Final answer:
To solve the probability of candidate A winning with the minimum majority in a committee of 5, we use binomial probability and find that the candidate wins if they get 3, 4, or 5 votes in favor. Calculating these probabilities yields a sum of 16/64 or 1/4, which is not listed among the answer choices provided.
Step-by-step explanation:
The question asks for the probability that candidate A wins the election with the minimum majority votes when there are 5 people in the committee, and each person has a 50% chance of voting in favor. In order for candidate A to win with the minimum majority, A must receive at least 3 out of 5 votes.
The scenario can be framed as a binomial probability problem, where the number of trials n is 5, the number of successes k is 3 (since we require a minimum majority), and the probability of success p for each trial is 0.5. We use the formula for binomial probability, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k).
Applying this formula to our case, we get P(X = 3) = (5 choose 3) * 0.5^3 * 0.5^(5-3). Calculating the combination (5 choose 3) yields 10. Thus, our probability is P(X = 3) = 10 * 0.5^3 * 0.5^2, which simplifies to P(X = 3) = 10 * 1/8. This results in 10/8 * 1/8, or 10/64. However, we also need to consider the cases where there are 4 or 5 votes in favor, which would also result in a majority win. Calculating these probabilities and adding them together (P(3) + P(4) + P(5)), we get a final probability of P(majority win) = 10/64 + 5/64 + 1/64, which sums to 16/64 or 1/4.
This final probability simplifies to 0.25, which is not listed in the options provided, so it seems there may be a misunderstanding or typo in the question or answer selections.