A In a certain city 30 percent of the people are Conservativ

A. In a certain city, 30 percent of the people are Conservatives 50 percent are Liberals, and 20 percent are Independents. Records show that in a particular election, 65 percent of the Conservatives voted, 82 percent of the Liberals voted, and 50 percent of the Independents voted. (This city must be in Canada!) If a person in the city is selected at random and it is learned that she did not vote in the last election, what is the probability that she is a Liberal?

B. Suppose that a box contains five coins, and that for each coin there is a different probability that a head will be obtained when the coin is tssed. Let probability of a head when the ith coin is tossed (i=1, . . . , 5), and suppose that p1 = 0, p2 = 1/4, p3 = 1/2, p4 = 3/4, and P5 = 1.

(a) Suppose that one coin is selected at random from the box and when it is tossed once, a head is obtained. What is the posterior probability that the ith coin was selected (i = 1, . . . , 5)?

(b) If the same coin were tossed again, what would be the probability of obtaining another head?

Solution

A)

Let

C = conservative
L = liberal
I = independent
V = voted
v\' = not voted

P(L|V\') = P(L and V\') / P(V\')

However,

P(L and V\') = P(L) P(V\'|L) = 0.50(1-0.82) = 0.09

Also,

P(V\') = P(C) P(V\'|C) + P(L) P(V\'|L) + P(I) P(V\'|I) = 0.30*(1-0.65) + 0.50*(1-0.82) + 0.20*(1-0.50) = 0.295

Thus,

P(L|V\') = P(L and V\') / P(V\') = 0.09/0.295 = 0.305084746 [ANSWER]

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