Question

A ball is taken at random from an urn containing 2 red balls and 3 blue balls. If the ball is red a fair coin is tossed three times. If the ball is blue non-fair coin is tossed three times; for this second coin the probability of heads is .6. In either case we count the number of heads in the three tosses and call that number X. (a) Compute the conditional probability that X-2 if it is known that the selected ball is red (b) Compute the conditional probability that X 2 if it is known that the selected ball is blue (c) Compute the probability that X 2 (d) Compute the conditional probability that the selected ball is blue given that X 2.

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Answer #1

Total number of balls : 2+3 = 5

Let R shows the event that red ball is selected and B shows the event that blue ball is selected. So we have

P(R) = 2/ 5 = 0.40, P(B) = 3/5 = 0.60

(a)

When a red ball is selected then fair coin is tossed so X has binomial distribution with parameters n=3 and p=0.50. So,

(0.5)2(1-0.5)3-2ー0.375

(b)

When a blue ball is selected then fair coin is tossed so X has binomial distribution with parameters n=3 and p=0.60. So,

P(X-21 B) ) (0.6)2(1-0.6)3-2 = 0.432

(c)

By the law of total probability

PlX-2) 0.6-0.4092 P(X-2|R)P(R) +PlX-2B)P(B) 0.375. 0.40+0.432

(d)

PlBIX = 2) = .P(X2B)P B) 0.432 0.6 = 0.6334 P(X 2) 0.4092

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