Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________