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

[swerve]

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.

[DavidG@VeritasPrep]

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

[raghuram123]

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