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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

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.

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.

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

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)

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

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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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

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.

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).

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

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

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