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15 May 2012, 17:52
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:54) correct
50%
(02:33)
wrong
based on 66
sessions
History
Date
Time
Result
Not Attempted Yet
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.
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.
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22 May 2014, 07:32
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Quicker Way I guess:
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Post 12_Dices.png [ 10.52 KiB | Viewed 21209 times ]
15 May 2012, 21:12
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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
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
