Probabiity

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Probabiity

by jayanti » Mon Aug 08, 2011 8:47 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 > 1/2?

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10


correct answer is E. though i think it should be A. Pls reply......

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by Brian@VeritasPrep » Mon Aug 08, 2011 3:03 pm
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.
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by sumgb » Mon Aug 08, 2011 3:06 pm
A is clearly not the answer,
Consider, if number of women = 6 then probability that both women are selected is 6c2 / 10c2 which is 1/3. so p < 1/2."NO"
Consider, if number of women = 9 then probability that both women are selected is 9c2 / 10c2 which is 36 / 45 = 4/5. so p > 1/2 "YES"

Insuff... move on to next statement :-)

hope this helps..

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by jayanti » Mon Aug 08, 2011 9:47 pm
Thanks a ton for the detailed explanation, really helpful... :)

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by gmatboost » Mon Aug 08, 2011 10:04 pm
All I can really say in addition to Brian's response is that I approached it with a bit of "translating the prompt," which is an oft-repeated but not often used strategy.

In order for p > 1/2, we need x/10*(x-1)/9 > 1/2 (Brian explained the left side of this formula)
x(x-1)/90 > 1/2
x(x-1) > 45

7 fails, but 8 works
So, I went into the question knowing that I needed to have 8 women out of 10 for the answer to be Sufficient. That way, rather than "testing" whatever numbers we come up with, I can just ask: is it necessarily 8 or more, or is not necessarily 8 or more.

I'm not claiming this made this particular question much easier, but I do think it's worth thinking about the approach, since it might come in handy on another question.
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