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If the probability that Mike can win a championship is 1/4, [#permalink]
26 Mar 2005, 19:43

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Question Stats:

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If the probability that Mike can win a championship is 1/4, that of Rob winning the championship is 1/3 and that of Ben winning is 1/6. What is the probability of either Mike or Rob winning the championship but not Ben?
A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

Re: PS -- Probability [#permalink]
26 Mar 2005, 21:20

If there can only be one winner for the championship then P=1/3+1/4=7/12.
If there can be multiple winners then
P=P(M)+P(R)-P(M&R)-P(M&B)-P(R&B)+P(M&R&B)
=1/3+1/4-1/12-1/24-1/18+1/72
=5/12

I think that they're playing in the same championship because the problem is given like this :

If the probability that Mike can win a championship is 1/4, that of Rob winning the championship is 1/3 and that of Ben winning is 1/6. What is the probability of either Mike or Rob winning the championship but not Ben?

I need some Verbal specialist but I assume that there is only one championship. Then I understood the question as : what is the probability that Mike or Rob win but not Ben .... What is the probability that the winner is Mike, Bob or anyone else but not Ben.

If it is not this meaning, I don't see the interest of notifying "not Ben"

If the probability of Mike winning the championship is 1/4, and the probability of Rob winning the championship is 1/3, then the probability that EITHER Mike OR Rob will win is 1/4+1/3, or 7/12. The information about Ben seems extraneous, since if either Mike or Rob win, Ben can't win.

You need to draw the diagram to see it more clearly. Basically when you do P(M)+P(R) you double counted P(M&R), P(M&B) and P(R&B). Do you need to take them out. But when you do -P(M&R)-P(M&B)-P(R&B) you have counted P(M&R&B) three times and when you do P(M)+P(R) you only double counted P(M&R&B) twice. So you need to add one P(M&R&B) back to cancel the third P(M&R&B) you took out. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Please don't ignore Ben. The question clearly says that Ben will not win.

Thus, 1/4 + 1/3 - 1/6 ( Probability that Ben will win)

Thus, B is the answer ( 5/12)

Well it seems that

probability that either one (except ben ) is 1/3+1/4

taking together

1/4 + 1/3 - 1/6 SUBSTRACT PROB THAT BEN WIN

or 1/4 + 1/3 +5/6 ADD THE probabilities that ben not win

both case we get 5/12 since those events are dependent does this make sense to anyone ? THANKS

That does not make sense to me. You can't take a number (1/4+1/3) and then say that either adding(5/6) or subtracting(1/6) a new number will give the same result unless that new number is 0. If you take 1/4+1/3+5/6=17/12 This is not a reasonable probability. If you take 1/4+1/3-1/6=5/12 Which is the OA, but it really doesn't matter to come up with the OA. It is only a helpful practice question if you know how to come up with the right answer, which means no monday-morning quarterbacking when you already know the answer is supposed to be 5/12 and then making up formulas to arrive at 5/12.

What seems most important to me about this question is whether or not I should expect actual GMAT questions to be in this format. To me it seemed that Ben's involvement was irrelevant. Do GMAT questions often include irrelevant information? I don't know but my guess is yes.

Do GMAT questions usually have so ambiguous an interpretation? If so, I should focus my study on resolving ambiguities at least as much as studying proper methods for determining probabilites. What does everyone think?

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