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

$\dfrac37$

$\dfrac{17}{36}$

$\dfrac12$

$\dfrac58$

$\dfrac{11}{12}$

Solution:

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 .

1992 AMC 8 Problem 23

Solution:

Problem 23 in Other Years