# Probability: Problems Based on Playing Cards – 04

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#### 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

• 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 13Cways

∴ n(A) = 13C4

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

Therefore the probability of getting all heart cards is (13C)/(52C4)

• 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 13Cways

∴ n(B) = 4 x 13C4

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

Therefore the probability of getting all the cards of the same suite is (4 x 13C)/(52C4)

• 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)

• 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)

• 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 4Cways

∴ 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)

• 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)

• 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)

• 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 39Cways

∴ n(H) = 39C4

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

Now P(F) = 1 – P(F) = 1 – (39C4/52C4)

Therefore the probability of getting at least one heart is (1 – (39C4/52C4))

• 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)

• 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)

• 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

• 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) = 4C148C4  = 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)

• 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))

• 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 (1 –  (48C5 )/(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 4Cways = 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)

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

all 4 kings (1/7735)

exactly 3 kings (9/1547)

at least three kings (46/7735)

#### Previous Topic: Problems Based on Drawing a Ball or Two Balls From Set of Coloured Balls

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