imoi wrote:
Hey everyone - I have a question about this problem here:
if-2-different-representatives-are-to-be-selected-at-random-128233.htmlI understand how to solve it - that's not the issue at all (thank you so much Bunuel)
My issue is this - can you really do all the calculations that Bunuel showed in under 2 min?
I mean after I read all the explanations and timed myself on the problem I was on 2.07, given that I pretty much knew the answer, the calculations and everything that is given. In a realistic world, even knowing HOW to solve it, would probably take me >3.5min.
Questions:
1. Is that normal or am I just bad with arithmetic?
2. Is there a "shortcut" for this problem?
3. You probably noted the topic is called "combinatorics vs probability" - I am curious - who prefers which solution to this problem? I find the combinatorics one to be even longer, is that lack of practice? Does anyone use combi for this?
Usually, the combinatorics approach will be easier and more clear, especially in "order matter or not" cases. It might take a few seconds extra but is well worth it in my opinion.
If done properly (with variables and all), the question will take a long time - no getting away from it. Everyone (except Bunuel, perhaps!) will take quite a bit of time here. So in such a question one could work with numbers provided one can easily handle fractions.
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.
If no. of women is 6,
p = 6C2/10C2 = 6*5/9*10 = (6/9)*(1/2) which is less than 1/2 because when you multiply a number by something between 0 and 1, the number becomes smaller. SO 1/2 multiplied by 6/9 will give less than 1/2.
If number of women is 7,
p = 7*6/10*9 = 7/15 which is less than 1/2
If no of women is 9,
p = 9*8/10*9 = 8/10 (more than 1/2)
Not sufficient
(2) The probability that both representatives selected will be men is less than 1/10.
Say no of men is 3 (and number of women 7)
P of selecting both men = 3*2/10*9 = 1/15. This is less than 1/10. We have seen that with 7 women, p is less than 1/2.
Number of men can be 2 or 1 also since in those cases, the probability of selecting 2 men will be even less than 1/15. We have seen that when number of women is 9, p is greater than 1/2
Not sufficient.
Using both together, we have already seen 2 cases:
number of women = 7 and men = 3. p is less than 1/2 in this case
number of women = 9 and men = 1. p is more than 1/2 in this case
Hence answer will be (E)
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