Probability

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pm; Followed by:6 members

Probability

by BTGmoderatorRO » Sun Dec 17, 2017 5:29 am

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 fiÂ…rst, 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) 1/1320
(B) 1/132
(C) 1/110
(D) 1/12
(E) 1/6

OA is E

I need an elaborate breakdown on why E is the correct answer. Can any expert assist here pls

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Source: Beat The GMAT — Problem Solving | Sun Dec 17, 2017 5:29 am

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by [email protected] » Sun Dec 17, 2017 1:28 pm

Hi Roland2rule,

We're told that a certain team has 12 members, including Joey - and a three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run fiÂ…rst, 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. Since there are two possible "successful" outcomes, we have to complete two calculations:

Probability of Joey running 2nd: = (Not Joey)(Joey)(Not Joey) = (11/12)(1/11)(10/10) = 1/12
Probability of Joey running 3rd: = (Not Joey)(Not Joey)(Joey) = (11/12)(10/11)(1/10) = 1/12

Probability of either of those options occurring = (1/12) + (1/12) = 2/12 = 1/6

Final Answer: E

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Rich

Contact Rich at [email protected]

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by Scott@TargetTestPrep » Mon Sep 16, 2019 10:21 am

BTGmoderatorRO 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 fiÂ…rst, 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) 1/1320
(B) 1/132
(C) 1/110
(D) 1/12
(E) 1/6

OA is E

I need an elaborate breakdown on why E is the correct answer. Can any expert assist here pls

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 be chosen to run second or third is:

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

Answer: E

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