Crowan throws 3 dice and records the product of the numbers

Question

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!

Keith@ThePrincetonReview · Answer

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

Case 1: all fives
P(1st roll is 5) = 1/6.
P(2nd roll is 5) = 1/6.
P(3rd roll is 5) = 1/6.
Since we want all of these events to happen together, we multiply the fractions:
1/6 * 1/6 * 1/6 = 1/216.

Case 2: two 5's and one roll that is 1 or 3
P(1st roll is 5) = 1/6.
P(2nd roll is 5) = 1/6.
P(3rd roll is 1 or 3) = 2/6.
Since we want all of these events to happen together, we multiply the fractions:
1/6 * 1/6 * 2/6 = 2/216.
Since the 1 or 3 could happen on the 1st roll, the 2nd roll, or the 3rd roll, we multiply by 3:
3 * 2/216 = 6/216.

Since either Case 1 or Case 2 will yield a good outcome, we add the results:
1/216 + 6/216 = 7/216.

The correct answer is A.

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regor60 · Answer

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. The point being, to the OP, you can't count 555 3 times as you do the other permutations, but only once, so that means two go away, hence 9 goes to 7

GMATGuruNY · Answer

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

Case 1: all fives
P(1st roll is 5) = 1/6.
P(2nd roll is 5) = 1/6.
P(3rd roll is 5) = 1/6.
Since we want all of these events to happen together, we multiply the fractions:
1/6 * 1/6 * 1/6 = 1/216.

Case 2: two 5's and one roll that is 1 or 3
P(1st roll is 5) = 1/6.
P(2nd roll is 5) = 1/6.
P(3rd roll is 1 or 3) = 2/6.
Since we want all of these events to happen together, we multiply the fractions:
1/6 * 1/6 * 2/6 = 2/216.
Since the 1 or 3 could happen on the 1st roll, the 2nd roll, or the 3rd roll, we multiply by 3:
3 * 2/216 = 6/216.

Since either Case 1 or Case 2 will yield a good outcome, we add the results:
1/216 + 6/216 = 7/216.

The correct answer is A.

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