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Re: Hard probability problem: 2 dice sum of 3 BEFORE sum of 7 [#permalink]
06 Oct 2010, 02:39

1

This post received KUDOS

Expert's post

catennacio wrote:

Anyone has idea about this?

You have two dice, what is the probability of rolling a sum of 3 BEFORE rolling a sum of 7?

Thanks..

First of all please note that this question is far beyond the GMAT scope.

\(P(sum=3)=\frac{2}{36}=\frac{1}{18}\): either (1,2) or (2,1) out of total of 36 different combinations of two dice;

\(P(sum=7)=\frac{6}{36}=\frac{1}{6}\): (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1);

\(P(other \ sum)=1-(\frac{1}{18}+\frac{1}{6})=\frac{7}{9}\), probability of sums: 4, 5, 6, 8, 9, 10, 11, and 12;

Winning scenarios: {sum=3} - we have sum of 3 on the first roll of two dice - \(P_1=\frac{1}{18}\); {other sum; sum=3} - on the first roll we have other sum and sum of 3 on the second roll - \(P_2=\frac{7}{9}*\frac{1}{18}\); {other sum; other sum; sum=3} - \(P_3=(\frac{7}{9})^2*\frac{1}{18}\); {other sum; other sum; other sum; sum=3} - \(P_4=(\frac{7}{9})^3*\frac{1}{18}\); ...

So probability of rolling a sum of 3 BEFORE rolling a sum of 7 would be the sum of the above infinite series: \(P=\frac{1}{18}+\frac{7}{9}*\frac{1}{18}+(\frac{7}{9})^2*\frac{1}{18}+(\frac{7}{9})^3*\frac{1}{18}+...=\frac{1}{18}(1+\frac{7}{9}+(\frac{7}{9})^2+(\frac{7}{9})^3+...)=\frac{1}{18}(1+\frac{\frac{7}{9}}{1-\frac{7}{9}})=\frac{1}{18}(1+\frac{7}{2})=\frac{1}{4}\) (for geometric progression with common ratio \(|q|<1\), the sum of the progression: \(b_1, b_2, ...\) is \(Sum=\frac{b_1}{1-q}\).).

OR:

As \(P(sum=3)=\frac{2}{36}\) and \(P(sum=7)=\frac{6}{36}\) then getting the sum of 7 is 3 times more likely than getting the sum of 3, so the sum of 3 has 1 chance out of 4 to get first out of any number of tries, so \(P=\frac{1}{4}\) or \(P=\frac{\frac{2}{36}}{\frac{2}{36}+\frac{6}{36}}=\frac{1}{4}\).

Re: Hard probability problem: 2 dice sum of 3 BEFORE sum of 7 [#permalink]
06 Oct 2010, 15:42

Thank you very much Buneul for your answer. However I have a few questions:

I understand up to P(other sum). How can you conclude the winning scenario is other sum then sum 3? I know it's a possible case, but why is it (other sum, sum 3) will lead to a (other sum, sum 3, then sum 7). It could be (other sum, sum 3, sum 8) - after 3 rollings, while what we expect is sum 7. Therefore the (other sum, other sum,..., sum 3) could be a wrong case.

On the other hand, if we can conclude other sum, other sum,..., sum 3 is a winning scenario, have we assume the probability of sum 7 and other sum but not 7 are the same, while there are different?

I don't understand this sentence too "So probability of rolling a sum of 3 BEFORE rolling a sum of 7 would be the sum of the above infinite series" - how can you conclude it's the sum that series, how about if we have an AFTER word? What is the difference between BEFORE and AFTER?

Edit: I like your second approach, by comparing the fraction of the probability of 2 events we can say P = 1/4, but I still don't understand the word BEFORE.. What if they ask for AFTER? Will we still compare the 2 probabilities?

Re: Hard probability problem: 2 dice sum of 3 BEFORE sum of 7 [#permalink]
07 Oct 2010, 02:12

Expert's post

catennacio wrote:

Thank you very much Buneul for your answer. However I have a few questions:

I understand up to P(other sum). How can you conclude the winning scenario is other sum then sum 3? I know it's a possible case, but why is it (other sum, sum 3) will lead to a (other sum, sum 3, then sum 7). It could be (other sum, sum 3, sum 8) - after 3 rollings, while what we expect is sum 7. Therefore the (other sum, other sum,..., sum 3) could be a wrong case.

On the other hand, if we can conclude other sum, other sum,..., sum 3 is a winning scenario, have we assume the probability of sum 7 and other sum but not 7 are the same, while there are different?

I don't understand this sentence too "So probability of rolling a sum of 3 BEFORE rolling a sum of 7 would be the sum of the above infinite series" - how can you conclude it's the sum that series, how about if we have an AFTER word? What is the difference between BEFORE and AFTER?

Edit: I like your second approach, by comparing the fraction of the probability of 2 events we can say P = 1/4, but I still don't understand the word BEFORE.. What if they ask for AFTER? Will we still compare the 2 probabilities?

When we roll 2 dice the sum can be 3, 7, or some other number (not 3 and not 7). The questions asks for the cases when we get the sum of 3 BEFORE the sum of 7.

For example why the cases I wrote are the winning scenarios: {sum=3} - means that right on the first throw we have sum of 3, so we have 3 before 7 (as no 7 at all); {other sum; sum=3} - first roll not 7 and not 3, so we can continue. On the second throw we have sum of 3, so again 3 before 7 - OK; {other sum; other sum; sum=3} - sum of 3 on the third roll; {other sum; other sum; other sum; sum=3} - sum of 3 on the fourth roll; .... {other sum; other sum; other sum; ..., other sum on the nth roll; sum of 3 on the (n+1)th roll} - out of n+1 rolls we have other sum for the rolls from 1 to n and sum of 3 on the (n+1)th roll, still 3 before 7 - OK; ...

The above can be continued infinitely, and all above case represent the scenario when we have the sum of 3 BEFORE we have the sum of 7 (which will eventually occur on some roll afterwards). So the probability of getting 3 before 7 would be the sum of the probabilities of the above events.

Hope it's clear. Anyway: you won't need this for GMAT, so don't worry too much. _________________

Re: Hard probability problem: 2 dice sum of 3 BEFORE sum of 7 [#permalink]
07 Oct 2010, 03:28

Thanks Bunuel now I understand it. However I think you miss the scenario where sum=7 occurs before sum=3. For example, 1st time sum=7, 2nd time sum=3, so the probability must be 6/36 * 2/36. You assumed that sum 7 never happens before sum 3 so you take the probability of other sum (7/9) to calculate. Am I wrong?

Re: Hard probability problem: 2 dice sum of 3 BEFORE sum of 7 [#permalink]
07 Oct 2010, 04:38

Expert's post

catennacio wrote:

Thanks Bunuel now I understand it. However I think you miss the scenario where sum=7 occurs before sum=3. For example, 1st time sum=7, 2nd time sum=3, so the probability must be 6/36 * 2/36. You assumed that sum 7 never happens before sum 3 so you take the probability of other sum (7/9) to calculate. Am I wrong?

Yes, you are wrong. I'm not assuming that 7 doesn't occur before 3 rather than I'm only interested to count the probability of the cases when 3 occur before 7 as this is what the question is asking, that's why I'm not considering the cases when 7 occur before 3. _________________

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