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A certain team has 12 members, including Joey. A three.....

This topic has 2 expert replies and 0 member replies

A certain team has 12 members, including Joey. A three.....

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A certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run first, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. What is the probability that Joey will be chosen to run second or third?

A. 1/1,320
B. 1/132
C. 1/110
D. 1/12
E. 1/6

The OA is the option E.

Experts, how can I solve this PS question? The first time, the probability that Joey wouldn't be chosen is 11/12? Or how should I start the solution?

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VJesus12 wrote:
A certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run first, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. What is the probability that Joey will be chosen to run second or third?

A. 1/1,320
B. 1/132
C. 1/110
D. 1/12
E. 1/6
The probability Joey will be chosen to run second is:

11/12 x 1/11 x 10/10 = 1/12

The probability Joey will be chosen to run third is:

11/12 x 10/11 x 1/10 = 1/12

Thus the probability that he will chosen to run second or third is:

1/12 + 1/12 = 2/12 = 1/6

Answer: E

_________________
Scott Woodbury-Stewart Founder and CEO

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Hi VJesus12,

We're told that a certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run first, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. We're asked for the probability that Joey will be chosen to run second or third. There are a couple of different ways to do this type of math; here's how you can break the calculation down into two 'pieces':

The probability that Joey is chosen to run second is:
(Not Joey 1st)(Joey 2nd)(Not Joey 3rd) = (11/12)(1/11)(10/10) = 1/12

The probability that Joey is chosen to run third is:
(Not Joey 1st)(Not Joey 2nd)(Joey 3rd) = (11/12)(10/11)(1/10) = 1/12

Thus, the total probability of either event occurring is 1/12 + 1/12 = 2/12 = 1/6

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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