16. Suppose a bag contains 12 poker chips (6 blue, 4 red and 2 yellow). Three ch

Question

16. Suppose a bag contains 12 poker chips (6 blue, 4 red and 2 yellow). Three chips are selected from the bag. Define the following events: A: Event A occurs when at least one of the three chips is blue B: Event B occurs when the three selected chips contain one of each color a. b. c. d. e. Compute P(A). Compute P(B). Compute P(AIB) Compute P(BIA) Are events A and B independent? Consider the following table representing all combinations of poker chips: BBR BBY BRR BYY -20 (0)-60 2(6)-30 36 BRY 6"4'2-48 RRY 22)-12 YYR Total +1244 a) PIA)-1.220,0.909. b) P(B)0.21818 c) PAIB) Piana)=48/220 d) PIBIA) Pn8/22024 e) No, events A and B are not independent: PIA|B) PIA) P(B) 48/220 P(A) 200/220

Explanation / Answer

6 blue, 4 red and 2 yellow

(a)
Consider an event where 3 chips are drawn and none is blue, probability of this event is 6C3/12C3 = 20/220 = 1/11

Hence probability that at least one of the three chips is blue, P(A) = 1 - 1/11 = 10/11 = 0.9091

(b)
One of each color chip selected, blue chip can be selected in 6C1 ways, 1 red chip can be selected in 4C1 ways and 1 yellow chip can be selected in 2C1 ways

Hence required probabiltiy P(B) = (6C1)*(4C1)*(2C1)/(12C3) = 0.2182

(c)
P(A and B) = 0, because both events can not occur simultaneously.

hence, P(A|B) = P(A and B)/P(B) = 0

(d)
P(A and B) = 0, because both events can not occur simultaneously.

hence, P(B|A) = P(A and B)/P(A) = 0

(e)
P(A) * P(B) = 0.9091 * 0.2182 = 0.1984
P(A and B) = 0

As P(A) * P(A) is not equal to P(A and B), event A and B are not independent.

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