Problems Based on Drawing a Playing Card

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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 playing cards.

Introduction to Playing Cards:

Before studying, the problems on playing cards, you should be thorough with the following facts:

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

Drawing a Single Playing Card From a Pack:

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 ⁵²C₁ ways

Hence n(S) = ⁵²C₁ = 52

a) a spade card

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 ¹³C₁ ways

∴ 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

b) 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 ²⁶C₁ ways

∴ 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

c) 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 ²⁶C₁ ways

∴ 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

d) 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 ⁴C₁ ways

∴ 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

e) a red ace

Let E 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 ²C₁ ways

∴ n(E) = ²C₁ =  2

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

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

f) a face card

Let F 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 ¹²C₁ ways

∴ n(F) = ¹²C₁ =  12

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

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

g) a card of denomination between 4 and 10

Let G 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 ²⁰C₁ ways

∴ n(G) = ²⁰C₁ =  20

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

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

h) a red face card

Let H 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 ⁶C₁ ways

∴ n(G) = ⁶C₁ =  6

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

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

i) a queen of hearts

Let J 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 1way

∴ n(J) = 1

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

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

j) a queen or a king

Let K 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 ⁸C₁ ways

∴ n(K) = ⁸C₁ =  8

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

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

k) a red card and a king

Let L 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 ²C₁ ways

∴ n(L) = ²C₁ =  2

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

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

l) a red card or a king  /a red king

Let M 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 ²⁸C₁ ways

∴ n(M) = ²⁸C₁ =  28

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

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

m) Neither the heart nor the king

Let N 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 ³⁶C₁ ways

∴ n(N) = ³⁶C₁ =  36

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

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

n) Neither an ace nor the king

Let Q 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 ⁴⁴C₁ ways

∴ n(Q) = ⁴⁴C₁ =  44

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

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

o) no diamond

Let R 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 ³⁹C₁ ways

∴ n(R) = ³⁹C₁ =  39

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

Therefore the probability of getting no diamond is 3/4

p) no ace

Let T 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 ⁴⁸C₁ ways

∴ n(T) = ⁴⁸C₁ =  48

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

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

q) not a black card

Let V 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 ²⁶C₁ ways

∴ n(V) = ²⁶C₁ =  26

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

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

In the next article, we shall study some basic problems of probability based on the drawing of two cards from a pack of playing cards.

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