1996 AMC 12 Problem 16

Question

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

2

is tossed?

\frac16

\frac{91}{216}

\frac12

\frac8{15}

\frac7{12}

Answer: D

Concepts:conditional probability case counting

Difficulty rating: 1730

Small hint:

Under the stated condition, count ordered pairs for the first two tosses whose sum is at most

6

Big hint:

For the favorable count, separate a third-toss

2

from pairs having a

2

among the first two tosses

Solution:

For third tosses

2,3,4,5,6,

the numbers of ordered first-two-toss pairs are

1,2,3,4,5,

for

15

equally likely conditional outcomes. A third toss of

2

contributes

(1,1,2).

A

2

in the first position gives

4

outcomes and a

2

in the second gives

4,

with

(2,2,4)

counted twice. Thus there are

1+4+4-1=8

favorable outcomes, and the correct answer is D.