Keith@ThePrincetonReview wrote:

BTGmoderatorLU wrote:Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?

A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8

The OA is A.

Unfortunately, I am unable to get the answer 7/216 as well.

What I do say though is that I believe the answer could be 9/216.

Given 3 dice, the outcome that gives an even product divisible by 25 would be x,5,5 where x is an even number. And i believe x,5,5 is different from 5,x,5 or 5,5,x. Since there are 3 possible combinations of dice sequence, I get 9/216 by multiplying (1/2*1/6*1/6)*3.

But I don't get why the OA has to be 7/216 also. Can anyone assist me with this PS question, please? Thanks in advance!

Hi there,

You're correct that the desired result can be reached only if the number 5 is rolled at least twice.
You're also correct that we're dealing with permutations.
In fact, you were very close to the answer. (Note that when x = 5, it's not the case that x, 5, 5 is different from 5, x, 5 or 5, 5, x).

When a probability or combinatorics question involves only a small number of (desired) outcomes, it's often profitable simply to list the outcomes.

In order for the result to be an odd multiple of 25, two of Crowan's rolls must be 5s, and one roll must be either 1, 3, or 5.
There are seven permutations of these integers:

1, 5, 5 ---------- 3, 5, 5 ---------- 5, 5, 5
5, 1, 5 ---------- 5, 3, 5
5, 5, 1 ---------- 5, 5, 3

Since there are 6*6*6 = 216 ways for Crowan to roll the dice, and only 7 of those ways result in an odd multiple of 25, the correct answer is choice A.

BTGmoderatorLU wrote:Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?

A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8