Nine women and two men sat I chairs. A male occupied seat 9 and 11. Eleven keys were drawn from a basket. Find the probability the woman in the 6th seat picked the correct key a.) if the keys were replaced and b) if the keys were not replaced. Find the probability the man in seat #9 selected the correct key.

So, you forgot to post your work and what you thought the answer was. We don't do your assignments for you... but we will help after you show us the work you did do.

The answer to part a) for both is 1/11. If the keys are replaced, each person regardless of position will have a 1/11 chance of selecting the correct key.For part b.), this would be a permutation because the order does make a difference. There would be 11! Ways to select a key. The woman in the sixth seat would have a 6!/11! Ways of selecting the correct key (? ).

The answer to part a) for both is 1/11. If the keys are replaced, each person regardless of position will have a 1/11 chance of selecting the correct key.For part b.), this would be a permutation because the order does make a difference. There would be 11! Ways to select a key. The woman in the sixth seat would have a 6!/11! Ways of selecting the correct key (? ).

Who WROTE this question? It assumes that the keys are numbered 1 -11 and that they must correspond to the numbers on the chairs, also 1 - 11, but mainly that 'if the keys are replaced' means put back into a random drawing the same as the first. (Or only those that didn't match a person the first time?)

This is really 3 questions. Are you copying it as written, or did you shorten it?

(The gender of the people is irrelevant as written.)

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Originally Posted by miamicane

The answer to part a) for both is 1/11. If the keys are replaced, each person regardless of position will have a 1/11 chance of selecting the correct key.For part b.), this would be a permutation because the order does make a difference. There would be 11! Ways to select a key. The woman in the sixth seat would have a 6!/11! Ways of selecting the correct key (? ).

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You are correct for Part A - with replacement the gender of the person doesn't matter, and there is a 1/11 probability that the correct key ends up in the correct position.

Your answer to part B though is incorrect. It is true that there are 11! ways that the 11 keys can be distributed without replacement, but the number of ways that key 6 ends up in position 6 is not 6!. Instead, consider this: if we know that key 6 is correctly placed how many ways can the other 10 keys be arranged among the other 10 people?