Problems Based on Drawing 2 Playing Cards

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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 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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