Crowan throws 3 dice and records the product of the numbers

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..

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

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)

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.

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

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

What is the source of this question?

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

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

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

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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.

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!

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.

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

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