# Probability: Problems Based on Playing Cards – 01

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#### Playing Cards:

• There are 52 playing cards in a pack of playing cards.
• There are four suites in a pack

# Spade (♠), Club (♣), Diamond (♦), Heart (♥)

• In each suite, there are 13 cards of different Denominations. viz. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King
• Thus there are 4 cards of each denomination in a pack, like 4 kings, 4 queens, 4 aces, 4 tens, 4 fives, etc.
• Spade and Club are black cards while Diamond and heart are red cards.
• There are 26 black cards and 26 red cards in a pack.
• Each card is unique in a pack.
• King, Queen and Jack cards are called picture cards or face cards.
• Thus there are total 12 face cards in a pack. 6 black face cards, 6 red black cards in a pack of playing cards
• There are 3 face cards in each suite.
• The Ace, King, Queen, and Jack of each suit are called honour cards
• The rest of the cards (2, 3, 4, 5, 6, 7, 8, 9, 10 )are called spot cards.
• Spades and Hearts are called the major suits and • Diamonds and Clubs are called the minor suits

#### Example – 01:

• A card is drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting
• Solution:

There are 52 cards in a pack.

one card out of 52 can be drawn by 52Cways

Hence n(S) = 52C1 = 52

Let A be the event of getting a spade card

There are 13 spade cards in a pack

one spade card out of 13 can be drawn by 13Cways

∴ n(A) = 13C1 =  13

By the definition P(A) = n(A)/n(S) = 13/52 = 1/4

Therefore the probability of getting a spade card is 1/4

• a red card

Let B be the event of getting a red card

There are 26 red cards in a pack

one red card out of 26 can be drawn by 26Cways

∴ n(A) = 26C1 =  26

By the definition P(B) = n(B)/n(S) = 26/52 = 1/2

Therefore the probability of getting a red card is 1/2

• a black card

Let C be the event of getting a black card

There are 26 black cards in a pack

one black card out of 26 can be drawn by 26Cways

∴ n(C) = 26C1 =  26

By the definition P(C) = n(C)/n(S) = 26/52 = 1/2

Therefore the probability of getting a black card is 1/2

• a king

Let D be the event of getting a king

There are 4 kings in a pack

one king out of 4 can be drawn by 4Cways

∴ n(D) = 4C1 =  4

By the definition P(D) = n(D)/n(S) = 4/52 = 1/13

Therefore the probability of getting a king is 1/13

Note: Probability of getting a card of a particular denomination is always 1/13

• a red ace

Let D be the event of getting a red ace

There are 2 red aces in a pack

one red ace out of 2 can be drawn by 2Cways

∴ n(D) = 2C1 =  2

By the definition P(D) = n(D)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red ace is 1/26

• a face card

Let E be the event of getting a face card

There are 12 face cards in a pack

one face card out of 12 can be drawn by 12Cways

∴ n(E) = 12C1 =  12

By the definition P(E) = n(E)/n(S) = 22/52 = 3/13

Therefore the probability of getting a face card is 3/13

• a card of denomination between 4 and 10

Let F be the event of getting a card 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

one such card out of 20 can be drawn by 20Cways

∴ n(F) = 20C1 =  20

By the definition P(F) = n(F)/n(S) = 20/52 = 5/13

Therefore the probability of getting a card of denomination between 4 and 10 is 5/13

• a red face card

Let G be the event of getting a red face card

There are 6 red face cards in a pack

one face card out of 6 can be drawn by 6Cways

∴ n(G) = 6C1 =  6

By the definition P(G) = n(G)/n(S) = 6/52 = 3/26

Therefore the probability of getting a red face card is 3/26.

• a queen of hearts

Let H be the event of getting a queen of hearts

There is only one queen of heart in a pack

one queen of hearts out of 1 can be drawn by 1 way

∴ n(H) = 1

By the definition P(H) = n(H)/n(S) = 1/52

Therefore the probability of getting a queen of hearts is 1/52

• a queen or a king

Let J be the event of getting a queen or a king

There 4 kings and 4 queens in a pack

Thus there are 4 + 4 = 8 favourable points.

one required card out of 8 favourable points can be drawn by 8Cways

∴ n(J) = 8C1 =  8

By the definition P(J) = n(J)/n(S) = 8/52 = 2/13

Therefore the probability of getting a queen or a king is 2/13

• a red card and a king

Let K be the event of getting a red card or a king

There 2 red cards which are king

Thus there are 2 favourable points.

one required card out of 2 favourable points can be drawn by 2Cways

∴ n(J) = 2C1 =  2

By the definition P(K) = n(K)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red card and king is 1/26

• a red card or a king  /a red king

Let L be the event of getting a red card or a king

There 26 red cards (including 2 red kings) and 2 black kings in a pack

Thus there are 26 + 2 = 28 favourable points.

one required card out of 28 favourable points can be drawn by 28Cways

∴ n(L) = 28C1 =  28

By the definition P(L) = n(L)/n(S) = 28/52 = 7/13

Therefore the probability of getting a red card or a king (a red king) is 7/13

• Neither the heart nor the king

Let M be the event of getting neither the heart nor the king

There 36 non-heart cards (excluding 3 kings) in a pack

one required card out of 36  favourable points can be drawn by 36Cways

∴ n(M) = 36C1 =  36

By the definition P(M) = n(M)/n(S) = 36/52 = 9/13

Therefore the probability of getting neither the heart nor the king is 9/13

• Neither an ace nor the king

Let N be the event of getting neither an ace nor a king

There are 4 aces and 4 kings in a pack

There 44 non-ace and non-king cards in a pack

one required card out of 44  favourable points can be drawn by 44Cways

∴ n(M) = 44C1 =  44

By the definition P(N) = n(N)/n(S) = 44/52 = 11/13

Therefore the probability of getting neither ace nor the king is 11/13

• no diamond

Let Q be the event of getting no diamond

There 39 non-diamond cards in a pack

one required card out of 39  favourable points can be drawn by 39Cways

∴ n(M) = 39C1 =  39

By the definition P(Q) = n(Q)/n(S) = 39/52 = 3/4

Therefore the probability of getting no diamond is 3/4

• no ace

Let R be the event of getting no ace

There are 4 aces in a pack

There 48 non-ace cards in a pack

one required card out of 48  favourable points can be drawn by 48Cways

∴ n(M) = 48C1 =  48

By the definition P(R) = n(R)/n(S) = 48/52 = 12/13

Therefore the probability of getting no ace is 12/13.

• not a black card

Let T be the event of getting no black card

There 26 non-black (red) cards in a pack

one required card out of 26  favourable points can be drawn by 26Cways

∴ n(T) = 26C1 =  26

By the definition P(T) = n(T)/n(S) = 26/52 = 1/2

Therefore the probability of getting no black card is 1/2.

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