If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than
1/10.
here my answer is C that is p<1/2 but OA is E. i think i am missing something while solving this, please explain..
probability
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Let's create the groups and test them
1) says that More than 1/2 of the 10 employees are women.
It's clear that if all are women the probability will be more than 1/2 so we need to prove that it can be also lower.
S = {W,W,W,W,W,W,M,M,M,M}
ways to pick two people from a set of 10 C(10,2) = 10!/8!2! = 45
ways to pick two women from a set of 6 C(6,2) = 6!/4!2! = 15
Probability of picking two women = 15/45 = 1/3 which is less than 1/2
2) we need to prove that even if the probability of picking two men is less than 1/2 we have two values for the probability of picking two women one more than 1/2 and one less than 1/2
S = {W,W,W,W,W,W,W,M,M,M}
ways to pick 2 men ----> C(3,2) = 3
probability of picking two men --> 3/45 = 1/15 which is less than 1/10
ways to pick 2 women ---> C(7,2) ---> 7!/5!2! = 21
probability of picking two women = 21/45 = 7/15 < 1/2
now we need to prove that we can find a set in which the probability is more than 1/2
S = {W,W,W,W,W,W,W,W,M,M}
ways to pick 2 women ---> C(8,2) ---> 8!/6!2! = 23
probability of picking two women = 23/45 > 1/2
if you pick
S={W,W,W,W,W,W,W,W,M,M} and S{W,W,W,W,W,W,W,M,M,M}
both 1 and 2 are correct and the probabilities are one more than 1/2 and the other less than 1/2
1) says that More than 1/2 of the 10 employees are women.
It's clear that if all are women the probability will be more than 1/2 so we need to prove that it can be also lower.
S = {W,W,W,W,W,W,M,M,M,M}
ways to pick two people from a set of 10 C(10,2) = 10!/8!2! = 45
ways to pick two women from a set of 6 C(6,2) = 6!/4!2! = 15
Probability of picking two women = 15/45 = 1/3 which is less than 1/2
2) we need to prove that even if the probability of picking two men is less than 1/2 we have two values for the probability of picking two women one more than 1/2 and one less than 1/2
S = {W,W,W,W,W,W,W,M,M,M}
ways to pick 2 men ----> C(3,2) = 3
probability of picking two men --> 3/45 = 1/15 which is less than 1/10
ways to pick 2 women ---> C(7,2) ---> 7!/5!2! = 21
probability of picking two women = 21/45 = 7/15 < 1/2
now we need to prove that we can find a set in which the probability is more than 1/2
S = {W,W,W,W,W,W,W,W,M,M}
ways to pick 2 women ---> C(8,2) ---> 8!/6!2! = 23
probability of picking two women = 23/45 > 1/2
if you pick
S={W,W,W,W,W,W,W,W,M,M} and S{W,W,W,W,W,W,W,M,M,M}
both 1 and 2 are correct and the probabilities are one more than 1/2 and the other less than 1/2
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Read Stuart's explanation of this problem if you want to see an easy and effective way to solve it:
https://www.beatthegmat.com/ds-question-t11853.html
https://www.beatthegmat.com/ds-question-t11853.html
The more you look, the more you see.
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- Master | Next Rank: 500 Posts
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- Joined: Tue Jan 20, 2009 10:16 pm
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