An exit poll in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that news organizations can try to figure out as soon as possible who won the election, before the votes are officially counted. This has been notoriously inaccurate in various elections, sometimes because of selection bias: the sample of people who are invited to and agree to participate in the survey may not be similar enough to the overall population of voters. Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let A be the event that they voted for A, and W be the event that they are willing to participate in the exit poll. Suppose that P(W|A) = 0.7 but P(W|A^c ) = 0.3. In the exit poll, 60% of the respondents say they voted for A (assume that they are all honest), suggesting a comfortable victory for A.

Required:
Find P(A), the true proportion of people who voted for A.

P(A) = 0.39

Step-by-step explanation:

We are given;

P(W|A) = 0.7

P(W|A^c ) = 0.3

We are told that 60% of the respondents said they voted for A. Thus;

P(A|W) = 60% = 0.6

Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).

Thus;

P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]

Now, P(A^c) can be expressed as 1 - P(A)

Thus, we now have;

P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]

Plugging in the relevant values gives;

P(W) = 0.7P(A) + 0.3(1 - P(A))

P(W) = 0.7P(A) + 0.3 - 0.3P(A)

P(W) = 0.3 + 0.4P(A)

Now,using Baye's theorem, we can find an expression for P(A|W)

Thus;

P(A|W) = [P(A ∩ W)]/P(W)

This can be further expressed as;

P(A|W) = [P(A) × P(W|A)]/P(W)

Plugging in relevant values, we have;

0.6 = 0.7P(A)/(0.3 + 0.4P(A))

Cross multiply to get;

0.6(0.3 + 0.4P(A)) = 0.7P(A)

0.18 + 0.24P(A) = 0.7P(A)

0.18 = 0.7P(A) - 0.24P(A)

0.46P(A) = 0.18

P(A) = 0.18/0.46

P(A) = 0.39

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

step-by-step explanation:

the answer is 128