Problem Solving

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!

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.

Hello BTGmoderatorLU.

The problem is that you considered the result as an even integer, and the question asks for an odd integer.

The way I'd solve it is as follows:

We have 3 dices. Then we have a total of 6*6*6=216 different combinations of results.

Now, as you said, the ways of getting an odd number divisible by 25 is getting two 5's and one odd number.

Hence, since the order matters we have the following options:
(i) Case 1:
1st dice ------- 2nd dice ------ 3rd dice
1 ------------------5--------------5
3------------------5--------------5
5------------------5--------------5

(ii) Case 2:
1st dice ------- 2nd dice ------ 3rd dice
5 ------------------1--------------5
5------------------3--------------5

(iii) Case 3:
1st dice ------- 2nd dice ------ 3rd dice
5 ------------------5--------------1
5------------------5--------------3

And there are no more cases. Note that 5, 5, 5 appears only once because changing the order will lead to the same result.

Therefore, there are 7 different option of getting an odd number divisible by 25. So, the probability is 7/216.

This is why the correct answer is the option A.

I hope it is clear enough.

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

A odd multiple of 25 will be yielded if the product is composed of two 5's and an odd integer.

Case 1: Crowan rolls three 5's
P(5 on the 1st roll) = 1/6. (Of the 6 possible rolls, one is 5.)
P(5 on the 2nd roll) = 1/6.
P(5 on the 3rd roll) = 1/6.
Since we want all of these events to happen, we MULTIPLY:
1/6 * 1/6 * 1/6 = 1/216.

Case 2: Crowan rolls two 5's and an odd integer other than 5
Let F = 5 and O = an odd integer other than 5.

P(exactly n times) = P(one way) * all possible ways.

P(one way):
One way to get exactly two 5's and an odd integer other than 5:
FFO.
P(F on the 1st roll) = 1/6. (Of the 6 possible rolls, one is 5.)
P(F on the 2nd roll) = 1/6.
P(O on the 3rd roll) = 2/6. (Of the 6 possible rolls, only 1 and 3 are odd integers other than 5.)
Since we want all of these events to happen, we MULTIPLY:
1/6 * 1/6 * 2/6 = 2/216.

All possible ways:
FFO is only ONE WAY to get two 5's and an odd integer other than 5.
Now we must account for ALL OF THE WAYS to get two 5's and an odd integer other than 5.
Any arrangement of the letters FFO represents one way to get two 5's and an odd integer other than 5.
Thus, to account for ALL OF THE WAYS to get two 5's and an odd integer other than 5, the result above must be multiplied by the number of ways to arrange the letters FFO.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical F's:
3!/2! = 3.

Multiplying the results above, we get:
P(two 5's and an odd integer other than 5) = 2/216 * 3 = 6/216.

Resulting probability:
Since a good outcome will yielded by Case 1 OR Case 2, we ADD the two cases:
1/216 + 6/216 = 7/216.

The correct answer is A.

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