Science > Mathematics > Statistics and Probability > Probability > Problems Based on Drawing 4 Playing Cards
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.