Question

A handful of cards contains 4 black cards and 4 red cards. 3 cards are drawn at random without replacement. The events A, B, C are defined as follows: • A ={more black cards than red cards are drawn} • B ={at least one black card is drawn} • C ={all the cards drawn are of the same color} (a) Compute P(A) (b) Compute P(B) (c) Compute P(C) (d) Compute P(A|B) (e) Are A and C independent?

Answer 1

Total Cards = 4 Black + 4 Red = 8. We are drawing 3 cards without replacement.

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(a) P(More Black than Red) i.e either 2 Black cards and 1 Red, or All 3 black cards

Case 1: 2 Black cards and 1 Red card: Let us Assume that we are picking the following card in order BBR.

P(The first card is Black) = (4/8)

P(The second card is Black) = (3/7)

P(The third card is Red) = (4/6)

Therefore the probability = (4/8) * (3/7) * (4/6) = 1/7

There are 3 ways of pulling out these cards i.e BBR (as done above), BRB and RBB. Each of the probabilities will be the same, and hence the required probability for Case 1 = 3 * 1/7 = 3/7

Case 2: All 3 Black cards which can be done in 1 way only

P(The first card is Black) = (4/8)

P(The second card is Black) = (3/7)

P(The third card is Red) = (2/6)

Therefore the probability for case 2 = (4/8) * (3/7) * (2/6) = 1/14

Therefore the probability = P(2 Blacks and 1 Red) + P( All 3 Blacks) = (3/7) + (1/14)

Therefore P(A) = 1/2

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(b) At least 1 black card is drawn. So we need to find the probabilities for 1 Black and 2 Red, 2 Blacks and 1 Red and 3 Blacks

From (a) we have the Probabilities for 2 Blacks and 1 Red and 3 Blacks as 3/7 and 1/14.

1 Black and 2 Reds: Let us Assume that we are picking the following card in order BRR.

P(The first card is Black) = (4/8)

P(The second card is Red) = (4/7)

P(The third card is Red) = (3/6)

Therefore the probability = (4/8) * (4/7) * (3/6) = 1/7

There are 3 ways of pulling out these cards i.e BRR (as done above), RBR and BBR. Each of the probabilities will be the same, and hence the required probability = 3 * 1/7 = 3/7

Therefore the Probability = P(1 Black and 2 Reds) + P(2 Blacks and 1 Red) + P( All 3 Blacks)

= (3/7) + (3/7) + (1/14) = 5/14

Therefore P(B) = 13/14

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(c) All cards drawn of the same colour, i.e all 3 are black or all 3 are red. Since the number of each colour is the same, their probabilities are also the same.

From (a) P(All 3 Black) = 1/14

Therefore P(All 3 red) = 1/14

Therefore the probability = 1/14 + 1/14

P(C) = 1/7

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(d) P(A/B) which means Probability of event A happening, given that B has happened.

By Bayes Theorem, P(A/B) = P(A B)/P(B)

P(A B) = common value between A (more black cards than red) and B (At least 1 black card)

= Common value between A(2B and 1R, 3B) and B(1B and 2R, 2B and 1R, 3B)

P(A B) = P(2B and 1R + 3B) = P(A) = 1/2

P(B) = 13/14

Therefore P(A/B) = (1/2) / (13/14) = 14/26 = 7/13

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(e) If 2 sets are independent then P(A C) = P(A) x P(C)

P(A) x P(C) = (1/2) x (1/7) = 1/14

P(A C) = Common value of A(2B and 1R, 3 Blacks) and B(3 Blacks or 3 Reds)

Therefore P(A C) = P(3 Blacks) = 1/14 (from (a)

Since P(A C) = P(A) x P(C) = 1/14, the 2 events are independent.

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