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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 ^{52}C_{1 }ways

Hence n(S) = ^{52}C_{1} = 52

**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 ^{13}C_{1 }ways

∴ n(A) = ^{13}C_{1} = 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 ^{26}C_{1 }ways

∴ n(A) = ^{26}C_{1} = 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 ^{26}C_{1 }ways

∴ n(C) = ^{26}C_{1} = 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 ^{4}C_{1 }ways

∴ n(D) = ^{4}C_{1} = 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 ^{2}C_{1 }ways

∴ n(D) = ^{2}C_{1} = 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 ^{12}C_{1 }ways

∴ n(E) = ^{12}C_{1} = 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 ^{20}C_{1 }ways

∴ n(F) = ^{20}C_{1} = 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 ^{6}C_{1 }ways

∴ n(G) = ^{6}C_{1} = 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 ^{8}C_{1 }ways

∴ n(J) = ^{8}C_{1} = 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 ^{2}C_{1 }ways

∴ n(J) = ^{2}C_{1} = 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 ^{28}C_{1 }ways

∴ n(L) = ^{28}C_{1} = 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 ^{36}C_{1 }ways

∴ n(M) = ^{36}C_{1} = 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 ^{44}C_{1 }ways

∴ n(M) = ^{44}C_{1} = 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 ^{39}C_{1 }ways

∴ n(M) = ^{39}C_{1} = 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 ^{48}C_{1 }ways

∴ n(M) = ^{48}C_{1} = 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 ^{26}C_{1 }ways

∴ n(T) = ^{26}C_{1} = 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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