Probabilities Problem - Crowan throwin' the Dice

[dellaboemia]

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

Source: Veritas Prep Combinatorics and Probability Page 102, #53

Correct Answer:
A

[GMATGuruNY]

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

Source: Veritas Prep Combinatorics and Probability Page 102, #53

Correct Answer:
A

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.

[dellaboemia]

Thanks Mitch. Your approach totally makes sense. I'm struggling to understand where my process went wrong. I rephrased the question to be the probability of getting a 5, a 5, and a 1 3 or 5. One scenario yields 1/6 * 1/6 * 3/6 or 3/216. I multiplied by 3 since these can happen in any order. I end up with 9/216. Why doesn't that approach work? What am I missing here?

[dellaboemia]

Thanks Mitch. Your approach totally makes sense. I'm struggling to understand where my process went wrong. I rephrased the question to be the probability of getting a 5, a 5, and a 1 3 or 5. One scenario yields 1/6 * 1/6 * 3/6 or 3/216. I multiplied by 3 since these can happen in any order. I end up with 9/216. Why doesn't that approach work? What am I missing here?

Ahh just got it. The scenario of 3 5s is over counted, ergo the importance of decomposing the problem into mutually exclusive events.

[GMATGuruNY]

Thanks Mitch. Your approach totally makes sense. I'm struggling to understand where my process went wrong. I rephrased the question to be the probability of getting a 5, a 5, and a 1 3 or 5. One scenario yields 1/6 * 1/6 * 3/6 or 3/216. I multiplied by 3 since these can happen in any order. I end up with 9/216. Why doesn't that approach work? What am I missing here?

Ahh just got it. The scenario of 3 5s is over counted, ergo the importance of decomposing the problem into mutually exclusive events.

Exactly right.
In your approach, 5-5-5 is counted three times.
Thus, from the 9 favorable outcomes that you counted, 2 must be subtracted:
9/216 - 2/216 = 7/216.

[bubbliiiiiiii]

My approach:

The product, when three dice are rolled, cannot exceed 216.

So, possible outcomes that are odd multiples of 25 are 25, 75, 125, and 175 (4 in all).

Total possible outcomes = 6*6*6

Thus, probability is 4/216!

What am I missing here?

[ronnie1985]

The first 2 dices are 5 and the last one is 1 or 3 or 5.
This can be done in any order. But then 5*5*5 is counted thrice, so subtract 2 times appearance of 5*5*5.
P = (1/6)*(1/6)*(3/6)*3 - (1/6)*(1/6)*1/6)*2 = 7/216.