1992 AJHSME Problems/Problem 23

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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 $\boxed{\text{(B)}\ \frac{17}{36}}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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