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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 ^{52}C_{3 }ways

Hence n(S) = ^{52}C_{3} = 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 ^{12}C_{3 }ways

∴ n(A) = ^{12}C_{3} = 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 ^{40}C_{3 }ways

∴ n(B) = ^{40}C_{3} = 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 ^{40}C_{3 }ways

∴ n(C) = ^{40}C_{3} = 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) = ^{12}C_{2} x ^{40}C_{1} + ^{12}C_{3} = 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) = ^{40}C_{3} +^{12}C_{1} x ^{40}C_{2} + ^{12}C_{2} x ^{40}C_{1} = 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 ^{26}C_{3 }ways

∴ n(E) = ^{26}C_{3} = 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) = ^{39}C_{3} +^{13}C_{1} x ^{39}C_{2} + ^{13}C_{2} x ^{39}C_{1} = 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 ^{39}C_{3 }ways

∴ n(F) = ^{39}C_{3} = 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 ^{4}C_{1} ways

∴ n(G) = ^{4}C_{1} x ^{4}C_{1} x ^{4}C_{1} = 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) = ^{4}C_{2} x ^{4}C_{1} = 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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