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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

10 Aug 2014, 03:58

11

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

58% (02:36) correct
42% (02:01) wrong based on 152 sessions

HideShow timer Statistics

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216

I could not understand how to approach this problem. Can someone throw some light on how to solve this.

Re: While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

10 Aug 2014, 07:46

3

This post received KUDOS

1

This post was BOOKMARKED

There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 1-6, 2-5, 3-4, 4-3, 5-2 and 6-1

Therefore, the chance to have a sum of 7 is \(\frac{1}{6}\).

Making a sum of 7 on the first time is \(\frac{1}{6}=\frac{36}{216}\) Making a sum of 7 on the second time is \(\frac{5}{6}*\frac{1}{6}=\frac{5}{36}=\frac{30}{216}\) Making a sum of 7 on the third time is \(\frac{5}{6}*\frac{5}{6}*\frac{1}{6}=\frac{25}{216}\)

While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

12 Aug 2014, 06:56

3

This post received KUDOS

1

This post was BOOKMARKED

Total outcomes possible: 36 Total outcomes possible with sum 7: 6 Probability to win when rolled once = 6/36 Probability not to win when rolled once = 30/36 = 5/6 Probability to win in three attempts= 1- Probability will not to win in all three attempts = 1- (5/6* 5/6*5/6) = 91/216

Re: While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

16 Aug 2014, 14:54

akhil911 wrote:

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216

I could not understand how to approach this problem. Can someone throw some light on how to solve this.

I understand the solution. But I have a meaning problem: How many dices could be rolled (max)? Three pairs (6) or three dices (3)

I understand that he rolls the First dice (from 1 to 6). No thing I can do. Second dice (1 to 6) = 1/6 of chance of getting a 7. If the sum is below 7: Third dice (1 to 6). I dont know how to calculate! =)

Re: While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

17 Aug 2014, 22:31

plaverbach wrote:

akhil911 wrote:

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216

I could not understand how to approach this problem. Can someone throw some light on how to solve this.

I understand the solution. But I have a meaning problem: How many dices could be rolled (max)? Three pairs (6) or three dices (3)

I understand that he rolls the First dice (from 1 to 6). No thing I can do. Second dice (1 to 6) = 1/6 of chance of getting a 7. If the sum is below 7: Third dice (1 to 6). I dont know how to calculate! =)

My final question is: is this problem ambiguous?

I think question sould be read as follows: Charlie has three rolls with two dices each of them. Hence every roll minimum score would be 2 (1+1) and maximum 12 (6+6). Probability of having 7 in the first roll is 1/6 (6 out of 36). You can follow one of the two ways explained here above. Both works. Hope it helps.

While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

26 Oct 2015, 23:35

1

This post was BOOKMARKED

wunsun wrote:

There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 1-6, 2-5, 3-4, 4-3, 5-2 and 6-1

Therefore, the chance to have a sum of 7 is \(\frac{1}{6}\).

Making a sum of 7 on the first time is \(\frac{1}{6}=\frac{36}{216}\) Making a sum of 7 on the second time is \(\frac{5}{6}*\frac{1}{6}=\frac{5}{36}=\frac{30}{216}\) Making a sum of 7 on the third time is \(\frac{5}{6}*\frac{5}{6}*\frac{1}{6}=\frac{25}{216}\)

Add them all together to get \(\frac{91}{216}\)

Answer is C.

I am not able to understand above approach. (please elaborate it)

however I have got answer using inclusion exclusion principle..... Considering 3 rolls of 2 dice A= event that I get sum equal to 7 in first roll B= event that I get sum equal to 7 in 2nd roll C= event that I get sum equal to 7 in 3rd roll

in every roll we have six ways to get sum equal to 7 (i.e. 1,6 or 6,1 and so on)

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216

I could not understand how to approach this problem. Can someone throw some light on how to solve this.

Probability of winning game = 1- Probability of losing game

Probability of losing game = (Probability of not getting sum 7 in any of the three attempts)

Ways of getting sum 7 = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)= 6 ways Total ways of getting outcome on two dice =6*6=36 Probability of getting sum as 7 in any attempt =6/36=1/6 Probability of NOT getting sum as 7 in any attempt = 1-(1/6)= 5/6

Probability of losing game =(5/6)*(5/6)*(5/6)=125/216 I.e. Probability of wining game = 1-(125/216) = 91/216

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

11 Mar 2016, 20:09

oh man..I understood that he has to throw 3 dice..not 3 times.... so i took.. probability that it won't work.. 1*5/6 x 5/6 = 25/36 and thus, 11/36 is probability that out of 3 dice, he will get 7..

Re: While playing a certain dice game, Chris wins if the sum of [#permalink]

Show Tags

13 Mar 2016, 08:48

akhil911 wrote:

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216

I could not understand how to approach this problem. Can someone throw some light on how to solve this.

There are 6 possibilities for the sum to be 7. (1,6) (2,5) (3.4) and the three reverse order pairs. The probability of a win is 6 over 36 or 1/6 and the probability of a loss is 5/6 Now, there are three possible cases: W or LW or LLW so, 1/6 + 5/6 * 1/6 + 5/6 * 5/6 * 1/6 = 91/216

gmatclubot

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
13 Mar 2016, 08:48

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...