1992 AJHSME Problems/Problem 23

Problem

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

$\text{(A)}\ \frac{3}{7}\qquad\text{(B)}\ \frac{17}{36}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{12}$

Solution

When the first dice is $1$ , there are no possibilities for the second dice.

When the first dice is $2$ , the second dice can be $6$ , giving one possibility.

When the first dice is $3$ , the second dice can be $4,5,6$ , giving three possibilities.

When the first dice is $4$ , the second dice can be $3,4,5,6$ , giving four possibilities.

When the first dice is $5$ , the second dice can be $3,4,5,6$ , giving four possibilities.

When the first dice is $6$ , the second dice can be $2,3,4,5,6$ , giving five possibilities.

The total number of ways for the product to greater than 10 is $1+3+4+4+5=17$ and the total number of possibilities is $(6)(6)=36$ , yielding a probability of $\boxed{\text{(B)}\ \frac{17}{36}}$ .

1992 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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1992 AJHSME Problems/Problem 23

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