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Problems Based on Drawing 4 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 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.

Getting both red cards

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 52Cways

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 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)

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 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)

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 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)

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 39Cways

∴ 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 52Cways

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 52Cways

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) = 4C148C4  = 4 x 48C4

By the definition P(A) = n(A)/n(S) = (4 x 48C)/(52C5)

Therefore the probability of getting all heart cards is

(4 x 48C)/(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) = (48C)/(52C5)

Now P(B) = 1 – P(B’) = 1 –  (48C)/(52C5)

Therefore the probability of getting atleast one ace is

(1 –  (48C)/(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) = (13C)/(52C5)

Therefore the probability of getting all hearts is

(13C)/(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 40Cways

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 52Cways

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) = 4C448C3  

By the definition P(A) = n(A)/n(S) = (4C448C)/(52C7)

Therefore the probability of getting all 4 kings is

(4C448C)/(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) = 4C348C4  

By the definition P(A) = n(A)/n(S) = (4C348C)/(52C7)

Therefore the probability of getting all 4 kings is

(4C348C)/(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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