Probability: Problems Based on Playing Cards – 03

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Example – 01:

  • Three 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.

Three cards out of 52 can be drawn by 52Cways

Hence n(S) = 52C3 = 26 x 17 x 50

  • all face cards

Let A be the event of getting all face cards



There are 12 face cards in a pack

three face cards out of 12 can be drawn by 12Cways

∴ n(A) = 12C3 = 4 x 11 x 5

By the definition P(A) = n(A)/n(S) = (4 x 11 x 5)/( 26 x 17 x 50) = 11/1105



Therefore the probability of getting all face cards is 11/1105

  • no face card

Let B be the event of getting no face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by 40Cways



∴ n(B) = 40C3 = 20 x 13 x 38

By the definition P(B) = n(B)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85

Therefore the probability of getting no face card is 38/85.

  • atleast one face card

Let C be the event of getting at least one face card

Thus C is an event of getting no face card



There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by 40Cways

∴ n(C) = 40C3 = 20 x 13 x 38

By the definition P(C) = n(C)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85



Now P(C) = 1 – P(C) = 1 – 38/85 = 47/85

Therefore the probability of getting at least one face card is 47/85.

  • at least two face cards

Let D be the event of getting at least two face cards

There are two possibilities



Case – 1: Getting two face cards and 1 non-face card

Case – 2: All three face cards

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(D) = 12C2 x  40C1 + 12C3  = 6 x 11 x 40 + 4 x 11 x 5 = 2860

By the definition P(D) = n(D)/n(S) = 3794/( 26 x 17 x 50) = 11/85



Therefore the probability of getting at least two face cards is 11/85.

  • at most two face cards

Let E be the event of getting at most two face cards

There are three possibilities

Case – 1: Getting no face card



Case – 2: Getting one face card and 2 non-face cards

Case – 3: Getting two face cards and 1 non-face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(E) = 40C3 +12C1 x  40C2 +  12C2 x  40C1 = 20 x 13 x 38 + 12 x 20 x 39 + 6 x 11 x 40 = 21880

By the definition P(E) = n(E)/n(S) = 21880/( 26 x 17 x 50) = 1094/1105



Therefore the probability of getting at most two face cards is 1094/1105

  • all red cards

Let F be the event of getting all red cards

There are 26 red cards in a pack

three red cards out of 26 red cards can be drawn by 26Cways

∴ n(E) = 26C3 = 4 x 11 x 5 = 13 x 25 x 8

By the definition P(E) = n(E)/n(S) = (13 x 25 x 8)/( 26 x 17 x 50) = 2/17

Therefore the probability of getting all red cards is 2/17

  • all are not heart

Let F be the event of getting draw such that all are not heart

Thus F is the event that the draw consists of atmost two heart

There are three possibilities

Case – 1: Getting no heart

Case – 2: Getting one heart and 2 non hearts

Case – 3: Getting two hearts and 1 non heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

∴ n(F) = 39C3 +13C1 x  39C2 +  13C2 x  39C1 = 13 x 19 x 37 + 13 x 39 x 19 + 13 x 6 x 39 = 21814

By the definition P(E) = n(E)/n(S) = 21814/( 26 x 17 x 50) = 839/850

Therefore the probability of getting all not heart is 839/850

  • atleast one heart

Let F be the event of getting at least one heart

Thus F is an event of getting no heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

three non-heart cards out of 39 non-heart cards can be drawn by 39Cways

∴ n(F) = 39C3 = 13 x 19 x 37

By the definition P(F) = n(F)/n(S) = (13 x 19 x 37)/( 26 x 17 x 50) = 703/1700

Now P(F) = 1 – P(F) = 1 – 703/1700 = 997/1700

Therefore the probability of getting at least one heart is 997/1700

  • a king,  a queen and a jack

Let G be the event of getting a king,  a queen and a jack

There are 4 kings, 4 queens and 4 jacks in a pack

Each specific selection can be done by 4C1 ways

∴ n(G) = 4C14C14C1 = 4 x 4 x 4 = 64

By the definition P(G) = n(G)/n(S) =64/( 26 x 17 x 50) = 16/5525

Therefore the probability of getting a king,  a queen and a jack is 16/5525

  • 2 aces and 1 king

Let H be the event of getting two aces and 1 king

There are 4 aces and 4 kings

∴ n(H) = 4C24C1  = 6 x 4  = 24

By the definition P(H) = n(H)/n(S) =24/( 26 x 17 x 50) = 6/5525

Therefore the probability of getting two aces and one king is 6/5525

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