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 \(\dfrac1{10}.\)

Answer: E

Source: Official Guide

## 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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Vincen wrote: ↑Fri Sep 03, 2021 10:07 amIf \(2\) 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 \(\dfrac1{10}.\)

Answer: E

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

Answer: E