1992 AMC 8 Problem 23

Question

If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than

10

is

\dfrac37

\dfrac{17}{36}

\dfrac12

\dfrac58

\dfrac{11}{12}

There are

6 \times 6 = 36

equally likely outcomes; count those whose product exceeds

10

Make a

6 \times 6

multiplication table and mark every product greater than

10

There are

36

equally likely outcomes. Counting the ordered pairs whose product exceeds

10

: with a first die of

2

there is

1

(namely

2 \times 6

); of

3

, there are

3

; of

4

, there are

4

; of

5

, there are

4

; of

6

, there are

5.

That is

1 + 3 + 4 + 4 + 5 = 17

favorable outcomes, so the probability is

\dfrac{17}{36}.

Thus, the correct answer is B .

Problem 23 in other years