⁷C₂ +  ⁴C₂ = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(K) = n(K)/n(S) = 63/(10 x 19) = 63/190

Now, P(K) = 1 – P(K) = 1 – 63/190 = 127/190

Therefore the probability of getting both balls of not of the same colour is 127/190.

k) both are red or both are black

Let L be the event of getting both red or both black balls.

There are 9 red and 4 black balls in the urn

∴ n(L) = ⁹C₂ +  ⁴C₂ = 9 x 4 + 2 x 3 = 36 + 6 = 42

By the definition P(L) = n(L)/n(S) = 42/(10 x 19) = 21/95

Therefore the probability of getting both red or both black balls is 21/95.

In the next article, we shall study some basic problems of probability based on the drawing of two or more balls from a collection of identical coloured balls.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Selection of Balls appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of four or more playing cards. For e.g. Four cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting all the cards of the same suite.

Problems based on the Draw of 4 Playing Cards:

Example – 01:

Four cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Four cards out of 52 can be drawn by ⁵²C₄ ways

Hence n(S) = ⁵²C₄

a) all are heart cards

Let A be the event of getting all heart cards

There are 13 heart cards in a pack

four heart cards out of 13 heart cards can be drawn by ¹³C₄ ways

∴ n(A) = ¹³C₄

By the definition P(A) = n(A)/n(S) = (¹³C₄ )/(⁵²C₄)

Therefore the probability of getting all heart cards is 

(¹³C₄ )/(⁵²C₄)

b) all the cards are of the same suite

Let B be the event of getting all the cards of the same suite

There are 4 suites in a pack

There are 13 cards in each suite

four cards of the same suite out of 13 cards of same suite can be drawn by ¹³C₄ ways

∴ n(B) = 4 x ¹³C₄

By the definition P(B) = n(B)/n(S) = (4 x ¹³C₄ )/(⁵²C₄)

Therefore the probability of getting all the cards of the same suite is 

(4 x ¹³C₄ )/(⁵²C₄)

c) all the cards of the same colour

Let C be the event of getting all the cards of the same colour.

There are 26 red and 26 black cards in a pack

Thus the selection is all red or all black.

∴ n(C) = ²⁶C₄ + ²⁶C₄ = 2(²⁶C₄)

By the definition P(C) = n(C)/n(S) = 2(²⁶C₄)/(⁵²C₄)

Therefore the probability of getting all the cards of the same colour is

2(²⁶C₄)/(⁵²C₄)

d) all the face cards

Let D be the event of getting all the face cards.

There are 12 face cards in a pack

∴ n(D) = ¹²C₄

By the definition P(D) = n(D)/n(S) = ¹²C₄/(⁵²C₄)

Therefore the probability of getting all face cards is

¹²C₄/(⁵²C₄)

e) all the cards are of the same number (denomination)

Let E be the event of getting all the cards of the same number

there are 4 cards of the same denomination in a pack and 1 in each suite.

There are such 13 sets

four  cards of the same number out of 4 cards can be drawn by ⁴C₄ ways

∴ n(E) = 13 x ⁴C₄  = 13 x 1 = 13

By the definition P(E) = n(E)/n(S) = (13)/(⁵²C₄)

Therefore the probability of getting all the cards of the same suite is 

(13)/(⁵²C₄)

f) Two red cards and two black cards

Let F be the event of getting two red cards and two black cards

There are 26 red and 26 black cards in a pack

∴ n(F) = (²⁶C₂) x (²⁶C₂) = (²⁶C₂)²

By the definition P(F) = n(F)/n(S) = (²⁶C₂)²/(⁵²C₄)

Therefore the probability of getting two red cards and two black cards is

(²⁶C₂)²/(⁵²C₄)

g) all honours of the same suite

Let G be the event of getting honours of the same suite

There are 4 honours (ace, king, queen, and jack) in a suite.

There are four suites

∴ n(G) = (⁴C₄) + (⁴C₄) + (⁴C₄) + (⁴C₄) = 1 + 1 + 1 + 1 = 4

By the definition P(G) = n(G)/n(S) = 4/(⁵²C₄)

Therefore the probability of getting all honours of the same suite is

4/(⁵²C₄)

h) atleast one heart

Let H be the event of getting at least one heart

Thus H’ is an event of getting no heart

Thus there are 39 non-heart cards in a pack

four non-heart cards out of 39 non-heart cards can be drawn by ³⁹C₄ ways

∴ n(H’) = ³⁹C₄

By the definition P(H’) = n(H’)/n(S) = ³⁹C₄/⁵²C₄

Now P(H) = 1 – P(H’) = 1 – (³⁹C₄/⁵²C₄)

Therefore the probability of getting at least one heart is

(1 – (³⁹C₄/⁵²C₄))

i) 3 kings and 1 jack

Let J be the event of getting 3 kings and 1 jack

There are 4 kings and 4 jacks in a pack

∴ n(J) = (⁴C₃ x ⁴C₁)

By the definition P(J) = n(J)/n(S) = (⁴C₃ x ⁴C₁)/(⁵²C₄)

Therefore the probability of getting 3 kings and one jack is

(⁴C₃ x ⁴C₁)/(⁵²C₄)

j) all clubs and one of them is a jack

Let K be the event of getting all clubs and one of them is a jack

There are 12 club  cards + 1 club jack i.e. total 13 club cards

∴ n(K) = (¹²C₃ x ¹C₁) = ¹²C₃

By the definition P(K) = n(K)/n(S) = (¹²C₃)/(⁵²C₄)

Therefore the probability of getting all clubs and one of them is a jack is

(¹²C₃)/(⁵²C₄)

k) 3 diamonds and 1 spade

Let L be the event of getting 3 diamonds and 1 spade

There are 13 diamond cards and 13 spade cards in a pack

∴ n(L) = (¹³C₃ x ¹³C₁)

By the definition P(L) = n(L)/n(S) = (¹³C₃ x ¹³C₁)/(⁵²C₄)

Therefore the probability of getting 3 diamonds and 1 spade is 

(¹³C₃ x ¹³C₁)/(⁵²C₄)

Example – 02:

In a shuffling a pack of 52 cards, four are accidentally dropped, find the probability that the missing cards should be one from each suite.

Solution:

There are 52 cards in a pack.

Four cards out of 52 can be drawn by ⁵²C₄ ways

Hence n(S) = ⁵²C₄

Let A be the event of getting one card from each suite

There are 13 cards in each suite.

∴ n(A) = (¹³C₁) x (¹³C₁) x (¹³C₁) x (¹³C₁)x = (¹³C₁)⁴

By the definition P(A) = n(A)/n(S) = (¹³C₁)⁴/(⁵²C₄)

Therefore the probability of getting one card from each suite is

(¹³C₁)⁴/(⁵²C₄)

Example – 03:

Five cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Five cards out of 52 can be drawn by ⁵²C₅ ways

Hence n(S) = ⁵²C₅

a) just one ace

Let A be the event of getting just one ace

There are 4 aces and 48 non-aces in a pack

∴ n(A) = ⁴C₁ x ⁴⁸C₄  = 4 x ⁴⁸C₄

By the definition P(A) = n(A)/n(S) = (4 x ⁴⁸C₄ )/(⁵²C₅)

Therefore the probability of getting all heart cards is

(4 x ⁴⁸C₄ )/(⁵²C₅)

b) atleast one ace

Let B be the event of getting atleast one ace

Hence B’ is the event of getting no ace

There are 4 aces and 48 non-aces in a pack

∴ n(B’) =  ⁴⁸C₅

By the definition P(B’) = n(B’)/n(S) = (⁴⁸C₅ )/(⁵²C₅)

Now P(B) = 1 – P(B’) = 1 –  (⁴⁸C₅ )/(⁵²C₅)

Therefore the probability of getting atleast one ace is

(1 –  (⁴⁸C₅ )/(⁵²C₅))

c) all cards are of hearts

Let C be the event of getting all hearts

There are 13 heart cards in a pack

∴ n(C) =  ¹³C₅

By the definition P(C) = n(C)/n(S) = (¹³C₅ )/(⁵²C₅)

Therefore the probability of getting all hearts is

(¹³C₅ )/(⁵²C₅)

Example – 04:

What is the probability of getting 9 cards of the same suite in one hand at a game of bridge?

Solution:

There are 52 cards in a pack.

In a game of bridge, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let B be the event of getting 9 cards of the same suite in one hand

There are 4 suites, thus the suite can be selected by ⁴C₁ ways = 4 ways

Now, in hand, there are 9 cards of same suite and 4 cards of other suites.

∴ n(B) = (⁴C₁) x (¹³C₉) x (³⁹C₄) =  4 x (¹³C₉) x (³⁹C₄)

By the definition P(B) = n(B)/n(S) = (4 x (¹³C₉) x (³⁹C₄) )/(⁵²C₄)

Therefore the probability of getting 9 cards of the same suite in one hand is

(4 x (¹³C₉) x (³⁹C₄) )/(⁵²C₄)

Example – 05:

What is the probability of getting 9 cards of the spade in one hand at a game of bridge?

Solution:

There are 52 cards in a pack.

In a game of bridge, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let C be the event of getting 9 cards of spade in one hand

Now, in hand, there are 9 cards of spade and 4 cards are non-spade.

∴ n(C) = (¹³C₉) x (³⁹C₄) =  (¹³C₉) x (³⁹C₄)

By the definition P(C) = n(C)/n(S) = (¹³C₉ x ³⁹C₄)/(⁵²C₄)

Therefore the probability of getting 9 cards of spade in one hand is

(¹³C₉ x ³⁹C₄)/(⁵²C₄)

Example – 06:

In a hand at whist, what is the probability that four kings are held by a specified player?

Solution:

There are 52 cards in a pack.

In a game, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let D be the event that four kings are held by a specified player

A particular player can be chosen by 1 way

Now, in hand, there are  4 kings and 48 non-king cards

∴ n(D) = (1) x (⁴C₄) x (⁴⁸C₉) =  ⁴⁸C₉

By the definition P(D) = n(D)/n(S) = (⁴⁸C₉ )/(⁵²C₄)

Therefore the probability of that four kings are held by a specified player is  (⁴⁸C₉ )/(⁵²C₄)

Example – 07:

The face cards are removed from a full pack. Out of 40 remaining cards, 4 are drawn at random. find the probability that the selection contains one card from each suite.

Solution:

There are 52 cards in a pack.

There are 12 face cards which are removed. Thus 40 cards remain

Four cards out of 40 can be drawn by ⁴⁰C₄ ways

Hence n(S) = ⁴⁰C₄

Let E be the event of getting one card from each suite

There are 10 cards in each suite.

∴ n(E) = (¹⁰C₁) x (¹⁰C₁) x (¹⁰C₁) x (¹⁰C₁)x = (¹⁰C₁)⁴

By the definition P(E) = n(E)/n(S) = (¹⁰C₁)⁴/(⁴⁰C₄)

Therefore the probability of getting one card from each suite is (¹⁰C₁)⁴/(⁴⁰C₄)

Example 08:

Find the probability that when a hand of 7 cards is dealt from a well-shuffled deck of 52 cards, it contains

Solution:

There are 52 cards in a pack.

Seven cards out of 52 can be drawn by ⁵²C₇ ways

Hence n(S) = ⁵²C₇

a) all 4 kings

Let A be the event of getting all 4 kings i.e. 4 kings and 3 non-king cards.

There are 4 kings and 48 non-king cards in a pack

∴ n(A) = ⁴C₄ x ⁴⁸C₃  

By the definition P(A) = n(A)/n(S) = (⁴C₄ x ⁴⁸C₃ )/(⁵²C₇)

Therefore the probability of getting all 4 kings is

(⁴C₄ x ⁴⁸C₃ )/(⁵²C₇)

b) exactly 3 kings

Let B be the event of getting exactly 3 kings i.e. 3 kings and 4 non-king cards.

There are 4 kings and 48 non-king cards in a pack

∴ n(A) = ⁴C₃ x ⁴⁸C₄  

By the definition P(A) = n(A)/n(S) = (⁴C₃ x ⁴⁸C₄ )/(⁵²C₇)

Therefore the probability of getting all 4 kings is

(⁴C₃ x ⁴⁸C₄ )/(⁵²C₇)

c) at least three kings

Let C be the event of getting at least three kings

There are two possibilities

Case – 1: Getting 3 kings and 4 non-king cards

Case – 2: Getting 4 kings and 3 non-king cards

There are 4 kings and 48 non-king cards in a pack

∴ n(C) = ⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃ 

By the definition P(C) = n(C)/n(S) = (⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃) / (⁵²C₇)

Therefore the probability of getting at least two face cards is

(⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃) / (⁵²C₇)

In the next article, we shall study some basic problems of probability based on the drawing of a single ball from a collection of identical coloured balls.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 4 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of three playing cards. For e.g. three cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting all red cards

Problems based on the Draw of 3 Playing Cards:

Example – 01:

Three cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Three cards out of 52 can be drawn by ⁵²C₃ ways

Hence n(S) = ⁵²C₃ = 26 x 17 x 50

a) all face cards

Let A be the event of getting all face cards

There are 12 face cards in a pack

three face cards out of 12 can be drawn by ¹²C₃ ways

∴ n(A) = ¹²C₃ = 4 x 11 x 5

By the definition P(A) = n(A)/n(S) = (4 x 11 x 5)/( 26 x 17 x 50) = 11/1105

Therefore the probability of getting all face cards is 11/1105

b) no face card

Let B be the event of getting no face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by ⁴⁰C₃ ways

∴ n(B) = ⁴⁰C₃ = 20 x 13 x 38

By the definition P(B) = n(B)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85

Therefore the probability of getting no face card is 38/85.

c) atleast one face card

Let C be the event of getting at least one face card

Thus C is an event of getting no face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by ⁴⁰C₃ ways

∴ n(C) = ⁴⁰C₃ = 20 x 13 x 38

By the definition P(C) = n(C)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85

Now P(C) = 1 – P(C) = 1 – 38/85 = 47/85

Therefore the probability of getting at least one face card is 47/85.

d) at least two face cards

Let D be the event of getting at least two face cards

There are two possibilities

Case – 1: Getting two face cards and 1 non-face card

Case – 2: All three face cards

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(D) = ¹²C₂ x  ⁴⁰C₁ + ¹²C₃  = 6 x 11 x 40 + 4 x 11 x 5 = 2860

By the definition P(D) = n(D)/n(S) = 3794/( 26 x 17 x 50) = 11/85

Therefore the probability of getting at least two face cards is 11/85.

e) at most two face cards

Let E be the event of getting at most two face cards

There are three possibilities

Case – 1: Getting no face card

Case – 2: Getting one face card and 2 non-face cards

Case – 3: Getting two face cards and 1 non-face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(E) = ⁴⁰C₃ +¹²C₁ x  ⁴⁰C₂ +  ¹²C₂ x  ⁴⁰C₁ = 20 x 13 x 38 + 12 x 20 x 39 + 6 x 11 x 40 = 21880

By the definition P(E) = n(E)/n(S) = 21880/( 26 x 17 x 50) = 1094/1105

Therefore the probability of getting at most two face cards is 1094/1105

f) all red cards

Let F be the event of getting all red cards

There are 26 red cards in a pack

three red cards out of 26 red cards can be drawn by ²⁶C₃ ways

∴ n(E) = ²⁶C₃ = 4 x 11 x 5 = 13 x 25 x 8

By the definition P(E) = n(E)/n(S) = (13 x 25 x 8)/( 26 x 17 x 50) = 2/17

Therefore the probability of getting all red cards is 2/17

f) all are not heart

Let F be the event of getting draw such that all are not heart

Thus F is the event that the draw consists of atmost two heart

There are three possibilities

Case – 1: Getting no heart

Case – 2: Getting one heart and 2 non hearts

Case – 3: Getting two hearts and 1 non heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

∴ n(F) = ³⁹C₃ +¹³C₁ x  ³⁹C₂ +  ¹³C₂ x  ³⁹C₁ = 13 x 19 x 37 + 13 x 39 x 19 + 13 x 6 x 39 = 21814

By the definition P(F) = n(F)/n(S) = 21814/( 26 x 17 x 50) = 839/850

Therefore the probability of getting all not heart is 839/850

g) atleast one heart

Let G be the event of getting at least one heart

Thus G’ is an event of getting no heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

three non-heart cards out of 39 non-heart cards can be drawn by ³⁹C₃ ways

∴ n(G’) = ³⁹C₃ = 13 x 19 x 37

By the definition P(G’) = n(G’)/n(S) = (13 x 19 x 37)/( 26 x 17 x 50) = 703/1700

Now P(G) = 1 – P(G’) = 1 – 703/1700 = 997/1700

Therefore the probability of getting at least one heart is 997/1700

h) a king,  a queen, and a jack

Let H be the event of getting a king,  a queen, and a jack

There are 4 kings, 4 queens and 4 jacks in a pack

Each specific selection can be done by ⁴C₁ ways

∴ n(H) = ⁴C₁ x ⁴C₁ x ⁴C₁ = 4 x 4 x 4 = 64

By the definition P(H) = n(H)/n(S) =64/( 26 x 17 x 50) = 16/5525

Therefore the probability of getting a king,  a queen and a jack is 16/5525

i) 2 aces and 1 king

Let J be the event of getting two aces and 1 king

There are 4 aces and 4 kings

∴ n(J) = ⁴C₂ x ⁴C₁  = 6 x 4  = 24

By the definition P(J) = n(J)/n(S) =24/( 26 x 17 x 50) = 6/5525

Therefore the probability of getting two aces and one king is 6/5525

In the next article, we shall study some basic problems of probability based on the drawing of four or more cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 3 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of two playing cards. For e.g. Two cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting both red cards

Problems based on the Draw of 2 Playing Cards:

Example – 01:

Two cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

two cards out of 52 can be drawn by ⁵²C₂ ways

Hence n(S) = ⁵²C₂ = 26 x 51

a) both club cards

Let A be the event of getting both club cards

There are 13 club cards in a pack

two club cards out of 13 club cards can be drawn by ¹³C₂ ways

∴ n(A) = ¹³C₂ =  13 x 6

By the definition P(A) = n(A)/n(S) = (13 x 6)/(26 x 51) = 1/17

Therefore the probability of getting both club cards is 1/17

b) both red cards

Let B be the event of getting both red cards

There are 26 red cards in a pack

two red cards out of 26 red cards can be drawn by ²⁶C₂ ways

∴ n(B) = ²⁶C₂ =  13 x 25

By the definition P(B) = n(B)/n(S) = (13 x 25)/(26 x 51) = 25/102

Therefore the probability of getting both red cards is 25/102

c) both black cards

Let C be the event of getting both black cards

There are 26 black cards in a pack

two black cards out of 26 black cards can be drawn by ²⁶C₂ ways

∴ n(C) = ²⁶C₂ =  13 x 25

By the definition P(C) = n(C)/n(S) = (13 x 25)/(26 x 51) = 25/102

Therefore the probability of getting both black cards is 25/102

d) both kings

Let D be the event of getting both kings

There are 4 kings in a pack

two kings out of four kings can be drawn by ⁴C₂ ways

∴ n(D) = ⁴C₂ =  2 x 3

By the definition P(D) = n(D)/n(S) = (2 x 3)/(26 x 51) = 1/221

Therefore the probability of getting both kings is 1/221

Note: The probability of getting two cards of a particular denomination is always 1/221

e) both red aces

Let E be the event of getting both red aces

There are 2 red aces in a pack

two red aces out of two red aces can be drawn by ²C₂ ways

∴ n(E) = ²C₂ =  1

By the definition P(E) = n(E)/n(S) = (1)/(26 x 51) = 1/1326

Therefore the probability of getting both red aces is 1/1326

f) both face cards

Let F be the event of getting both face cards

There are 12 face cards in a pack

two face cards out of 12 face cards can be drawn by ¹²C₂ ways

∴ n(F) = ¹²C₂ =  6 x 11

By the definition P(F) = n(F)/n(S) = (6 x 11)/(26 x 51) = 11/221

Therefore the probability of getting both red aces is 1/1326

g) cards of denomination between 4 and 10

Let G be the event of getting cards of denomination between 4 and 10

Denominations between 4 and 10 are 5, 6, 7, 8, 9 (total 5 denominations)

Each denomination has 4 cards

Thus there are 5 x 4 = 20 cards of denomination between 4 and 10

two such cards out of 20 can be drawn by ²⁰C₂ ways

∴ n(G) = ²⁰C₂ =  10 x 19

By the definition P(G) = n(G)/n(S) = (10 x 19)/(26 x 51) = 95/663

Therefore the probability of getting cards of denomination between 4 and 10 is 95/663

h) both red face cards

Let H be the event of getting both red face cards

There are 6 red face cards in a pack

two red face cards out of 6 red face cards can be drawn by ⁶C₂ ways

∴ n(H) = ⁶C₂ =  3 x 5

By the definition P(H) = n(H)/n(S) = (3 x 5)/(26 x 51) = 5/442

Therefore the probability of getting both red face cards is 5/442.

i) a queen and a king

Let J be the event of getting a queen and a king

There 4 kings and 4 queens in a pack

one king out of 4 can be selected by  ⁴C₁ ways and

one queen out of 4 can be selected by  ⁴C₁ ways

∴ n(J) = ⁴C₁ x ⁴C₁ =  4 x 4 = 16

By the definition P(J) = n(J)/n(S) = 16/(26 x 51) = 6/663

Therefore the probability of getting a queen and a king is 6/663

j) one spade card and another non-spade card.

Let K be the event of getting one spade card and another non-spade card.

There 13 spade cards and 39 non-spade cards in a pack

one spade card out of 13 spade cards can be selected by  ¹³C₁ ways and

one non-spade card out of 39 non-spade cards can be selected by  ³⁹C₁ ways

∴ n(K) = ¹³C₁ x ³⁹C₁ =  13 x 39

By the definition P(K) = n(K)/n(S) = (13 x 39)/(26 x 51) = 13/34

Therefore the probability of getting one spade card and another non-spade card is 13/34

l) both cards from the same suite

Let L be the event of getting both cards from the same suite

There 13 cards in each suite

two cards out of 13 cards of the same suite can be selected by  ¹³C₂ ways

There are four suites in a pack

Event M = both are spade cards or both are club cards or both are diamond cards or both are heart cards

∴ n(M) = ¹³C₂ + ¹³C₂ + ¹³C₂ + ¹³C₂ = 4 x ¹³C₂  = 4 x 13 x 6

By the definition P(M) = n(M)/n(S) = (4 x 13 x 6)/(26 x 51) = 4/17

Therefore the probability of getting both cards of the same suite is 4/17

m) both are of the same denomination

Let N be the event of getting both cards of the same denomination

There 4 cards of the same denomination

two cards out of 4 cards of the same denomination can be selected by  ⁴C₂ ways

There are 13 sets of the same denomination

∴ n(N) =  13 x ⁴C₂  = 13 x 2 x 3

By the definition P(N) = n(N)/n(S) = (13 x 2 x 3)/(26 x 51) = 1/17

Therefore the probability of getting both cards of the same denomination is 1/17

o) One is spade and other is ace

Let Q be the event of getting one spade and another ace

There are two possibilities

Case – 1: When the first card is spade with spade ace included and another is ace from remaining three aces

Case – 2: When the first card is spade with ace excluded and another is ace from four aces

∴ n(Q) =  ¹³C₁ x ³C₁  +  ¹²C₁ x ⁴C₁  = 13 x 3 + 12 x 4 = 39 + 48 = 87

By the definition P(N) = n(N)/n(S) = 87/(26 x 51) = 29/442

Therefore the probability of getting one spade and other ace is 29/442

In the next article, we shall study some basic problems of probability based on the drawing of three cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 2 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving playing cards.

Introduction to Playing Cards:

Before studying, the problems on playing cards, you should be thorough with the following facts:

• There are 52 playing cards in a pack of playing cards.
• There are four suites in a pack viz: Spade (♠), Club (♣), Diamond (♦), Heart (♥)
• In each suite, there are 13 cards of different Denominations. viz. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King
• Thus there are 4 cards of each denomination in a pack, like 4 kings, 4 queens, 4 aces, 4 tens, 4 fives, etc.
• Spade and Club are black cards while Diamond and Heart are red cards.
• There are 26 black cards and 26 red cards in a pack.
• Each card is unique in a pack.
• King, Queen, and Jack cards are called picture cards or face cards.
• Thus there are total 12 face cards in a pack. 6 black face cards, 6 red face cards in a pack of playing cards
• There are 3 face cards in each suite.
• The Ace, King, Queen, and Jack of each suit are called honour cards
• The rest of the cards (2, 3, 4, 5, 6, 7, 8, 9, 10 ) are called spot cards.
• Spades and Hearts are called the major suits and Diamonds and Clubs are called the minor suits

Drawing a Single Playing Card From a Pack:

Example – 01:

A card is drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

one card out of 52 can be drawn by ⁵²C₁ ways

Hence n(S) = ⁵²C₁ = 52

a) a spade card

Let A be the event of getting a spade card

There are 13 spade cards in a pack

one spade card out of 13 can be drawn by ¹³C₁ ways

∴ n(A) = ¹³C₁ =  13

By the definition P(A) = n(A)/n(S) = 13/52 = 1/4

Therefore the probability of getting a spade card is 1/4

b) a red card

Let B be the event of getting a red card

There are 26 red cards in a pack

one red card out of 26 can be drawn by ²⁶C₁ ways

∴ n(A) = ²⁶C₁ =  26

By the definition P(B) = n(B)/n(S) = 26/52 = 1/2

Therefore the probability of getting a red card is 1/2

c) a black card

Let C be the event of getting a black card

There are 26 black cards in a pack

one black card out of 26 can be drawn by ²⁶C₁ ways

∴ n(C) = ²⁶C₁ =  26

By the definition P(C) = n(C)/n(S) = 26/52 = 1/2

Therefore the probability of getting a black card is 1/2

d) a king

Let D be the event of getting a king

There are 4 kings in a pack

one king out of 4 can be drawn by ⁴C₁ ways

∴ n(D) = ⁴C₁ =  4

By the definition P(D) = n(D)/n(S) = 4/52 = 1/13

Therefore the probability of getting a king is 1/13

Note: Probability of getting a card of a particular denomination is always 1/13

e) a red ace

Let E be the event of getting a red ace

There are 2 red aces in a pack

one red ace out of 2 can be drawn by ²C₁ ways

∴ n(E) = ²C₁ =  2

By the definition P(E) = n(E)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red ace is 1/26

f) a face card

Let F be the event of getting a face card

There are 12 face cards in a pack

one face card out of 12 can be drawn by ¹²C₁ ways

∴ n(F) = ¹²C₁ =  12

By the definition P(F) = n(F)/n(S) = 22/52 = 3/13

Therefore the probability of getting a face card is 3/13

g) a card of denomination between 4 and 10

Let G be the event of getting a card of denomination between 4 and 10

Denominations between 4 and 10 are 5, 6, 7, 8, 9 (total 5 denominations)

Each denomination has 4 cards

Thus there are 5 x 4 = 20 cards of denomination between 4 and 10

one such card out of 20 can be drawn by ²⁰C₁ ways

∴ n(G) = ²⁰C₁ =  20

By the definition P(G) = n(G)/n(S) = 20/52 = 5/13

Therefore the probability of getting a card of denomination between 4 and 10 is 5/13

h) a red face card

Let H be the event of getting a red face card

There are 6 red face cards in a pack

one face card out of 6 can be drawn by ⁶C₁ ways

∴ n(G) = ⁶C₁ =  6

By the definition P(H) = n(H)/n(S) = 6/52 = 3/26

Therefore the probability of getting a red face card is 3/26.

i) a queen of hearts

Let J be the event of getting a queen of hearts

There is only one queen of heart in a pack

one queen of hearts out of 1 can be drawn by 1way

∴ n(J) = 1

By the definition P(J) = n(J)/n(S) = 1/52

Therefore the probability of getting a queen of hearts is 1/52

j) a queen or a king

Let K be the event of getting a queen or a king

There 4 kings and 4 queens in a pack

Thus there are 4 + 4 = 8 favourable points.

one required card out of 8 favourable points can be drawn by ⁸C₁ ways

∴ n(K) = ⁸C₁ =  8

By the definition P(K) = n(K)/n(S) = 8/52 = 2/13

Therefore the probability of getting a queen or a king is 2/13

k) a red card and a king

Let L be the event of getting a red card or a king

There 2 red cards which are king

Thus there are 2 favourable points.

one required card out of 2 favourable points can be drawn by ²C₁ ways

∴ n(L) = ²C₁ =  2

By the definition P(L) = n(L)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red card and king is 1/26

l) a red card or a king  /a red king

Let M be the event of getting a red card or a king

There 26 red cards (including 2 red kings) and 2 black kings in a pack

Thus there are 26 + 2 = 28 favourable points.

one required card out of 28 favourable points can be drawn by ²⁸C₁ ways

∴ n(M) = ²⁸C₁ =  28

By the definition P(M) = n(M)/n(S) = 28/52 = 7/13

Therefore the probability of getting a red card or a king (a red king) is 7/13

m) Neither the heart nor the king

Let N be the event of getting neither the heart nor the king

There 36 non-heart cards (excluding 3 kings) in a pack

one required card out of 36  favourable points can be drawn by ³⁶C₁ ways

∴ n(N) = ³⁶C₁ =  36

By the definition P(N) = n(N)/n(S) = 36/52 = 9/13

Therefore the probability of getting neither the heart nor the king is 9/13

n) Neither an ace nor the king

Let Q be the event of getting neither an ace nor a king

There are 4 aces and 4 kings in a pack

There 44 non-ace and non-king cards in a pack

one required card out of 44  favourable points can be drawn by ⁴⁴C₁ ways

∴ n(Q) = ⁴⁴C₁ =  44

By the definition P(Q) = n(Q)/n(S) = 44/52 = 11/13

Therefore the probability of getting neither ace nor the king is 11/13

o) no diamond

Let R be the event of getting no diamond

There 39 non-diamond cards in a pack

one required card out of 39  favourable points can be drawn by ³⁹C₁ ways

∴ n(R) = ³⁹C₁ =  39

By the definition P(R) = n(R)/n(S) = 39/52 = 3/4

Therefore the probability of getting no diamond is 3/4

p) no ace

Let T be the event of getting no ace

There are 4 aces in a pack

There 48 non-ace cards in a pack

one required card out of 48  favourable points can be drawn by ⁴⁸C₁ ways

∴ n(T) = ⁴⁸C₁ =  48

By the definition P(T) = n(T)/n(S) = 48/52 = 12/13

Therefore the probability of getting no ace is 12/13.

q) not a black card

Let V be the event of getting no black card

There 26 non-black (red) cards in a pack

one required card out of 26  favourable points can be drawn by ²⁶C₁ ways

∴ n(V) = ²⁶C₁ =  26

By the definition P(V) = n(V)/n(S) = 26/52 = 1/2

Therefore the probability of getting no black card is 1/2.

In the next article, we shall study some basic problems of probability based on the drawing of two cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing a Playing Card appeared first on The Fact Factor.

In the last article, we have studied problems on the throwing of dice. In this article, we shall study to solve problems to find probability involving numbered tickets.

Drawing Two or More Numbered tickets:

Example – 01:

Tickets numbered from 1 to 50 are mixed up together and then two tickets are drawn at random what is the probability that

Solution:

The sample space is S = {1, 2, 3, …….., 50}.

Two tickets are drawn at random.

Hence n(S) = ⁵⁰C₂ = 1225

a) both the tickets bear an even number

Let A be the event that both the tickets bear an even number

Favourable points are 2, 4, 6, 8, 10, … , 50

There are 25 favourable points

∴ n(A) = ²⁵C₂ =  300

By the definition P(A) = n(A)/n(S) = 300/1225 = 12/49

Therefore the probability that both the tickets bear an even number is 12/49

b) both the tickets bear an odd number

Let B be the event that both the tickets bear an odd number

Favourable points are 21, 3, 5, 7, ….., 49

There are 25 favourable points

∴ n(B) = ²⁵C₂ =  300

By the definition P(B) = n(B)/n(S) = 300/1225 = 12/49

Therefore the probability that both the tickets bear an odd number is 12/49

c) both the tickets bear a perfect square

Let C be the event that both the tickets bear a perfect square

Favourable points are 1, 4, 9, 16, 25, 36, 49

There are 7 favourable points

∴ n(C) = ⁷C₂ =  21

By the definition P(C) = n(C)/n(S) = 21/1225 = 3/175

Therefore the probability that both the tickets bear a perfect square is 3/175

d) Both the tickets bear a number multiple of four (or divisible by four)

Let D be the event that both the tickets bear a number multiple of four

Favourable points are 4, 8, 12, 16, …., 48

There are 12 favourable points

∴ n(D) = ¹²C₂ =  66

By the definition P(D) = n(D)/n(S) = 66/1225

Therefore the probability that both the tickets bear a number multiple of four is 66/1225

e) both the tickets bear a number multiple of three (or divisible by three)

Let E be the event that both the tickets bear a number multiple of three

Favourable points are 3, 6, 9, …., 48

There are 16 favourable points

∴ n(E) = ¹⁶C₂ =  120

By the definition P(E) = n(E)/n(S) = 120/1225 = 24/245

Therefore the probability that both the tickets bear a number multiple of three is 24/245

f) both the tickets bear a number greater than 44

Let F be the event that both the tickets bear a number greater than 44

Favourable points are 45, 46, 47, 48, 49, 50

There are 6 favourable points

∴ n(F) = ⁶C₂ =  15

By the definition P(F) = n(F)/n(S) = 15/1225 = 3/245

Therefore the probability that both the tickets bear a number greater than 44 is 3/245

g) both the tickets bear a number less than 11

Let G be the event that both the tickets bear a number less than 11

Favourable points are 1, 2, 3, ……, 10

There are 10 favourable points

∴ n(G) = ¹⁰C₂ =  45

By the definition P(G) = n(G)/n(S) = 45/1225 = 9/245

Therefore the probability that both the tickets bear a number less than 11 is 9/245

h) both the tickets bear perfect square or number less than 10

Let H be the event that both the tickets bear perfect square or number less than 10

Favourable points are 1, 4, 9, 16, 25, 36, 49, 2, 3, 5, 6, 7, 8

There are 13 favourable points

∴ n(H) = ¹³C₂ =  78

By the definition P(H) = n(H)/n(S) = 45/1225 = 9/245

Therefore the probability that both the tickets bear perfect square or a number less than 5 is 78/1225

i) both the tickets bear a prime number

Let J be the event that both the tickets bear a prime number

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

There are 15 favourable points

∴ n(J) = ¹⁵C₂ =  105

By the definition P(J) = n(J)/n(S) = 105/1225 = 3/35

Therefore the probability that both the tickets bear a prime number is 3/35

j) both the tickets bear a prime number or a perfect square

Let K be the event that both the tickets bear a prime number or a perfect square

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,

41, 43, 47, 1, 4, 9, 16, 25, 36, 49

There are 22 favourable points

∴ n(K) = ²²C₂ =  231

By the definition P(K) = n(K)/n(S) = 231/1225 = 33/175

Therefore the probability that both the tickets bear a prime number or a perfect square is 33/175

both the tickets bear a number greater than 35 and an even number.

Let L be the event that both the tickets bear a number greater than 35 and an even number

Favourable points are 36, 38, 40, 42, 44, 46, 48, 50

There are 8 favourable points

∴ n(L) = ⁸C₂ =  28

By the definition P(L) = n(L)/n(S) = 28/1225 = 4/175

Therefore the probability that both the tickets bear a number greater than 35 and an even number is 4/175

k) both the tickets bear an even number and multiple of 5

Let M be the event that both the tickets bear an even number and multiple of 5

Favourable points are 10, 20, 30, 40, 50

There are 5 favourable points

∴ n(L) = ⁵C₂ =  10

By the definition P(M) = n(M)/n(S) = 10/1225 = 2/245

Therefore the probability that both the tickets bear an even number and multiple of 5 is 2/245

In the next article, we shall study some basic problems of probability based on the drawing of a single card from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Numbered Cards 02 appeared first on The Fact Factor.

In the last article, we have studied problems on the throwing of dice. In this article, we shall study to solve problems to find probability involving numbered cards.

Drawing a Single Numbered Card / Ticket:

Example – 01:

Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random what is the probability of getting a ticket bearing

Solution:

The sample space is S = {1, 2, 3, …….., 20}.

One ticket is drawn at random.

Hence n(S) = ²⁰C₁ = 20

a) an even number

Let A be the event of getting ticket bearing an even number

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20

∴ n(A) = ¹⁰C₁ = 10

By the definition P(A) = n(A)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing even number is 1/2

b) an odd number

Let B be the event of getting ticket bearing an odd number

Favourable points are , 3, 5, 7, 9, 11, 13, 15, 17, 19

∴ n(B) = ¹⁰C₁ = 10

By the definition P(B) = n(B)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing odd number is 1/2

c) a perfect square

Let C be the event of getting ticket bearing a perfect square

Favourable points are 1, 4, 9,16

∴ n(C) = ⁴C₁ = 4

By the definition P(C) = n(C)/n(S) = 4/20 = 1/5

Therefore the probability of getting ticket bearing a perfect square is 1/5

d) multiple of four (or divisible by four)

Let D be the event of getting ticket bearing a number multiple of 4

Favourable points are 4, 8, 12, 16, 20

∴ n(D) = ⁵C₁ = 5

By the definition P(D) = n(D)/n(S) = 5/20 = 1/4

Therefore the probability of getting ticket bearing a number multiple of 4 is 1/4

e) multiple of three (or divisible by three)

Let E be the event of getting ticket bearing a number multiple of 3

Favourable points are 3, 6, 8, 12, 15, 18

∴ n(E) = ⁶C₁ = 6

By the definition P(E) = n(E)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a number multiple of 3 is 3/10

f) a number greater than 4

Let F be the event of getting ticket bearing a number greater than 4

Favourable points are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

∴ n(F) = ¹⁷C₁ = 17

By the definition P(F) = n(F)/n(S) = 17/20

Therefore the probability of getting ticket bearing a number greater than 4 is 17/20

g) a number less than 11

Let G be the event of getting ticket bearing a number less than 11

Favourable points are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

∴ n(G) = ¹⁰C₁ = 10

By the definition P(G) = n(G)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing a number less than 11 is 1/2

h) perfect square or less than 5

Let H be the event of getting ticket bearing a number a perfect square or less than 5

Favourable points are 1, 4, 9, 16, 2, 3

∴ n(H) = ⁶C₁ = 6

By the definition P(H) = n(H)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a perfect square or less than 5 is 3/10

i) a prime number

Let J be the event of getting ticket bearing a number a prime number

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19

∴ n(J) = ⁸C₁ = 8

By the definition P(J) = n(J)/n(S) = 8/20 = 2/5

Therefore the probability of getting ticket bearing a prime number is 2/5

j) a prime number or a perfect square

Let K be the event of getting ticket bearing a number a prime number or a perfect square

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 1, 4, 9, 16

∴ n(K) = ¹²C₁ = 12

By the definition P(K) = n(K)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

k) an even number or a perfect square

Let L be the event of getting ticket bearing an even number or a perfect square

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 1, 9

∴ n(L) = ¹²C₁ = 12

By the definition P(L) = n(L)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

l) an even number or a number divisible by 5

Let M be the event of getting ticket bearing an even number or a number divisible by 5

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 5, 15

∴ n(M) = ¹²C₁ = 12

By the definition P(M) = n(M)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

m) a perfect square or a number multiple of 3

Let N be the event of getting ticket bearing a perfect square or a number multiple of 3

Favourable points are 1, 4, 9, 16, 3, 6, 12,15, 18

∴ n(N) = ⁹C₁ = 9

By the definition P(N) = n(N)/n(S) = 9/20

Therefore the probability of getting ticket bearing a perfect square or a number multiple of 3 is 9/20

n) a number greater than 9 and an even number.

Let Q be the event of getting ticket bearing a number greater than 9 and an even number

Favourable points are 10, 12, 14, 16, 18, 20

∴ n(Q) = ⁶C₁ = 6

By the definition P(Q) = n(Q)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a number greater than 9 and an even number is 3/10

o) an even number and multiple of 3

Let R be the event of getting ticket bearing an even number and multiple of 3

Favourable points are 6, 12, 18

∴ n(R) = ³C₁ = 3

By the definition P(R) = n(R)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

In the next article, we shall study some basic problems of probability based on drawing two or more cards from the collection of numbered cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Numbered Cards 01 appeared first on The Fact Factor.