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

Understanding the Question

We need to determine whether the probability p of selecting 2 women from 10 employees is greater than \(\frac{1}{2}\).

What We Need to Determine

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).

Key Insight

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.

Analyzing Statement 1

Statement 1: More than \(\frac{1}{2}\) of the 10 employees are women.

What Statement 1 Tells Us

This means we have at least 6 women (since more than 5 out of 10).

Testing Different Scenarios

Let's examine the range of possibilities:

• Minimum case (6 women, 4 men): With a barely-more-than-half split, there are still many ways to select mixed pairs (1 woman + 1 man) or 2 men. The substantial number of men keeps other outcomes common.
• Maximum case (10 women, 0 men): With all women, selecting 2 women is guaranteed (\(\mathrm{p} = 1\)).

The critical question: Can we get different answers to "Is \(\mathrm{p} > \frac{1}{2}\)?"

Yes! Consider:

• With 6-7 women: There are enough men to make non-women pairs common, likely keeping \(\mathrm{p} < \frac{1}{2}\)
• With 8+ women: The female majority becomes overwhelming enough that \(\mathrm{p} > \frac{1}{2}\)

Conclusion

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.

Analyzing Statement 2

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 \(\frac{1}{10}\).

What Statement 2 Tells Us

If selecting 2 men is very unlikely (\(< \frac{1}{10}\)), there must be very few men in the group.

Logical Analysis

To make the "2 men" probability less than \(\frac{1}{10}\):

• If there were 4 or more men, selecting 2 men wouldn't be rare enough
• This constraint forces us to have at most 3 men
• Therefore, we must have at least 7 women

But here's the crucial question: Does having 7+ women guarantee \(\mathrm{p} > \frac{1}{2}\)?

Let's consider our range:

• With exactly 7 women (minimum): We need to check if this gives \(\mathrm{p} > \frac{1}{2}\)
• With 10 women (maximum): Definitely \(\mathrm{p} > \frac{1}{2}\)

Since the answer could vary between YES and NO depending on the exact count, we cannot determine a definitive answer.

Conclusion

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.

Combining Statements

Combined Information

From both statements together:

• Statement 1: At least 6 women
• Statement 2: At most 3 men (so at least 7 women)
• Combined constraint: We have at least 7 women

The Critical Test

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:

• Women pairs: Choose 2 from 7 women = 21 ways
• Mixed pairs: Choose 1 woman and 1 man = \(7 \times 3 = 21\) ways
• Men pairs: Choose 2 from 3 men = 3 ways

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 \(\mathrm{p} > \frac{1}{2}\).

Conclusion

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 Answer: E

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."

Strategic Takeaway

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 \(\frac{1}{2}\).