In this article, we shall study to check whether the given function is a probability mass function or not.

Example – 01: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 1 ≤ x ≤ 4) = P(1) + P(2) + P(3) + P(4)

∴   ∑ P(X = x, 1 ≤ x ≤ 4) = 0.1 + 0.2 + ).3 + 0.4 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 02: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 3) = P(0) + P(1) + P(2) + P(3)

∴   ∑ P(X = x, 0 ≤ x ≤ 3) = 0.5 + 0.2 + 0.18 + 0.12 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 03: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.2 < 0 Hence the first condition is not satisfied.

Hence given function is not a p.m.f.

Example – 04: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 1 = P(0) + P(1) + P(2)

∴   ∑ P(X = x, 0 ≤ x ≤ 2) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Thus sum of all probabilities is not equal to1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 05: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 2, 4, 6, 8) = P(2) + P(4) + P(6) + P(8)

∴   ∑ P(X = x, x = 2, 4, 6, 8) = 0.2 + 0.4 + 0.6 + 0.8 = 2 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 06: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.1 < 0 Hence the first condition is not satisfied.

Hence given function is not a p.m.f.

Example – 07: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 0) = 0²/5 = 0/5 = 0

∴   P(X = 1) = 1²/5 = 1/5

∴   P(X = 2) = 2²/5 = 4/5

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 0, 1, 2) = P(0) + P(1) + P(2)

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

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 08: 

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 0) = (1- 1)/3 = 0/3 = 0

∴   P(X = 1) = (2 – 1)/3 = 1/3

∴   P(X = 2) = (3 – 1)/3 = 2/3

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 1, 2, 3) = P(1) + P(2) + P(3)

∴   ∑ P(X = x, x = 1, 2, 3) = 0 + 1/3 + 2/3 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 09: 

The function is P( X = x) = (x – 5)/4, x = 5.5, 6.5, 7.5

Verify whether the function can be regarded as a probability mass function (p.m.f.)

Solution:

∴   P(X = 5.5) = (5.5 – 5)/4 = 0.5/4 = 1/8

∴   P(X = 6.5) = (6.5 – 5)/4 = 1.5/4 = 3/8

∴   P(X = 7.5) = (7.5 – 5)/4 = 2.5/4 = 5/8

The probability distribution is

Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 5.5, 6.5, 7.5) = P(5.5) + P(6.5) + P(7.5)

∴   ∑ P(X = x, x = 5.5, 6.5, 7.5) = 1/8 + 3/8 + 5/8 = 9/8 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

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In this article, we shall study to write probability mass function and to write probability distribution for the given event.

Example – 01:

If a coin is tossed two times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed two times. The sample space for the experiment is as follows

S = {HH, HT, TH, TT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails)

Probability mass function is

P( X = 0) = P(0) = 1/4

P( X = 1) = P(1) = 2/4 = 1/2

P( X = 2) = P(02) = 1/4
Hence, The probability distribution for the number of tails is as follows.

Example – 02:

If a coin is tossed three times and X denotes the number of tails. Find the probability mass function of X. Also write the probability distribution of X.
Solution:

If a coin is tossed three times. The sample space for the experiment is as follows

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8

The probability distribution for the number of tails is as follows.

Example – 03:

If a coin is tossed four times and X denotes the number of tails. Find the probability distribution of X.

Solution:

METHOD- I:

If a coin is tossed four times. The sample space for the experiment is as follows

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, TTHT, TTTH, TTTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8
P( X = 4) = P(4) = 3/8

The probability distribution for the number of tails is as follows.

METHOD- II

If a coin is tossed four times. The sample space for the experiment is as follows

S = 2⁴ = 16

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁴C₀ /16 = 1/16
P( X = 1) = P(1) = ⁴C₁ /16 = 4/16 = 1/4
P( X = 2) = P(2) = ⁴C₂ /16 = 6/16 = 3/8
P( X = 3) = P(3) = ⁴C₃ /16 = 4/16 = 1/4
P( X = 4) = P(4) = ⁴C₄ /16 = 1/16

The probability distribution for the number of tails is as follows.

Example – 04:

If a coin is tossed five times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed five times. The sample space for the experiment is as follows

S = 2⁵ = 32

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁵C₀ /32 = 1/32
P( X = 1) = P(1) = ⁵C₁ /32 = 5/32
P( X = 2) = P(2) = ⁵C₂ /32 = 10/32 = 5/16
P( X = 3) = P(3) = ⁵C₃ /32 = 10/32 = 5/16
P( X = 4) = P(4) = ⁵C₄ /32 = 5/32
P( X = 5) = P(5) = ⁵C₅ /32 = 1/32

The probability distribution for the number of tails is as follows.

Example – 05:

If a coin is tossed six times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed six times. The sample space for the experiment is as follows

S = 2⁶ = 64

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails) or 6 (six tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁶C₀ /64 = 1/64
P( X = 1) = P(1) = ⁶C₁ /64 = 6/64 = 3/32
P( X = 2) = P(2) = ⁶C₂ /64 = 15/64
P( X = 3) = P(3) = ⁶C₃ /64 = 20/64 = 5/16
P( X = 4) = P(4) = ⁶C₄ /64 = 15/64
P( X = 5) = P(5) = ⁶C₅ /64 = 6/64 = 3/32
P( X = 6) = P(6) = ⁶C₆ /64 = 1/64

The probability distribution for the number of tails is as follows.

In the next article, we shall study problems in which we will be checking, whether the distribution given is probability mass function or not.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

In the last article, we have studied to solve problems to calculate probability when a single die is thrown. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice.

Algorithm:

1. Study experiment and write the sample space
2. Find favourable point and write event space
3. Use the definition of probability and find it.

Throwing of Two Dice:

Two fair dice are thrown Or a die is thrown twice

S = {1,2,3,4,5,6} ×  {1,2,3,4,5,6}

• The sum of the two numbers on two dice is called the score on two dice.
• The minimum score on two dice is 2 and the maximum score on two dice is 12.
• The cases favourable to a particular score can be read along the diagonal of that score.

Example – 01:

Two fair dice are tossed. Find the probability in the following cases:

Solution:

Two fair dice are thrown

The sample space is

S = {1,2,3,4,5,6} ×  {1,2,3,4,5,6}

∴ n (S) =  36

a) the sum of the scores is even

Let A be the event of that the sum of the scores is even i.e. 2, 4, 6, 8, 10, 12

∴ A = { (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1),

(3, 3), (3, 5), (4, 2), (4, 4), (4, 6),

(5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6) }

∴ n(A) = 18

By the definition P(A) = n(A)/n(S) = 18/36 = 1/2

Ans: the probability that the sum of the scores is even is 1/2

b) the sum of the scores is odd

Let B be the event of that the sum of the scores is odd i.e. 3, 5, 7, 9, 11

∴ B = { (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6),

(2, 5), (3, 4), (4, 3), (5, 2), (6, 1),

(3, 6), (4, 5), (5, 4), (6, 3), (6, 5), (5, 6) }

∴ n(A) = 18

By the definition P(B) = n(B)/n(S) = 18/36 = 1/2

Ans: the probability that the sum of the scores is odd is 1/2

c) the sum of the scores is a perfect square

Let C be the event of that the sum of the scores is a perfect square i.e. 4, 9.

∴ C = { (1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3) }

∴ n(C) = 7

By the definition P(C) = n(C)/n(S) = 7/36

Ans: the probability that the sum of the scores is a perfect square is 7/36

d) the sum of the score is a multiple of four or the score is divisible by 4

Let D be the event of that the sum of the score is a multiple of four i.e. 4, 8, 12

∴ D = { (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6) }

∴ n(D) = 9

By the definition P(D) = n(D)/n(S) = 9/36 = 1/4

Ans: the probability that the sum of the score is a multiple of four is 1/4

e) the sum of the scores is a multiple of 3

Let E be the event of that the sum of the scores is a multiple of 3 i.e. 3, 6, 9, 12

∴ E = { (1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2),

(5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

∴ n(E) = 12

By the definition P(E) = n(E)/n(S) = 12/36 = 1/3

Ans: the probability that the sum of the score is a multiple of 3 is 1/3..

f) the sum of the points obtained is greater than 4

Let F be the event of that the sum of the points obtained is

greater than 4 i.e. 5, 6, 7, 8, 9, 10, 11, 12.

∴ F = { (1, 4), (1, 5), (1, 6) (2. 3), (2, 4), (2, 5). (2, 6), (3, 2),

(3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3),

(4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4),

(5, 5), (5, 6), (6, 1I), (6, 2), (6, 3). (6, 4), (6, 5), (6, 6) }

∴ n(F) = 30

By the definition P(F) = n(F)/n(S) = 30/36 = 5/6

Ans: the probability that the sum of the points obtained is greater than 4 is 5/6

g) the sum of the points is at least 11

Let G be the event of that the sum of the points is at least 11 i.e. 11, 12

∴ G ={ (5, 6), (6, 5), (6, 6) }

∴ n(G) = 3

By the definition P(G) = n(G)/n(S) = 3/36 = 1/12

Ans: the probability that the sum of the points is at least 11 is 1/12.

h) the same score on the first die and second die

Let H be the event of that the same score on the first die and second die.

∴ H = { (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) }

∴ n(H) = 6

By the definition P(H) = n(H)/n(S) = 6/36 = 1/6

Ans: the probability that the same score on the first die and second die is 1/6.

i) the score on the second die is greater than the score on the first die

Let J be the event of that the score on the second die is greater than the score on the first die

∴ J = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2. 3), (2, 4), (2, 5).

(2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) }

∴ n(J) = 15

By the definition P(J) = n(J)/n(S) = 15/36 = 5/12

Ans: the probability that the score on the second die is greater than the score on the first die is 5/12.

j) the sum of the numbers on their faces obtained is either a perfect square or their sum is less than 5.

Let K be the event of that the sum of the numbers on their faces

obtained is either a perfect square or their sum is less than 5.

∴ K = { (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3. 1), (3, 6), (4, 5), (5, 4), (6, 3) }

∴ n(K) = 10

By the definition P(K) = n(K)/n(S) = 10/36 = 5/18

Ans: the probability that the sum of the numbers on their faces

obtained is either a perfect square or their sum is less than 5 is 5/18.

l) the sum of numbers shown is 7 or product is 12

Let M be the event of that the sum of numbers shown is 7 or product is 12

∴ M = { (1, 6), (2, 5), (2, 6), (3, 4), (4, 3), (5, 2), (6, 1), (6, 2) }

∴ n(M) = 8

By the definition P(M) = n(M)/n(S) = 8/36 = 2/9

Ans: the probability that the sum of numbers shown is 7 or the product is 12 is 2/9.

m) the sum of these scores is either a perfect square or a prime number

Let N be the event of that the sum of these scores is either

a perfect square or a prime number i.e. 4, 9, 2, 3, 5, 7, 11

∴ N = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 2),

(2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5),

(3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5) }

∴ n(N) = 22

By the definition P(N) = n(N)/n(S) = 22/36 = 11/18

Ans: the probability that the sum of numbers a perfect square or a prime number is 11/18.

n) the product of the scores is 12

Let Q be the event of that the product of the scores is 12

∴ Q = { (2, 6), (3, 4), (4, 3), (6, 2) }

∴ n(Q) = 4

By the definition P(Q) = n(Q)/n(S) = 4/36 = 1/9

Ans: the probability that the product of the scores is 12 is 1/9

o) the sum of these scores is either a perfect square or an even number

Let R be the event of that the sum of these scores is either

a perfect square or an even number i.e. 4, 9, 2, 6, 8,10, 12

∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 5),

(2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5), (4, 4),

(5, 3), (6, 2), (4, 6), (5, 5), (6, 4), (6, 6) }

∴ n(R) = 22

By the definition P(R) = n(R)/n(S) = 22/36 = 11/18

Ans: the probability that the sum of these scores is either a perfect square or an even number is 11/18.

p) the sum of these scores is either an even number or a number divisible by 5

Let T be the event of that the sum of these scores is either

an even number or a number divisible by 5 i.e. 2, 4, 6, 8,10, 12, 5

∴ T = {(1. 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3),

(4, 2), (5, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2),

(4, 6), (5, 5), (6, 4), (6, 6), (1, 4), (2, 3), (3, 2), (4, 1) }

∴ n(T) = 22

By the definition P(T) = n(T)/n(S) = 22/36 = 11/18

Ans: the probability that the sum of these scores is either

an even number or a number divisible by 5 is 11/18.

q) the sum of these scores is either a perfect square or a multiple of 3

Let T be the event of that the sum of these scores is either

a perfect square or a multiple of 3 i.e. 4, 9, 3, 6, 12

∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 2), (2, 1),

(2, 4), (3, 3), (4, 2), (5, 1), (6, 6) }

∴ n(T) = 14

By the definition P(T) = n(T)/n(S) = 14/36 = 7/18

Ans: the probability that the sum of these scores is either a perfect square or a multiple of 3 is 7/18.

r) the sum of the scores is either greater than 9 and an even number.

Let U be the event of that the sum of these scores is either

greater than 9 i.e. 10, 12

∴ U = {(4, 6), (5, 5), (6, 4), (6, 6) }

∴ n(U) = 4

By the definition P(U) = n(U)/n(S) = 4/36 = 1/9

Ans: the probability that the sum of these scores is either greater than 9 and an even number is 1/4.

s) the product of the scores is a perfect square

Let V be the event of that the product of the scores is

a perfect square. i.e. 1, 4, 9, 16, 25, 36

∴ V = {(1, 1), (1, 4), (2, 2), (4, 1), (3, 3), (4, 4), (5, 5), (6, 6) }

∴ n(V) = 8

By the definition P(V) = n(V)/n(S) = 8/36 = 2/9

Ans: the probability that the product of the scores is a perfect square. is 2/9.

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

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