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OG2015 PS A committee is composed

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by lionsshare » Wed Sep 13, 2017 7:42 pm
A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by

(A) w/m
(B) w/(w+m)
(C) (w+3)/(m+2)
(D) (w+3)/(w+m+3)
(E) (w+3)/(w+m+5)

OA: E

Hello, Experts! Please explain how to solve the problem. Thanks a lot.
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by Jay@ManhattanReview » Wed Sep 13, 2017 9:22 pm
lionsshare wrote:A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by

(A) w/m
(B) w/(w+m)
(C) (w+3)/(m+2)
(D) (w+3)/(w+m+3)
(E) (w+3)/(w+m+5)

OA: E

Hello, Experts! Please explain how to solve the problem. Thanks a lot.
After the inclusion of 3 women and 2 men, there are (w + 3) women and (m + 2) men in the committee.

So there are a total of (w + 3 + m + 2) = (w + m +5) persons in the committee

We have to select one person such that the selected person is a woman.

Probability that the person is woman = (Total number of woman) / (Total number of persons)

Probability that the person is woman = (w + 3) / (w + m + 5)

The correct answer: E

Hope this helps!

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by Brent@GMATPrepNow » Thu Sep 14, 2017 6:43 am
lionsshare wrote:A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by

(A) w/m
(B) w/(w+m)
(C) (w+3)/(m+2)
(D) (w+3)/(w+m+3)
(E) (w+3)/(w+m+5)
A committee is composed of w women and m men. If 3 women and 2 men are added to the committee . . .
After the 3 women and 2 men are added, we have...
w+3 women and m+2 men

. . . and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by
P(select woman) = (# of women in committee)/(TOTAL # of people on committee)
= (w+3)/(w+3 + m+2 )
= (w+3)/(w+m+5)

Answer: E

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Brent
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by [email protected] » Thu Sep 14, 2017 11:10 am
Hi lionsshare,

This question is perfect for TESTing VALUES. We're told that a committee is composed of W women and M men.

IF...
W = 2
M = 3

3 women and 2 men are then ADDED to the committee.....

Women = 2+3 = 5
Men = 3+2 = 5

The question asks for the probability of selecting a woman from this larger group:

5 women and 10 total people --> 5/10

So we're looking for an answer that = 5/10 = 1/2 when W=2 and M=3

Answer A: 2/3 NOT a match
Answer B: 2/(2+3) = 2/5 NOT a match
Answer C: (2+3)/(3+2) = 5/5 NOT a match
Answer D: (2+3)/(2+3+3) = 5/8 NOT a match
Answer E: (2+3)/(2+3+5) = 5/10 This IS a match

Final Answer: E

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by Jeff@TargetTestPrep » Tue Sep 19, 2017 2:56 pm
lionsshare wrote:A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by

(A) w/m
(B) w/(w+m)
(C) (w+3)/(m+2)
(D) (w+3)/(w+m+3)
(E) (w+3)/(w+m+5)

OA: E
To solve the problem, we recall that:

probability = (number of favorable outcomes)/(total number of outcomes)

In the problem, we are asked the probability of selecting a woman. In the enlarged committee, there are (w + 3) women and there are (m + 2) men, making the total number of people (w + 3 + m + 2) = (w + m + 5).

Thus, the probability of selecting a woman is:

(w + 3)/(w + m + 5)

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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