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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 probability that both representatives selected will be women, is \(\mathrm{p} > \frac{1}{2}\)?
We need to determine whether the probability p of selecting 2 women from 10 employees is greater than \(\frac{1}{2}\).
The question asks: Is \(\mathrm{p} > \frac{1}{2}\)?
This is a yes/no question. To have \(\mathrm{p} > \frac{1}{2}\), selecting 2 women must be MORE likely than all other possible outcomes combined (selecting 2 men or selecting 1 man and 1 woman).
For the probability of selecting 2 women to exceed \(\frac{1}{2}\), women must significantly outnumber men in the group—not just be a simple majority. The favorable outcome (2 women) needs to dominate all other outcomes combined.
Statement 1: More than \(\frac{1}{2}\) of the 10 employees are women.
This means we have at least 6 women (since more than 5 out of 10).
Let's examine the range of possibilities:
The critical question: Can we get different answers to "Is \(\mathrm{p} > \frac{1}{2}\)?"
Yes! Consider:
Statement 1 is NOT sufficient because we get different answers depending on the exact number of women.
[STOP - Not Sufficient!] This eliminates choices A and D.
Now let's forget Statement 1 completely and analyze Statement 2 independently.
Statement 2: The probability that both representatives selected will be men is less than \(\frac{1}{10}\).
If selecting 2 men is very unlikely (\(< \frac{1}{10}\)), there must be very few men in the group.
To make the "2 men" probability less than \(\frac{1}{10}\):
But here's the crucial question: Does having 7+ women guarantee \(\mathrm{p} > \frac{1}{2}\)?
Let's consider our range:
Since the answer could vary between YES and NO depending on the exact count, we cannot determine a definitive answer.
Statement 2 is NOT sufficient because it narrows our range but doesn't determine whether \(\mathrm{p} > \frac{1}{2}\).
[STOP - Not Sufficient!] This eliminates choice B.
From both statements together:
Is 7 women enough to guarantee \(\mathrm{p} > \frac{1}{2}\)?
Let's think about the possible outcomes with exactly 7 women and 3 men:
Total favorable outcomes (2 women): 21
Total other outcomes: 21 + 3 = 24
Since 21 < 24, we have \(\mathrm{p} < \frac{1}{2}\) with exactly 7 women.
But what if we have 8 or more women? The women pairs would increase dramatically while mixed pairs decrease, giving us \(\mathrm{p} > \frac{1}{2}\).
Since we could have exactly 7 women (answer: NO) or 8+ women (answer: YES), we still cannot definitively answer the question.
[STOP - Not Sufficient!] This eliminates choice C.
The statements together are not sufficient because we cannot determine whether we have exactly 7 women (where \(\mathrm{p} < \frac{1}{2}\)) or 8+ women (where \(\mathrm{p} > \frac{1}{2}\)).
Answer Choice E: "The statements together are not sufficient."
When asked whether a probability exceeds \(\frac{1}{2}\), remember that the favorable outcome must outnumber all other outcomes combined. A simple majority in the underlying population isn't enough—you need a significant imbalance for the probability to tip past \(\frac{1}{2}\).