## If $$2$$ different representatives are to be selected at random from a group of $$10$$ employees and if $$p$$ is the pro

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### If $$2$$ different representatives are to be selected at random from a group of $$10$$ employees and if $$p$$ is the pro

by Vincen » Fri Sep 03, 2021 10:07 am

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## Global Stats

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 $$p > \dfrac12?$$

(1) More than $$\dfrac12$$ of the $$10$$ employees are women.
(2) The probability that both representatives selected will be men is less than $$\dfrac1{10}.$$

Source: Official Guide

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### Re: If $$2$$ different representatives are to be selected at random from a group of $$10$$ employees and if $$p$$ is the

by [email protected] » Sat Sep 04, 2021 6:27 am

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## Global Stats

Vincen wrote:
Fri Sep 03, 2021 10:07 am
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 $$p > \dfrac12?$$

(1) More than $$\dfrac12$$ of the $$10$$ employees are women.
(2) The probability that both representatives selected will be men is less than $$\dfrac1{10}.$$

Source: Official Guide
Target question: Is the probability that both representatives selected will be women > 1/2?

This is a good candidate for rephrasing the target question.

Let's examine some scenarios and see which ones yield situation where the probability that both representatives selected will be women > 1/2
Scenario #1 - 5 women & 5 men: P(both selected people are women) = (5/10)(4/9) = 20/90 (NOT greater than 1/2)
Scenario #2 - 6 women & 4 men: P(both selected people are women) = (6/10)(5/9) = 30/90 (NOT greater than 1/2)
Scenario #3 - 7 women & 3 men: P(both selected people are women) = (7/10)(6/9) = 42/90 (NOT greater than 1/2)
Scenario #4 - 8 women & 2 men: P(both selected people are women) = (8/10)(7/9) = 56/90 (PERFECT - greater than 1/2)

IMPORTANT: So, if there are 8 or more women, the probability will be greater than 1/2
We can even REPHRASE the target question...
REPHRASED target question: Are there 8 or more women?

Statement 1: More than 1/2 of the 10 employees are women.
This is not enough information. Consider these two conflicting cases:
Case a: there are 7 women, in which case there are NOT 8 or more women
Case b: there are 8 women, in which case there ARE 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statement 2: The probability that both representatives selected will be men is less than 1/10.
This is not enough information. Consider these two conflicting cases:
Case a: there are 8 women & 2 men. Here P(both are men) = (2/10)(1/9) = 2/90, which is less than 1/2. In this case there ARE 8 or more women
Case b: there are 7 women & 3 men. Here P(both are men) = (3/10)(2/9) = 6/90, which is less than 1/2. In this case there are NOT 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Even when we combine the statements, we can see that it's possible to have 7 women in the group OR 8 women in the group.
Since we still cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT