** I understand that A, B, and C are all 1/6, I just need help with D ** Barbara
ID: 3023261 • Letter: #
Question
** I understand that A, B, and C are all 1/6, I just need help with D **
Barbara is interviewing 6 candidates for a position. As she interviews candidates, she can determine their relative rank but not their true rank. For example, if there are 5 candidates and their true rank is 4, 2, 5, 1, 3 and she has interviewed the first three, she would rank them 2, 1, 3. She interviews the candidates one at a time, and after each interview she must either offer the candidate the position or reject him. Once she rejects a candidate he is lost forever. She wants to decide on a strategy for deciding when to stop and accept a candidate that will maximize the probability that she gets the best candidate. Assume the candidates arrive in random order.
(A) What is the probability that she gets the best candidate if she chooses the first one she interviews?
(B) What is the probability that she gets the best candidate if she chooses the fifth one she interviews?
(C) What is the probability that she gets the best candidate if she interviews all of the candidates?
(D) Suppose the interviewer adopts the following strategy. She rejects the first 4 candidates. If the second to last candidate is better than all of the previous 4 candidates, then she chooses that candidate. Otherwise, she chooses the last candidate. What is the probability that she gets the best candidate with this strategy?
Hint: Think about when the best candidate arrives. In particular, if the best candidate arrives last, what will happen if the second-best candidate arrives second to last?
Explanation / Answer
The situations where the interviewer gets the best candidate are:
1) Best candidate arrives second to last. Here, he would be better than all the previous 4 candidates and hence will be chosen.
2) When the best candidate does not arrive at the 5th position then, he can only be chosen if he arrives last. Also, for him to be chosen, the second best guy should not arrive second to last position.
The slightly involved part of the question is finding out the number of favourable cases in situation 2.
Situation 1: Fix best candidate in the 5th position. The remaining 5 candidates can arrive at any of the remaining positions. Hence the number of favourable cases is 5! .
Situation 2 : Fix best candidate at the 6th place. Now, we need to bother about the second best candidate. Our strategy is: find total permutations for remaining candidates and then subtract from this the permuations in which the second best is in the 5th place. This evaluates to 5!-4! = 4!(5-1) = 4x4!
Total Probability = P(situation 1) + P(situation 2) = (5! + (4x4!))/6! = 3/10 = 0.3
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