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

Question

A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on 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

I do not know if this kind of question will appear on the GMAT even at the most difficult level. For me is too cumbersome.

I would like to know how to solve it in less than 2 minutes.

Thanks.

rajatjain14 · Accepted Answer

Quicker Way I guess:

GyanOne · Answer

The least sum of the numbers of the two dice in the right hand is 2. The maximum sum of the numbers on these two dice can be 5, for the left hand dice to be able to have a greater number.

Therefore the number on the dice in the left hand must be 3 or more.

Number of ways to get a sum of 2 with the two dice in the right hand= 1
Number of ways to get a sum of 3 with the two dice in the right hand= 2
Number of ways to get a sum of 4 with the two dice in the right hand= 3
Number of ways to get a sum of 5 with the two dice in the right hand= 4

Therefore required probability = P(two right dice have a sum of 2)*P(left dice has 3) + P(right dice have sum of 2 or 3)*P(left dice has 4) + P(two right dice have a sum of 2,3, or 4)*P(left dice has 5) + P(two right dice have a sum of 2,3,4,or 5)*P(left dice has 6)
= (1/36)*(1/6) + (3/36)*(1/6) + (6/36)*(1/6) + (10/36)*(1/6)
= (1/216) * (1+3+6+10)
= 20/216
= 5/54

Option B

Carcass · Answer

this is a pretty nice approach to the question

abhishekshannu · Answer

is der any other easier method?

Carcass · Answer

I do not think.

the question is tough...........this method for me is quite fast due to the cumbersome question

cyberjadugar · Answer

Why are you considering (1,2) different from (2,1)?

PiyushK · Answer

We can simplify two dice problem by referring following grid for most of the questions.

Attachment:

Grid.jpg [ 56.76 KiB | Viewed 19909 times ]

If we have 6 in left hand then 10 combinations of two dice provide sum less than 6.
If we have 5 in left hand then 6 combinations.
If we have 4 then 3 combinations.
Similarly, If we have 3, then 1 combination.

=P(6 in left hand)*P(sum less than 6) + P(5 in left hand)*P(sum less than 5) + P(4 in left hand)*P(sum less than 4) + P(3 in left hand)*P(sum less than 4)
\(= \frac{1}{6}*\frac{10}{36} + \frac{1}{6}*\frac{6}{36} + \frac{1}{6}*\frac{3}{1} + \frac{1}{6} * \frac{1}{36}\)
\(=\frac{5}{54}\)

JusTLucK04 · Answer

1- 0 cases
2- 0 cases
3- 2,2 - 1
4- 1,3 3,1 -2 
5- 1,3 3,1,2,2 - 3
5 - see the pattern..4 cases
6- 5 cases

Possibility of each case is 1/6*0 + 1/6*0 + 1/6 * 1/36 + 1/6 * 3/36 + 1/6* 6/36 + 1/6* 10/36 
= 20/216
= 5/54

I wonder how can this be completed in less than 2 mins...

PiyushK · Answer

The grid(table) I posted above can be drawn in few seconds... and it really helps to solve dice problem without worrying about any overlapping or confusion.

Musicat · Answer

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

Possible answers are:

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

Anyone can explain the whole solving pattern?

Bunuel · Answer

Merging topics. Please search before posting and provide the OA along with a question.

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

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