Probability - Three Dice

[siliconfish]

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?

My problem with the above question is whether we need to consider each set of result as a distinct possibility such as 5 5 3 and 5 3 5 . If so can you please explain the reasoning.

Thanks in advance.

[putneyswope]

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?

My problem with the above question is whether we need to consider each set of result as a distinct possibility such as 5 5 3 and 5 3 5 . If so can you please explain the reasoning.

Thanks in advance.

Think about a place holder of three spots for each dice:

_ _ _

155 (you could throw a 1 first, then a 5, then a 5)
515 (you could throw a 5 first, then a 1, then a 5)
551 ...
355 ...
535
553
555

So probability is 7/6^3.

but i'm no math guru!

[siliconfish]

Isn't it that three dice are thrown at once?

[gmatfundoo]

Yes you need to consider them as separate, since two 5's can be obtained on any of the three dice, and would hence increase the probability.

The answer IMO is 1/24 as explained below (Kindly confirm if this is the OA)

For getting a product divisible by 25, we need a 5 on atleast two of the three dice

So probability of getting a product divisible by 25 is 1/6 *1/6 *3 = 1/12
{We need to multiply by three since there are three ways of getting a 5 on atleast two of the three die namely, (1st 2nd) (1st 3rd ) (3rd 1st)}

Probability of getting an odd multiple is half of 1/12, which is 1/24 (since there is an equal probability of getting an even multiple and odd multiple, because there are an equal no. of odd and even numbers between 1 and 6)

So answer is 1/24.

Alternatively, Probability of getting 5 on two die and an odd number on the third die

= 1/6 *16* ½ *3 = 1/24 ( Multiplied by 3 for the same reason as mentioned above)

[deltasquare]

7/216 is the correct answer. The simplest way to do it is as putneyswope mentioned.

gmatfundoo: your approach is a good example for how complicated thinking will frequently result in wrong answers. in your answer 1/24 (= 9/216 = 7/216 + 2/216) you've somehow counted the occurence of 5,5,5 combination thrice. If all the possible products are listed, 125 will occur only once and not thrice as you've calculated. hence your answer exceeds the actual probability by 2/216.

[putneyswope]

Isn't it that three dice are thrown at once?

hey siliconfish

I know what you mean and this does seem confusing. I've learned the best thing to do is imagine place holders for the dice.

Like a slot machine.

_ _ _

The first place holder is for dice #1, the second for dice #2, etc.

Don't think about it like throwing 3 "at once"

imagine the slot machine in vegas

slot 1 slot 2 slot 3

and the outcomes of the dice show up accordingly

whats the way of getting a 531 product? only three ways 531, 513, etc

whats the way of getting a 555 product? only ONE way