BTGmoderatorDC wrote: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
OA A
Source: Magoosh
The minimum sum the two blue dice can have is 2, and the maximum sum they can have such that the red die can exceed that sum is 5.
Let's find out the probabilities of finding the sum of 2, 3, 4, and 5 for the two blue dice.
1. Sum = 2 => Ways: {1, 1} => Probaility = 1/36
2. Sum = 3 => Ways: {1, 2} & {2, 1} => Probaility = 2/36
3. Sum = 4 => Ways: {1, 3}, {2, 2}, & {3, 1} => Probaility = 3/36
4. Sum = 5 => Ways: {1, 4}, {2, 3}, {3, 2} => Probaility = 4/36
Let's find out the probability of getting 3, 4, 5, & 6 for the red dice.
1. Probability of getting a 3, 4, 5, or 6 = 4/6
2. Probability of getting a 4, 5, or 6 = 3/6
3. Probability of getting a 5 or 6 = 2/6
4. Probability of getting a 6 = 1/6
Thus,
The probability that the number on the red die will be greater than the sum of the two blue dice
= (4/6 * 1/36) + (3/6 * 2/36) +(2/6 * 3/36) +(1/6 * 4/36) = (1/36 *1/6) [4*1 + 3*2 + 2*3 + 1*4] = [4 + 6 + 6 + 4]/216 = 20/216 = 5/54
The correct answer:
A
Hope this helps!
-Jay
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