December 15, 2018 December 15, 2018 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299)
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 11 Aug 2011
Posts: 165
Location: United States
Concentration: Economics, Finance
Schools: Booth '17, Sloan '17, CBS '17, Ross '17 (S), Haas '17, Stern '17, Yale '17, Anderson '17, Kelley '17 (S), KenanFlagler '17 (S), HEC Dec '17, HKUST '16, Jones '16, NUS '17
GMAT Date: 10162013
GPA: 3
WE: Analyst (Computer Software)

While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
10 Aug 2014, 02:58
Question Stats:
58% (02:29) correct 42% (02:35) wrong based on 226 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kudos me if you like my post !!!!




Intern
Joined: 16 Jul 2014
Posts: 2

While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
12 Aug 2014, 05:56
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




Intern
Joined: 14 May 2014
Posts: 26
Concentration: General Management, Operations
GPA: 3.15

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
10 Aug 2014, 06:46
There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 16, 25, 34, 43, 52 and 61
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.



Intern
Joined: 25 Mar 2014
Posts: 31

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
16 Aug 2014, 13:54
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! =) My final question is: is this problem ambiguous?



Intern
Joined: 13 Sep 2013
Posts: 1

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
17 Aug 2014, 21:31
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.



Manager
Joined: 30 Jul 2013
Posts: 125
Concentration: Strategy, Sustainability

While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
31 Aug 2014, 23:59
Why dont be consider a 4th option where on none of the 3 rolls he gets a 7? shouldnt we be adding [5][/6]^3 to the 91/216?



Intern
Joined: 25 Oct 2015
Posts: 1

While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
26 Oct 2015, 22:35
wunsun wrote: There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 16, 25, 34, 43, 52 and 61
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) A∪B∪C = A + B + C  A∩B  B∩C  A∩C + A∩B∩C = \(1/6+1/6+1/61/6*1/61/6*1/61/6*1/6+1/6*1/6*1/6\) = \(91/216\) IS this way correct?



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2711
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
27 Oct 2015, 06:01
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. 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 Answer : Option C
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Board of Directors
Joined: 17 Jul 2014
Posts: 2618
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
11 Mar 2016, 19:09
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..



Manager
Joined: 09 Jun 2015
Posts: 93

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
13 Mar 2016, 07:48
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



Intern
Joined: 30 Apr 2018
Posts: 7
Location: United States
GPA: 3.92

Re: While playing a certain dice game, Chris wins if the sum of
[#permalink]
Show Tags
22 Aug 2018, 12:00
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. The sum of two dice can be 7 in six ways  (1,6) , (2,5), (3,4), (4,3) , (5,2) and (6,1) Total combinations of the numbers on throwing two dice = 36 Thus probability of getting a sum of 3 = 6/36 = 1/6 Probability of not getting a sum of 7 = 11/6 = 5/6 Now Chris has 3 attempts to get the desired sum. If he gets sum 7 in 1st attempt than the result is achieved and the other two attempts are not required > P1= 1/6 If Chris does not gets sum 7 in 1st attempt, then he needs another attempt and if he gets six then prob > P2= 5/6*1/6 (P(not getting sum 7 in 1st attempt)* P(get sum 7 in 2nd attempt)) Similarly, a 3rd attempt would be needed if Chris fails to gets sum 7 in previous attempts > P3 = 5/6*5/6*1/6 Total prob = P1 +P2+ P3 = 1/6 + 5/6*1/6 * 5/6*5/6*1/6 = (36+30+25)*6^3 = 91/126




Re: While playing a certain dice game, Chris wins if the sum of &nbs
[#permalink]
22 Aug 2018, 12:00






