A magician holds one six-sided die in his left hand...

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swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Sun Jan 07, 2018 6:39 am

A magician holds one six-sided die in his left hand and two in his right. What is the probability the number of the dice in his left hand is greater than the sum of the dice in his right?

A. 7/108
B. 5/54
C. 1/9
D. 2/17
E. 1/4

The OA is B.

Please, can any expert explain this PS question for me? I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

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Source: Beat The GMAT — Problem Solving | Sun Jan 07, 2018 6:39 am

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Mon Jan 08, 2018 9:01 am

swerve wrote:A magician holds one six-sided die in his left hand and two in his right. What is the probability the number of the dice in his left hand is greater than the sum of the dice in his right?

A. 7/108
B. 5/54
C. 1/9
D. 2/17
E. 1/4

The OA is B.

Please, can any expert explain this PS question for me? I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

P(x) = # Desired Outcomes/# Total possible outcomes

# Total possible outcomes when rolling 3 dice = 6^3 = 216.

Notice that for the single die to be greater than the sum of the other two, the single die must be at least 3.

Single die: 3; Two other dice sum to less than 3: [1,1] --> 1 desired outcome
Single die: 4; Two other dice sum to less than 4: [1,1; 2,1; 1,2] --> 3 desired outcomes
Single die: 5; Two other dice sum to less than 5:[1,1; 2,1; 1,2, 3,1; 1,3; 2,2] --> 6 desired outcomes
Single die: 6; Two other dice sum to less than 6: [1,1; 2,1; 1,2, 3,1; 1,3; 2,2; 2,3; 3,2; 1,4; 4,1} --> 10 desired outcomes

# Desired outcomes: 1 + 3 + 6 + 10 = 20

20/216 = 5/54. the answer is B

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raghuram123: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Mon Jun 27, 2016 10:47 pm

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by raghuram123 » Tue Jan 09, 2018 10:29 pm

DavidG@VeritasPrep wrote:
swerve wrote:A magician holds one six-sided die in his left hand and two in his right. What is the probability the number of the dice in his left hand is greater than the sum of the dice in his right?

A. 7/108
B. 5/54
C. 1/9
D. 2/17
E. 1/4

The OA is B.

Please, can any expert explain this PS question for me? I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.
P(x) = # Desired Outcomes/# Total possible outcomes

# Total possible outcomes when rolling 3 dice = 6^3 = 216.

Notice that for the single die to be greater than the sum of the other two, the single die must be at least 3.

Single die: 3; Two other dice sum to less than 3: [1,1] --> 1 desired outcome
Single die: 4; Two other dice sum to less than 4: [1,1; 2,1; 1,2] --> 3 desired outcomes
Single die: 5; Two other dice sum to less than 5:[1,1; 2,1; 1,2, 3,1; 1,3; 2,2] --> 6 desired outcomes
Single die: 6; Two other dice sum to less than 6: [1,1; 2,1; 1,2, 3,1; 1,3; 2,2; 2,3; 3,2; 1,4; 4,1} --> 10 desired outcomes

# Desired outcomes: 1 + 3 + 6 + 10 = 20

20/216 = 5/54. the answer is B

is there any other possible solution rather than counting manually
thanks in advance

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