A conjuror will roll one red, six-sided die in his right...

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

The OA is A.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advance.

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by [email protected] » Sat Nov 04, 2017 11:26 am
Hi AAPL,

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: A

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Rich
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