GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2019, 16:22 ### GMAT Club Daily Prep

#### 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

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. ### Request Expert Reply # While playing a certain dice game, Chris wins if the sum of

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager  Joined: 11 Aug 2011
Posts: 158
Location: United States
Concentration: Economics, Finance
GMAT Date: 10-16-2013
GPA: 3
WE: Analyst (Computer Software)
While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

2
21 00:00

Difficulty:   65% (hard)

Question Stats: 58% (02:29) correct 42% (02:38) wrong based on 179 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.

_________________
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

5
1
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
##### General Discussion
Intern  Joined: 14 May 2014
Posts: 25
Concentration: General Management, Operations
GPA: 3.15
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

4
1
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 $$\frac{91}{216}$$

Intern  Joined: 13 Sep 2013
Posts: 1
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

1
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.
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2972
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

1
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
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One 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
Intern  B
Joined: 30 Apr 2018
Posts: 7
Location: United States
GMAT 1: 690 Q47 V38 GPA: 4
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

1
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 = 1-1/6 = 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
Intern  Joined: 25 Mar 2014
Posts: 29
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

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?
Manager  B
Joined: 30 Jul 2013
Posts: 125
Concentration: Strategy, Sustainability
GMAT 1: 710 Q49 V39 While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

Why dont be consider a 4th option where on none of the 3 rolls he gets a 7? shouldnt we be adding [/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

1
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 $$\frac{91}{216}$$

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|
= $$1/6+1/6+1/6-1/6*1/6-1/6*1/6-1/6*1/6+1/6*1/6*1/6$$
= $$91/216$$

IS this way correct?
Board of Directors P
Joined: 17 Jul 2014
Posts: 2523
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

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: 91
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

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
Non-Human User Joined: 09 Sep 2013
Posts: 12399
Re: While playing a certain dice game, Chris wins if the sum of  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: While playing a certain dice game, Chris wins if the sum of   [#permalink] 16 Sep 2019, 14:54
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  