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# 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 [#permalink]

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12 Feb 2011, 04:45
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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

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!
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Jul 2013, 04:38, edited 1 time in total.
Renamed the topic and edited the question.

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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12 Feb 2011, 05:57
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..

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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12 Feb 2011, 09:45
I get 9/216 too.

1/6*1/6*1/2, but there are 3 combinations of this, the 1/2 can be in slot 1, 2 or 3.

(1/6*1/6*1/2)*3 = 9/216

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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12 Feb 2011, 15:27
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?
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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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12 Feb 2011, 17:19
almostfamous wrote:
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?

Nope, i am pretty sure its 9/216. Btw, does the OA contain explanations as well? Also mind sharing what the answer choices are since this should be a PS question

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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12 Feb 2011, 18:53
almostfamous wrote:
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?

x3 not to the power of 3. There are 3 different ways it can happen.

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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13 Feb 2011, 05:46
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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)

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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17 Feb 2011, 09:26
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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.

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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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17 Feb 2011, 11:59
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almostfamous wrote:
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!

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?
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Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

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17 Feb 2011, 13:23
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almostfamous wrote:
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!

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.

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.
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Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

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01 Sep 2014, 20:57
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7 favorable events (551, 553, 555, 515, 535, 155 & 355) out of 216 events. As only odd products of 25 are asked for.

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Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

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06 Oct 2015, 23:42
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

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Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

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Re: Crowan throws 3 dice and records the product of the numbers   [#permalink] 23 Nov 2017, 21:12
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