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