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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 ^{52}C_{4}_{ }ways

Hence n(S) = ^{52}C_{4}

**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 ^{13}C_{4 }ways

∴ n(A) = ^{13}C_{4}

By the definition P(A) = n(A)/n(S) = (^{13}C_{4 })/(^{52}C_{4})

Therefore the probability of getting all heart cards is (^{13}C_{4 })/(^{52}C_{4})

**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 ^{13}C_{4 }ways

∴ n(B) = 4 x ^{13}C_{4}

By the definition P(B) = n(B)/n(S) = (4 x ^{13}C_{4 })/(^{52}C_{4})

Therefore the probability of getting all the cards of the same suite is (4 x ^{13}C_{4 })/(^{52}C_{4})

**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) = ^{26}C_{4} + ^{26}C_{4} = 2(^{26}C_{4})

By the definition P(C) = n(C)/n(S) = 2(^{26}C_{4})/(^{52}C_{4})

Therefore the probability of getting all the cards of the same colour is 2(^{26}C_{4})/(^{52}C_{4})

**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) = ^{12}C_{4}

By the definition P(D) = n(D)/n(S) = ^{12}C_{4}/(^{52}C_{4})

Therefore the probability of getting all face cards is ^{12}C_{4}/(^{52}C_{4})

**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 ^{4}C_{4 }ways

∴ n(E) = 13 x ^{4}C_{4 }= 13 x 1 = 13

By the definition P(E) = n(E)/n(S) = (13)/(^{52}C_{4})

Therefore the probability of getting all the cards of the same suite is (13)/(^{52}C_{4})

**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) = (^{26}C_{2}) x (^{26}C_{2}) = (^{26}C_{2})^{2}

By the definition P(F) = n(F)/n(S) = (^{26}C_{2})^{2}/(^{52}C_{4})

Therefore the probability of getting two red cards and two black cards is (^{26}C_{2})^{2}/(^{52}C_{4})

**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) = (^{4}C_{4}) + (^{4}C_{4}) + (^{4}C_{4}) + (^{4}C_{4}) = 1 + 1 + 1 + 1 = 4

By the definition P(G) = n(G)/n(S) = 4/(^{52}C_{4})

Therefore the probability of getting all honours of the same suite is 4/(^{52}C_{4})

**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 ^{39}C_{4 }ways

∴ n(H) = ^{39}C_{4}

By the definition P(H) = n(H)/n(S) = ^{39}C_{4}/^{52}C_{4}

Now P(F) = 1 – P(F) = 1 – (^{39}C_{4}/^{52}C_{4})

Therefore the probability of getting at least one heart is (1 – (^{39}C_{4}/^{52}C_{4}))

**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) = (^{4}C_{3} x ^{4}C_{1})

By the definition P(J) = n(J)/n(S) = (^{4}C_{3} x ^{4}C_{1})/(^{52}C_{4})

Therefore the probability of getting 3 kings and one jack is (^{4}C_{3} x ^{4}C_{1})/(^{52}C_{4})

**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) = (^{12}C_{3} x ^{1}C_{1}) = ^{12}C_{3}

By the definition P(K) = n(K)/n(S) = (^{12}C_{3})/(^{52}C_{4})

Therefore the probability of getting all clubs and one of them is a jack is (^{12}C_{3})/(^{52}C_{4})

**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) = (^{13}C_{3} x ^{13}C_{1})

By the definition P(L) = n(L)/n(S) = (^{13}C_{3} x ^{13}C_{1})/(^{52}C_{4})

Therefore the probability of getting 3 diamonds and 1 spade is (^{13}C_{3} x ^{13}C_{1})/(^{52}C_{4})

**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 ^{52}C_{4}_{ }ways

Hence n(S) = ^{52}C_{4}

Let A be the event of getting one card from each suite

There are 13 cards in each suite.

∴ n(A) = (^{13}C_{1}) x (^{13}C_{1}) x (^{13}C_{1}) x (^{13}C_{1})x = (^{13}C_{1})^{4}

By the definition P(A) = n(A)/n(S) = (^{13}C_{1})^{4}/(^{52}C_{4})

Therefore the probability of getting one card from each suite is (^{13}C_{1})^{4}/(^{52}C_{4})

#### 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 ^{52}C_{5}_{ }ways

Hence n(S) = ^{52}C_{5}

**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) = ^{4}C_{1} x ^{48}C_{4} = 4 x ^{48}C_{4}

By the definition P(A) = n(A)/n(S) = (4 x ^{48}C_{4}_{ })/(^{52}C_{5})

Therefore the probability of getting all heart cards is (4 x ^{48}C_{4}_{ })/(^{52}C_{5})

**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) = ^{48}C_{5}

By the definition P(B) = n(B)/n(S) = (^{48}C_{5}_{ })/(^{52}C_{5})

Now P(B) = 1 – P(B) = 1 – (^{48}C_{5}_{ })/(^{52}C_{5})

Therefore the probability of getting atleast one ace is (1 – (^{48}C_{5}_{ })/(^{52}C_{5}))

**all cards are of hearts**

Let C be the event of getting all hearts

There are 13 heart cards in a pack

∴ n(C) = ^{13}C_{5}

By the definition P(C) = n(C)/n(S) = (^{13}C_{5}_{ })/(^{52}C_{5})

Therefore the probability of getting all hearts is (1 – (^{48}C_{5}_{ })/(^{52}C_{5}))

**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 ^{52}C_{13}_{ }ways

Hence n(S) = ^{52}C_{13}

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 ^{4}C_{1 }ways = 4 ways

Now, in hand, there are 9 cards of same suite and 4 cards of other suites.

∴ n(B) = (^{4}C_{1}) x (^{13}C_{9}) x (^{39}C_{4}) = 4 x (^{13}C_{9}) x (^{39}C_{4})

By the definition P(B) = n(B)/n(S) = (4 x (^{13}C_{9}) x (^{39}C_{4}) )/(^{52}C_{4})

Therefore the probability of getting 9 cards of the same suite in one hand is (4 x (^{13}C_{9}) x (^{39}C_{4}) )/(^{52}C_{4})

**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 ^{52}C_{13}_{ }ways

Hence n(S) = ^{52}C_{13}

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) = (^{13}C_{9}) x (^{39}C_{4}) = (^{13}C_{9}) x (^{39}C_{4})

By the definition P(C) = n(C)/n(S) = (^{13}C_{9} x ^{39}C_{4})/(^{52}C_{4})

Therefore the probability of getting 9 cards of spade in one hand is (^{13}C_{9} x ^{39}C_{4})/(^{52}C_{4})

**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 ^{52}C_{13}_{ }ways

Hence n(S) = ^{52}C_{13}

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 (^{4}C_{4}) x (^{48}C_{9}) = ^{48}C_{9}

By the definition P(D) = n(D)/n(S) = (^{48}C_{9} )/(^{52}C_{4})

Therefore the probability of that four kings are held by a specified player is (^{48}C_{9} )/(^{52}C_{4})

**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 ^{40}C_{4}_{ }ways

Hence n(S) = ^{40}C_{4}

Let E be the event of getting one card from each suite

There are 10 cards in each suite.

∴ n(E) = (^{10}C_{1}) x (^{10}C_{1}) x (^{10}C_{1}) x (^{10}C_{1})x = (^{10}C_{1})^{4}

By the definition P(E) = n(E)/n(S) = (^{10}C_{1})^{4}/(^{40}C_{4})

Therefore the probability of getting one card from each suite is (^{10}C_{1})^{4}/(^{40}C_{4})

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 of a Three Cards From Pack of Playing Cards

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

Science > Mathematics > Probability > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |