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Two combinatorics problems that are solved differently, help

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Two combinatorics problems that are solved differently, help [#permalink] New post 05 Nov 2013, 09:27
In a diving competition, each diver has a 20% chance of a perfect dive. The first per­fect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn).

A) 1/5
B) 1/15
C) 4/25
D) 16/125
E) 61/125

So that's the first problem, and here's the OE:

[Reveal] Spoiler: OE
D. 16/125: In order for Janet to receive a perfect score, neither of the previous two divers can receive one. Therefore, you are finding the probability of a chain of three events: that diver one will not get a perfect score AND diver two will not get a perfect score AND Janet will get a perfect score. Multiply the probabilities: 4/5 x 4/5 x 1/5 = 16/125.The probability is 16/125 that Janet will receive a perfect score.



But here's another problem that's identical and is solved completely differently:

Molly is rollling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?


A) 1/6
B) 5/36
C)1/216
D)91/216
E) 1/2


[Reveal] Spoiler: OE
The setup for this problem is actually more complex than it may initially appear. First, you need to sep­arate three possibilities from each other. If Molly rolls less than 4 times before stopping, then she rolls a 1 on her first roll OR on her second roll OR on her third roll. That means that you will ultimately have to ADD the three probabilities:

P(lst roll end) + P(2nd roll) + P(3rd roll)

The probability of rolling a 1 on her first roll is 1/6. P(lst roll end) = 1/6. If Molly does not roll a 1 until her second roll, that means she did not roll a 1 on her first roll. So to get
P(2nd roll end), you need to calculate the probability of not rolling a 1 on the first roll AND rolling a 1 on the second roll.

P(2nd roll end) = P(Not 1 on 1st) x P(1 on 2nd)

Similarly, if Molly does not roll a 1 until her third roll, you need the probability of not 1 on her first roll AND not 1 on her second roll AND 1 on her third roll:

P(3rd roll end) = P(Not 1 on 1st) x P(Not 1 on 2nd) x P(1 on 3rd)

The probability that she rolls a 1 on her first roll is 1/6. There are 6 possible outcomes, but only 1 of them is desirable. The probability of not rolling a 1 is then equal to 1 minus the probability of rolling a 1.1- 1/6 = 5/6. You can fill in the appropriate probabilities and solve:

P(lst roll) + P(2nd roll) + P(3rd roll) = (1/6) + (1/6)(5/6)+(1/6)(5/6)(5/6)= 91/216




So here's what I don't get:

In the first problem you only calculated the odds of the diver NOT getting a perfect score, and in the second one you calculated the odds of NOT getting a 1 and then added to it the odds of not getting a 1, multiplied by the odds of getting something else, and then added to that the odds of not getting a 1 again, and multiplied it by the odds of rolling something else twice.

Using that logic the first problem should have been solved by multiplying the odds of the first diver getting a perfect score (1/5) and then adding to that the odds of the 2nd diver getting a perfect score, multiplied by their odds of getting something else (1/5)(4/5) and then adding to that the odds that the first and second divers don't get a perfect score, multiplied by the odds that she does get a perfect score (1/5)(4/5)(4/5), so the answer should be 61/125.
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 05 Nov 2013, 19:36
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Just a brief thought on this one.

In the first problem, we are only worried about the chances of Jane's perfect dive, assuming that she does dive. So, the outcomes in which either of the first two divers gets a perfect score are ignored.

In the second problem, any outcome in which a 1 is rolled within the first 3 rolls counts. So, we don't ignore the outcomes in which a 1 is rolled first or second.

I hope that helps!
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 19:31
WillEconomistGMAT wrote:
Just a brief thought on this one.

In the first problem, we are only worried about the chances of Jane's perfect dive, assuming that she does dive. So, the outcomes in which either of the first two divers gets a perfect score are ignored.

In the second problem, any outcome in which a 1 is rolled within the first 3 rolls counts. So, we don't ignore the outcomes in which a 1 is rolled first or second.

I hope that helps!


I'm confused on the exact two problems above as well. Even after readying your explanation Will, I'm still not quite sure why the methodology to solve the two problems is different.
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 19:37
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NvrEvrGvUp wrote:
WillEconomistGMAT wrote:
Just a brief thought on this one.

In the first problem, we are only worried about the chances of Jane's perfect dive, assuming that she does dive. So, the outcomes in which either of the first two divers gets a perfect score are ignored.

In the second problem, any outcome in which a 1 is rolled within the first 3 rolls counts. So, we don't ignore the outcomes in which a 1 is rolled first or second.

I hope that helps!


I'm confused on the exact two problems above as well. Even after readying your explanation Will, I'm still not quite sure why the methodology to solve the two problems is different.


Perhaps I am misinterpreting you when you say "methodology." I would say the methodology is the same in both problems. However, in the first you are solving for one chain of events (two less-than-perfect dives followed by a perfect), and in the second you are solving for three separate chains (1 on the first roll, 1 on the second roll, or 1 on the third roll). Each of the chains in the second problem is mutually exclusive (e.g. if you roll a 1 on the first roll, you never make the second or third roll).

Does that help? If not, maybe you can elaborate on the methodology question. :)
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 19:39
WillEconomistGMAT wrote:
NvrEvrGvUp wrote:
WillEconomistGMAT wrote:
Just a brief thought on this one.

In the first problem, we are only worried about the chances of Jane's perfect dive, assuming that she does dive. So, the outcomes in which either of the first two divers gets a perfect score are ignored.

In the second problem, any outcome in which a 1 is rolled within the first 3 rolls counts. So, we don't ignore the outcomes in which a 1 is rolled first or second.

I hope that helps!


I'm confused on the exact two problems above as well. Even after readying your explanation Will, I'm still not quite sure why the methodology to solve the two problems is different.


Perhaps I am misinterpreting you when you say "methodology." I would say the methodology is the same in both problems. However, in the first you are solving for one chain of events (two less-than-perfect dives followed by a perfect), and in the second you are solving for three separate chains (1 on the first roll, 1 on the second roll, or 1 on the third roll). Each of the chains in the second problem is mutually exclusive (e.g. if you roll a 1 on the first roll, you never make the second or third roll).

Does that help? If not, maybe you can elaborate on the methodology question. :)


But can't you make that exact same argument for the first problem? That, if someone has a perfect dive (1/5 probability) on the first dive, that you never make the second or third dive?
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 19:46
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NvrEvrGvUp wrote:
WillEconomistGMAT wrote:

Perhaps I am misinterpreting you when you say "methodology." I would say the methodology is the same in both problems. However, in the first you are solving for one chain of events (two less-than-perfect dives followed by a perfect), and in the second you are solving for three separate chains (1 on the first roll, 1 on the second roll, or 1 on the third roll). Each of the chains in the second problem is mutually exclusive (e.g. if you roll a 1 on the first roll, you never make the second or third roll).

Does that help? If not, maybe you can elaborate on the methodology question. :)


But can't you make that exact same argument for the first problem? That, if someone has a perfect dive (1/5 probability) on the first dive, that you never make the second or third dive?


Indeed. But the question asks "If Jane is the third diver to dive..." So, to calculate her chances of success when she dives third, you must assume that she dives third. The antecedent of a conditional like this is always treated as true on the GMAT.
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 19:52
WillEconomistGMAT wrote:
NvrEvrGvUp wrote:
WillEconomistGMAT wrote:

Perhaps I am misinterpreting you when you say "methodology." I would say the methodology is the same in both problems. However, in the first you are solving for one chain of events (two less-than-perfect dives followed by a perfect), and in the second you are solving for three separate chains (1 on the first roll, 1 on the second roll, or 1 on the third roll). Each of the chains in the second problem is mutually exclusive (e.g. if you roll a 1 on the first roll, you never make the second or third roll).

Does that help? If not, maybe you can elaborate on the methodology question. :)


But can't you make that exact same argument for the first problem? That, if someone has a perfect dive (1/5 probability) on the first dive, that you never make the second or third dive?


Indeed. But the question asks "If Jane is the third diver to dive..." So, to calculate her chances of success when she dives third, you must assume that she dives third. The antecedent of a conditional like this is always treated as true on the GMAT.


Right, I had started to see this antecedent is what set the former apart from the latter problem.

Still, I'm missing "something" because I still don't understand why the diving problem shouldn't be solved as the OP said:

Quote:
Using that logic the first problem should have been solved by multiplying the odds of the first diver getting a perfect score (1/5) and then adding to that the odds of the 2nd diver getting a perfect score, multiplied by their odds of getting something else (1/5)(4/5) and then adding to that the odds that the first and second divers don't get a perfect score, multiplied by the odds that she does get a perfect score (1/5)(4/5)(4/5), so the answer should be 61/125.


If this 61/125 is not the probability of her getting that perfect score, is it just a meaningless number? Or what is this?

or

Maybe this would help: how could we manipulate the diving problem SO THAT we have to use the above approach?

Sorry if this is confusing as hell, my brain is in pain from this chapter...
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 07 Nov 2013, 22:36
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AccipiterQ wrote:
In a diving competition, each diver has a 20% chance of a perfect dive. The first per­fect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn).

A) 1/5
B) 1/15
C) 4/25
D) 16/125
E) 61/125

So that's the first problem, and here's the OE:

[Reveal] Spoiler: OE
D. 16/125: In order for Janet to receive a perfect score, neither of the previous two divers can receive one. Therefore, you are finding the probability of a chain of three events: that diver one will not get a perfect score AND diver two will not get a perfect score AND Janet will get a perfect score. Multiply the probabilities: 4/5 x 4/5 x 1/5 = 16/125.The probability is 16/125 that Janet will receive a perfect score.



But here's another problem that's identical and is solved completely differently:

Molly is rollling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?


A) 1/6
B) 5/36
C)1/216
D)91/216
E) 1/2


[Reveal] Spoiler: OE
The setup for this problem is actually more complex than it may initially appear. First, you need to sep­arate three possibilities from each other. If Molly rolls less than 4 times before stopping, then she rolls a 1 on her first roll OR on her second roll OR on her third roll. That means that you will ultimately have to ADD the three probabilities:

P(lst roll end) + P(2nd roll) + P(3rd roll)

The probability of rolling a 1 on her first roll is 1/6. P(lst roll end) = 1/6. If Molly does not roll a 1 until her second roll, that means she did not roll a 1 on her first roll. So to get
P(2nd roll end), you need to calculate the probability of not rolling a 1 on the first roll AND rolling a 1 on the second roll.

P(2nd roll end) = P(Not 1 on 1st) x P(1 on 2nd)

Similarly, if Molly does not roll a 1 until her third roll, you need the probability of not 1 on her first roll AND not 1 on her second roll AND 1 on her third roll:

P(3rd roll end) = P(Not 1 on 1st) x P(Not 1 on 2nd) x P(1 on 3rd)

The probability that she rolls a 1 on her first roll is 1/6. There are 6 possible outcomes, but only 1 of them is desirable. The probability of not rolling a 1 is then equal to 1 minus the probability of rolling a 1.1- 1/6 = 5/6. You can fill in the appropriate probabilities and solve:

P(lst roll) + P(2nd roll) + P(3rd roll) = (1/6) + (1/6)(5/6)+(1/6)(5/6)(5/6)= 91/216




Actually the problems are not identical. Here is how we would have to change the second problem to make it identical to the first one:

Molly is rollling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die 3 times before stopping?

Mind you, 'less than 4 times' and '3 times' are not the same.
less than 4 implies 1 or 2 or 3
3 times implies just 3

In the first question, we want Janet, the third diver to get the perfect score.
In the second modified question, we want to get 1 on the third roll of the die.
Now the logic to be used in the two questions will be the same - We want the first and second attempts to fail and the third attempt to succeed in both questions - so we find the probability of chain of 3 events in both cases -> Fail Prob * Fail Prob * Pass Prob


Alternatively, you can change the first question to match the second one like this:
In a diving competition, each diver has a 20% chance of a perfect dive. The first per­fect dive of the competition, but no subsequent dives, will receive a perfect score. What is the probability that one of the first three divers receives a perfect score? (Assume that each diver can perform only one dive per turn).

Here you are saying that you want one of the first three to get the perfect score. THis can happen in 3 ways:
First diver gets a perfect score and then no one else does.
First diver doesn't get a perfect score, second does and then no one else does.
First diver doesn't get a perfect score, second doesn't get but third one does.

You will find the probability in each of these chains of events and then add them up.
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 18 Nov 2013, 05:46
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NvrEvrGvUp wrote:

If this 61/125 is not the probability of her getting that perfect score, is it just a meaningless number? Or what is this?

or

Maybe this would help: how could we manipulate the diving problem SO THAT we have to use the above approach?

Sorry if this is confusing as hell, my brain is in pain from this chapter...


Karishma provided a very nice analysis.

I will add that it might help to think of the probability in the diving question as containing a history. Since the event in question is dependent on earlier events, the probability of that particular outcome tells a sort of story about the competition thus far.
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 18 Nov 2013, 23:44
Thanks Will and Karishma, I've actually been thinking about this problem for a long time now and I finally see why it's different.

My main struggle was that in the first problem, Janet MUST dive and dive 3rd. Given that she MUST dive, the problem is asking us what is the probability that she'll succeed. Therefore, given that she must dive, the two previous divers MUST have failed (4/5 x 4/5) and then we need to take into consideration that now that Janet is up, she's not guaranteed a successful dive (1/5). So, the probability of Janet getting to dive and then having a successful dive = 4/5 * 4/5 * 1/5 = 16/125.

In the second problem, it's actually different because we don't know whether Molly is stopping once, twice, OR three times. Therefore, we have to calculate the individual probabilities of the cases (getting 1 on the first roll, second roll, or 3rd roll) and add them all up.

Hope this helps anyone else who was struggling with this problem, I know I was!

-Rich
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Re: Two combinatorics problems that are solved differently, help [#permalink] New post 23 Nov 2013, 09:29
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NvrEvrGvUp wrote:
Thanks Will and Karishma, I've actually been thinking about this problem for a long time now and I finally see why it's different.

My main struggle was that in the first problem, Janet MUST dive and dive 3rd. Given that she MUST dive, the problem is asking us what is the probability that she'll succeed. Therefore, given that she must dive, the two previous divers MUST have failed (4/5 x 4/5) and then we need to take into consideration that now that Janet is up, she's not guaranteed a successful dive (1/5). So, the probability of Janet getting to dive and then having a successful dive = 4/5 * 4/5 * 1/5 = 16/125.

In the second problem, it's actually different because we don't know whether Molly is stopping once, twice, OR three times. Therefore, we have to calculate the individual probabilities of the cases (getting 1 on the first roll, second roll, or 3rd roll) and add them all up.

Hope this helps anyone else who was struggling with this problem, I know I was!

-Rich


Happy to help, Rich. Way to stick with it!
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Re: Two combinatorics problems that are solved differently, help   [#permalink] 23 Nov 2013, 09:29
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