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06 Mar 2015, 07:26
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A group of N students will be randomly seated in a row of N chairs. What is the probability that Beth, one of the students, will be at the extreme right-hand end of the row?
(1) N is an odd prime number
(2) the probability that Steve, another one of the students, is at the extreme left-hand end of the row is 1/13
(1) N is an odd prime number
(2) the probability that Steve, another one of the students, is at the extreme left-hand end of the row is 1/13
06 Mar 2015, 08:34
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1) This statement, alone and by itself, is insufficient as there can be infinite prime numbers possible.
2) when we fix one seat then we are arranging n-1 ! students, n! was the original arrangements possible, so probability = n-1!/n! = 1/n
left and right will be similar in probability and hence we can calculate the probability. Answer is B.
2) when we fix one seat then we are arranging n-1 ! students, n! was the original arrangements possible, so probability = n-1!/n! = 1/n
left and right will be similar in probability and hence we can calculate the probability. Answer is B.
08 Mar 2015, 16:27
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Bunuel
A group of N students will be randomly seated in a row of N chairs. What is the probability that Beth, one of the students, will be at the extreme right-hand end of the row?
(1) N is an odd prime number
(2) the probability that Steve, another one of the students, is at the extreme left-hand end of the row is 1/13
Kudos for a correct solution.
(1) N is an odd prime number
(2) the probability that Steve, another one of the students, is at the extreme left-hand end of the row is 1/13
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:
Statement #1: the chance that Beth will be on the right end depends on the number of students. Obviously, as the number of students increases, it becomes less and less likely that any given student is at the end of the row. With this statement, we know only that N is an odd prime number — it could be 3 or 5 or 7, or it could be 109. It could be N = 524,287 (you do not need to know how to find prime number this big!!) Obviously, if N is very big the chances that Beth will be on the right end of this half-million row would be just about zero. Different values of N give different probabilities, and different answers to the prompt question. This statement, alone and by itself, is insufficient.
Statement #2: If the probability that Steve is on the left end is 1/13, this must mean that N = 13. Thus, the probability that any given student is on any given end is 1/13. This allows us to give a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.
First not sufficient, second sufficient.
Answer = B.
