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#### 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 ^{52}C_{2 }ways

Hence n(S) = ^{52}C_{2} = 26 x 51

**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 ^{13}C_{2 }ways

∴ n(A) = ^{13}C_{2} = 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

**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 ^{26}C_{2 }ways

∴ n(B) = ^{26}C_{2} = 13 x 25

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

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

**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 ^{26}C_{2 }ways

∴ n(B) = ^{26}C_{2} = 13 x 25

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

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

**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 ^{4}C_{2 }ways

∴ n(D) = ^{4}C_{2} = 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:** Probability of getting two cards of a particular denomination is always 1/221

**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 ^{2}C_{2 }ways

∴ n(E) = ^{2}C_{2} = 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

**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 ^{12}C_{2 }ways

∴ n(F) = ^{12}C_{2} = 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

**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 ^{20}C_{2 }ways

∴ n(G) = ^{20}C_{2} = 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

**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 ^{6}C_{2 }ways

∴ n(H) = ^{6}C_{2} = 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.

**a queen and a king**

Let K 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 ^{4}C_{1 }ways and

one queen out of 4 can be selected by ^{4}C_{1 }ways

∴ n(K) = ^{4}C_{1} x ^{4}C_{1} = 4 x 4 = 16

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

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

**one spade card and another non-spade card.**

Let L 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 ^{13}C_{1 }ways and

one non-spade card out of 39 non-spade cards can be selected by ^{39}C_{1 }ways

∴ n(K) = ^{13}C_{1} x ^{39}C_{1} = 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

**both cards from the same suite**

Let M 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 ^{13}C_{2 }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) = ^{13}C_{2} + ^{13}C_{2} + ^{13}C_{2} + ^{13}C_{2} = 4 x ^{13}C_{2} = 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

**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 ^{4}C_{2 }ways

There are 13 sets of the same denomination

∴ n(N) = 13 x ^{4}C_{2} = 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

**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) = ^{13}C_{1} x ^{3}C_{1} + ^{12}C_{1} x ^{4}C_{1} = 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

#### Previous Topic: Problems Based on Drawing of a Single Card From Pack of Playing Cards

#### Previous Topic: Problems Based on Drawing of Three Cards From Pack of Playing Cards

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