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A certain club has 10 members, including Harry. One of the [#permalink]

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24 Jul 2012, 18:55

2

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E

Difficulty:

55% (hard)

Question Stats:

64% (00:59) correct
36% (02:31) wrong based on 45 sessions

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D7 GMAT 12th ed Review (I know this question is already explained, I searched)

A certain club has 10 members, including Harry. One of the 10 members is chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chose to be secretary or the member chose to be treasurer?

A 1/720 B 1/80 C 1/10 D 1/9 E 1/5

The answer is E. 1/5 And the explanation is: (9/10x1/9)+(9/10x8/9x1/8)

Molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. What is the probability that Molly will roll the die less than four times before stopping?

Re: I am having problem with how to handle these probability q's [#permalink]

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24 Jul 2012, 23:34

In order to become a secretary the person cannot be chosen as a president. This can be done in 9/10 ways. Then to become a secretary this person's probability is 1/9

Probability to become a secretary = (9/10 * 1/9) = 1/10

Similar concept for treasurer. Probability of not president = 9/10, Probability of not secretary = 8/9, probability of treasurer 1/8

probability to become a treasurer = (9/10 * 8/9 * 1/8) = 1/10

Since, the problem is saying secretary OR Treasurer it is addition: 1/10 + 1/10 = 1/5

Re: I am having problem with how to handle these probability q's [#permalink]

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25 Jul 2012, 00:31

Quote:

So what I am confused on is why do we not use this strategy to solve for D7

P(not being chosen president) + P(not being chosen president+being chosen secretary) + P(not being chose president+not being chosen secretary+being chosen treasurer) = (9/10)+(9/10x1/9)+(9/10x8/9x1/8)

What is the difference between these two problems??

Ok so first, you are not using at all the same "strategy" as what was described by MGMAT. You have absolutely no reason to sum P(not being chosen president). By the way, the result of your calculus is equal to 11/10, there's no such thing as a probability higher than one, it should tell you you are mistaken right away.

If you just remove P(not being chosen president) from your sum then you'll have exactly the OA.

Re: I am having problem with how to handle these probability q's [#permalink]

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25 Jul 2012, 09:50

LalaB wrote:

All of these 10 guys have an equal chance to be chosen (1/10)

so, to be chosen as a secretary or treasure is euqal to (1/10)+(1/10)=2/10=1/5

Don't we have to take into account that it is being done without replacement?

duriangris wrote:

By the way, the result of your calculus is equal to 11/10, there's no such thing as a probability higher than one, it should tell you you are mistaken right away.

If you just remove P(not being chosen president) from your sum then you'll have exactly the OA.

I know that it adds to 11/10 and I know that if I remove the P(not being chosen president) I have the answer. But the reason why Im asking is why. Why do we not add that P(not being chosen president)? If I know why it will prevent future mistakes...otherwise im likely to repeat in another problem

Re: I am having problem with how to handle these probability q's [#permalink]

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25 Jul 2012, 19:14

akuma86 wrote:

Don't we have to take into account that it is being done without replacement?

You can, and will get the same result. The thing is MGMAT strategy you quote tells you that actually you don't need to and can save 10 seconds by skipping the factorization. I suggest you to forget this strategy if it is not natural to you, or keep it in case you are in a hurry during the exam and need a quick answer before time's up.

akuma86 wrote:

I know that it adds to 11/10 and I know that if I remove the P(not being chosen president) I have the answer. But the reason why Im asking is why. Why do we not add that P(not being chosen president)? If I know why it will prevent future mistakes...otherwise im likely to repeat in another problem

Try to draw a tree to this problem and you will see that P(not being chosen president) appears nowhere at the end of that tree. If still need to be convinced then explain to us why you think P(not being chosen president) should be added (and tell us also why P(the president doesn't like pizza) is not also in the sum)

Re: I am having problem with how to handle these probability q's [#permalink]

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25 Jul 2012, 21:13

duriangris wrote:

Try to draw a tree to this problem and you will see that P(not being chosen president) appears nowhere at the end of that tree. If still need to be convinced then explain to us why you think P(not being chosen president) should be added (and tell us also why P(the president doesn't like pizza) is not also in the sum)

Rude Comment. Thanks but I was able to figure out the problem on my own. I just don't think you should be disrespectful to people who are trying to learn as that is the point of this forum.

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