1996 AHSME Problems/Problem 16

Problem

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?

$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$

Solution

The third die cannot be $1$ , since the minimal sum on the other two dice is $2$ .

If the third die is $2$ , then the first two dice must be $(1,1)$ .

If the third die is $3$ , then the first two dice must be $(1,2)$ or $(2,1)$ .

If the third die is $4$ , then the first two dice must be $(1,3)$ , $(2,2)$ , or $(3,1)$ .

If the third die is $5$ , then the first two dice must be $(1,4)$ , $(2,3)$ , $(3,2)$ , or $(4,1)$ .

If the third die is $6$ , then the first two dice must be $(1,5)$ , $(2,4)$ , $(3,3)$ , $(4,2)$ , or $(5,1)$ .

There are $15$ possibilities for the three dice. Of those possibiltiies, $7$ have a $2$ in the first two dice, and $1$ has a $2$ in the third die. Therefore, the answer is $\boxed{8}{15}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1996 AHSME Problems/Problem 16

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