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A conjuror will roll one red, six-sided die in his right hand and two

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A conjuror will roll one red, six-sided die in his right hand and two [#permalink]

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New post 20 Jul 2017, 21:01
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A conjuror will roll one red, six-sided die in his right hand and two blue, six-sided dice. What is the probability that the number on the red die will be greater than the sum of the two blue dice?

(A) 5/54
(B) 5/108
(C) 11/216
(D) 7/36
(E) 5/18
[Reveal] Spoiler: OA

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A conjuror will roll one red, six-sided die in his right hand and two [#permalink]

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In total, there are (\(6^3 = 216\)) combinations.

The minimum sum on the 2 blue die is 2(1+1).
Since we have been asked the probability of red die's sum
to be greater than the blue die's, the minimum number we can have on red die is 3. (1 combination)

When we get a 4 on the red die, there are 3 combinations for the blue die{(1,2),(2,1),(1,1)}

When we get a 5 on the red die, there are 6 combinations for the blue die{(1,2),(2,1),(1,1),(2,2),(1,3),(3,1)}

When we get a 6 on the red die, there are 10 combinations for the blue die{(1,2),(2,1),(1,1),(2,2),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1)}

The probability that the number on the red die will be greater than the sum of the two blue dice = \(\frac{1+3+6+10}{216} = \frac{20}{216} = \frac{5}{54}\)(Option A)
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Re: A conjuror will roll one red, six-sided die in his right hand and two [#permalink]

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New post 22 Nov 2017, 19:22
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Hi All,

This prompt asks for the probability that the number on one 6-sided die will be greater than the SUM of the numbers on two 6-sided dice. This question requires us to consider multiple possible situations.

To start, we only have to consider a few possible sums for the 'pair' of dice: 2, 3, 4 and 5 (in all other situations, the sum of the two dice CANNOT be less than the total on the one die).

Probability of rolling a total of 2 on two dice (1 and 1) and higher than 2 on one die = (1/6)(1/6)(4/6) = 4/216
Probability of rolling a total of 3 on two dice (1 and 2 in some order) and higher than 3 on one die = (2/6)(1/6)(3/6) = 6/216
Probability of rolling a total of 4 on two dice (1 and 3 or 2 and 2, in some order) and higher than 4 on one die = (3/6)(1/6)(2/6) = 6/216
Probability of rolling a total of 5 on two dice (1 and 4 or 2 and 3, in some order) and higher than 5 on one die = (4/6)(1/6)(1/6) = 4/216

Total = 4/216 + 6/216 + 6/126 + 4/216 = 20/216 = 5/54

Final Answer:
[Reveal] Spoiler:
A


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Re: A conjuror will roll one red, six-sided die in his right hand and two [#permalink]

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New post 26 Nov 2017, 01:09
EMPOWERgmatRichC wrote:
Hi All,

This prompt asks for the probability that the number on one 6-sided die will be greater than the SUM of the numbers on two 6-sided dice. This question requires us to consider multiple possible situations.

To start, we only have to consider a few possible sums for the 'pair' of dice: 2, 3, 4 and 5 (in all other situations, the sum of the two dice CANNOT be less than the total on the one die).

Probability of rolling a total of 2 on two dice (1 and 1) and higher than 2 on one die = (1/6)(1/6)(4/6) = 4/216
Probability of rolling a total of 3 on two dice (1 and 2 in some order) and higher than 3 on one die = (2/6)(1/6)(3/6) = 6/216
Probability of rolling a total of 4 on two dice (1 and 3 or 2 and 2, in some order) and higher than 4 on one die = (3/6)(1/6)(2/6) = 6/216
Probability of rolling a total of 5 on two dice (1 and 4 or 2 and 3, in some order) and higher than 5 on one die = (4/6)(1/6)(1/6) = 4/216

Total = 4/216 + 6/216 + 6/126 + 4/216 = 20/216 = 5/54

Final Answer:
[Reveal] Spoiler:
A


GMAT assassins aren't born, they're made,
Rich


if it isnt mentioned that the dice are different do we assume that they are in gmat?

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Re: A conjuror will roll one red, six-sided die in his right hand and two   [#permalink] 26 Nov 2017, 01:09
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A conjuror will roll one red, six-sided die in his right hand and two

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