Wow, great question! Here it's important to keep in mind the principles of conditional probability. Once one woman is selected, there is one fewer woman in the pool to be drawn from, and we need to calculate the probability of two consecutive events both happening. So for statement 1, we know that the minimum number of women in the group is 6 (more than 1/2 of the 10). Which means that, at least, the probability of drawing a woman first is 6/10. But if that woman is picked, then there would only be 5 women left out of the 9 remaining. So the probability of drawing two women in a row is:
6/10 * 5/9 = 30/90 = 1/3
Since that isn't greater than 50%, but we could have all 10 employees be women in which case it's a 100% certainty, statement 1 is not sufficient.
For statement 2, we know that the probability of two men is less than 1/10. And we know from the above that the probability of two consecutive men, if we call the number of men "m", is:
m/10 * (m-1)/9
So we have the relationship:
m/10 * (m-1)/9 < 1/10
We know that the denominator on the left will multiply to 90, so we can call the minimum probability on the right hand side of the inequality 9/90. So we want:
m(m-1)/90 < 9/90
So:
m(m-1) < 9
The smallest value of m for which this works is 3. 4*3 = 12 which is too big; 3*2 is 6 which is less than 9, so m could be 3 men. That means that the group has at least 7 women. So repeating the calculation from statement 1 with a minimum of 7 women, we have:
7/10 * 6/9 = 42/90. Since 42/90 is just less than half, we still don't know whether the probability is greater than 1/2. And since statement 2 only raises the minimum # of women higher than statement 1 had, there's no added value to using both together. That's why the answer is E.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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