silvia928 wrote:If 2 different representative 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.
All that math looks way too complicated to me - let's use probability instead of combinations.
(1) If 6 are women, then the chance of 2 women is (6/10)(5/9) = 30/90.
Is 1/3 > 1/2? NO.
If all 10 are women, then the chance of 2 women is 100%. Is 100% > 50%? YES.
(2) Let's see the maximum number of men that would make this true:
If there were 2 men, prob of MM would be (2/10)(1/9) = 2/90... allowed.
If there were 3 men, prob of MM would be (3/10)(2/9) = 6/90... allowed.
If there were 4 men, prob of MM would be (4/10)(3/9) = 12/90 which is more than 1/10, so not allowed!
So, the maximum # of men is 3, which means that the minumum # of women is 7.
If we had 7 women, the chance of WW would be (7/10)(6/9) = 42/90. That's less than half, so "NO" answer.
Again, if we had 10 women, chance would be 100%, which gives us a "YES" answer.
So, each statement by itself can give us a "yes" or "no". Since the scenarios we examined for statement (2) also fit statement (1), we're no better off after combination: choose (E).
