Problems Based on Drawing 3 Playing Cards

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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 the draw of three playing cards. For e.g. three cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting all red cards

Problems based on the Draw of 3 Playing Cards:

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 ⁵²C₃ ways

Hence n(S) = ⁵²C₃ = 26 x 17 x 50

a) 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 ¹²C₃ ways

∴ 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

b) 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 ⁴⁰C₃ ways

∴ 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.

c) 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 ⁴⁰C₃ ways

∴ 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.

d) 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.

e) 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

f) 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 ²⁶C₃ ways

∴ 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

f) 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(F) = n(F)/n(S) = 21814/( 26 x 17 x 50) = 839/850

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

g) atleast one heart

Let G be the event of getting at least one heart

Thus G’ 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 ³⁹C₃ ways

∴ n(G’) = ³⁹C₃ = 13 x 19 x 37

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

Now P(G) = 1 – P(G’) = 1 – 703/1700 = 997/1700

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

h) a king,  a queen, and a jack

Let H 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(H) = ⁴C₁ x ⁴C₁ x ⁴C₁ = 4 x 4 x 4 = 64

By the definition P(H) = n(H)/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

i) 2 aces and 1 king

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

There are 4 aces and 4 kings

∴ n(J) = ⁴C₂ x ⁴C₁  = 6 x 4  = 24

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

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

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

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