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

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

Could anyone enlighten me on this tricky probability problem:

"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"

My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72

(Source: Veritas Prep, Combinatorics p.102)

However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.

Thanks!

Bunuel · Accepted Answer

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

In this case as 555 can occur only one way, then in nominator we'll have 3+3+1=7:
551 - 515 - 155;
553 - 535 - 355;
555.

P(55O)=\frac{7}{216}  (if you want to calculate as in previous case then:  P(55O)=P(551)+P(553)+P(555)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{7}{216} )

Note that above 16 scenarios describe all cases of having at least two 5 on 3 dice: 7 cases give odd sum and 9 give even sum.

Answer: A.

If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an  EVEN  number divisible by 25 at any attempt?

P(55E)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{3}{6}=\frac{9}{216} , we multiply by  \frac{3!}{2!}  or simply by 3 because 5-5-Even can occur in 3 different ways: 5-5-Even, 5-Even-5, or Even-5-5 (for example: 552, 525, or 255).

Check all 9 scenarios:
552 - 525 - 255;
554 - 545 - 455;
556 - 565 - 655.

Hope it helps.

ScottTargetTestPrep · Answer

The total number of ways to throw 3 dice is 6 x 6 x 6 = 216

The only way to get a product of the numbers appearing on the top of 3 dice to be an odd number that is divisible by 25 is if two of the numbers are both 5 and the third number is an odd number. Thus, we have:

{1, 5, 5}, {5, 1, 5}, {5, 5, 1}, {3, 5, 5}, {5, 3, 5}, {5, 5, 3} and {5, 5, 5}

Thus the probability is 7/216.

Answer: A

drifting · Answer

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..

eragotte · Answer

I have a theory. There is another question going similar to this but slightly different where:

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

In this situation the answer is 7/216 because 5x5x5 only occurs once, not 3 times.

Where as in our example 552 255 525 455 545 554 655 565 556 all occur (9 options)

miguelmick · Answer

I also get 9/216: (1/6 x 1/6 x 3/6) x 3. The probabilities would be:

2 x 5 x 5
4 x 5 x 5
6 x 5 x 5
5 x 2 x 5
5 x 4 x 5
5 x 6 x 5
5 x 5 x 2
5 x 5 x 4
5 x 5 x 6

So, we have 216 possible outcomes and 9 desired results, as shown above. I strongly believe the OA is wrong.

gmatpapa · Answer

Hi, the question on p102 is:

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

Answer is 7/216 and the explanation to me is quite convincing.

What is the source of this question?

IshanVas · Answer

7 favorable events (551, 553, 555, 515, 535, 155 & 355) out of 216 events. As only odd products of 25 are asked for.

ramajha · Answer

if i am in hurry ..i will mark A as answer as Denominator has to a factor of 216 so option A and E are only valid ones..
in three throws 5 have to occur twice so permutation will be 3!/2! x 3(for three odd numbers 1,3,5) - 2(as for the case when 5 will appear in all three throws)=7 so 7/216

arya251294 · Answer

How do we know that crowan is going to write the different result for 551 and 155 and 515, as all three are going to be 25, isn't crowan going to write 25 only once. Please help

Posted from my mobile device

Bunuel · Answer

The point is that the dice are different and 5 on first die, 5 on the second die and 1 on the third die; is different case from 5 on first die, 1 on the second die and 5 on the third die; and from 1 on first die, 5 on the second die and 5 on the third die.

vishy001 · Answer

We can start with the maximum product of 3 dice. Max side = 6. So, max product of 3 dice = 6x6x6 = 218.
Now take all multiples of 25 < 218. That will be 25,50,75,100,125,150,175,200 (225 > 218 so, eliminate)
Now consider only odd values. That will be 25,75,125,175.
Factorizing them, we get 1x5x5, 3x5x5, 5x5x5, 7x5x5. Again, 7x5x5 is not possible coz max side on dice = 6
which leaves us with 3 possibilities: 1x5x5, 3x5x5x, 5x5x5. Which means, (two 5s, one 1)  OR  (two 5s, one 3)  OR  three 5s.
Since all sides are equally likely, probability of any side = 1/6. Three dices, total probability = 1/6x1/6x1/6 = 1/216
Now consider the number of combinations of 1x5x5, 3x5x5x, 5x5x5:
For two 5s, one 1: 3 (515, 551, 155) (or 3!/2!)
For two 5s, one 2: 3 (535,553, 355)
For three 5s: 1 (555) (all are identical, so changing order doesn't have effect).
So, total combinations = 3 +3 + 1 = 7
The probability of event = 1/216 * 7 = 7/216!

Jay1594 · Answer

You almost solved it perfectly!
The only problem is that you didn't take into account the sequence 5-5-5. 
This is the only outcome that will not have 3 possible combinations, instead it has only one. So your numerators decreases by two and your answer is A.

Vibhatu · Answer

No Read the question carefully,   any attempt is an odd integer divisible by 25

Mananjain · Answer

Probability = P(Favourable outcomes) / P(Total outcomes)

P(Total outcomes) = 6*6*6 = 216
Possible favorable outcomes are 25, 75, 125
No. of ways of getting 25: (5*5*1) = 3!/2! = 3 { 5,5,1 ; 5,1,5 ; 1,5,5 }
No. of ways of getting 75: (5*5*3) = 3!/2! = 3 { 5,5,3 ; 5,3,5 ; 3,5,5 }
No. of ways of getting 125: (5*5*5) = 3!/3! = 1 { 5,5,5 }

Probability = 7/216

BrushMyQuant · Answer

Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt.

Since Crowan is rolling three dice => Number of cases =  6^3  = 216

What is the probability that the result of any attempt is an odd integer divisible by 25?

For the sum to be an odd integer divisible by 25 (5*5), we need to have two die show 5 each and one die show an odd number

Lets list down the possible cases as there wont be many cases:

First die gets a 5
  (5,5,Odd number) - 3 cases
(5,Odd Number,5) - 2 cases (ignoring (5,5,5) case which was counted above)
Total cases = 5

Second Die gets a 5  
(5,5,Odd number) - 0 cases - counted above
(Odd number,5,5) - 2 cases (ignoring (5,5,5) case)
 Total cases = 2

Third Die gets a 5  
(5,Odd Number,5) - 0 cases - counted above
(Odd Number,5,5) - 0 cases - counted above

Total Cases  = 5 + 2 + 0 = 7

=> P(Product = Odd multiple of 25) =  \frac{7}{216}

So,  Answer will be A 
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

PSKhore · Answer

Maximum number is 6*6*6 = 216

Hence, the odd multiples of 25 will be : 25,75,125,175
25 = {1,5,5}
75 = {5,5,3}
125 = {5,5,5}

{1,5,5} and {5,5,3} can be arranged amongst themselves in 3!/2! ways since 5,5 are identical and {5,5,5} can be arranged in 1 way

Therefore, total ways are 2+2+1 and in the denominator 6*6*6 = 216

= 7/216

Crowan throws 3 dice and records the product of the numbers

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Crowan throws 3 dice and records the product of the numbers

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