A gambler has 3 coins in his pocket two of which are fair an

A gambler has 3 coins in his pocket, two of which are fair and one is two headed. He selects a coin. If he tosses the selected coin

(a)   and it turns up a head, what is the probability that the coin is a fair?
(b)   5 times, what is the probability that he gets five heads?
(c)   6 times, what is the probability of getting 5 heads followed by a tail?

Solution

A)

P(head) = P(fair) P(head|fair) + P(not fair) P(head|not fair) = (2/3)(1/2) + (1/3)(1)

P(head) = 2/3

P(fair|head) = P(head and fair) / P(head) = P(fair) P(head|fair) / P(head)

= (2/3)(1/2) / (2/3)

= 1/2 [ANSWER]

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b)

P(5 heads) = P(fair) P(head|fair)^5 + P(not fair) P(head|not fair)^5

= (2/3)(1/2)^5 + (1/3)(1^5)

= 17/48 [ANSWER]

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c)

This cannot happen when you have a two headed coin. Thus, he has to draw the fair coins,

P(5 heads then tail) = P(fair) P(heads|fair)^5 P(tail|fair)

= (2/3)(1/2)^5 (1/2)

= 1/96 [ANSWER]