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 Post subject: Combinatorics and Probability (Vol. XII), Question #53.Posted: Sat Oct 30, 2010 8:42 pm

Joined: Thu Jul 01, 2010 12:05 am
Posts: 35
This question has the following situation: "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

The correct answer is (A). I suspected that this question is asking for complimentary events. That is, what is the probability of occurrence? This is solved by subtracting the probability of non-occurrence from 1. However, I don't understand how we're left with 7 favorable outcomes, because my answer came to be 116/216. Can you please explain how we're left with 7 favorable outcomes and provide a comment on my approach to this question?

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 Post subject: Re: Combinatorics and Probability (Vol. XII), Question #53.Posted: Mon Nov 08, 2010 12:10 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Odd integer divisible by 25 means we have 5, 5, and either 1, 3, or 5.
This one may be easier to do without using complementary events...

We can have
5, 5, 1
5, 1, 5
1, 5, 5,
Then, replace the 1 with a 3, and then add 5, 5, 5, to the mix.
This gives us 7 different ways we can get our desired outcome.

There are a total of 6*6*6 possible outcomes, so we have
7/216

Veritas Help

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