Science > Mathematics > Statistics and Probability > Probability > Binomial Distribution
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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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) = nCr pr qn – 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