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May 17th, 2010, 02:59 PM   #1
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Probability question about right college choice

I have to solve this problems and after a couple of weeks trying to answer it I have failed to come up with the right answer.

Here is the problems.

There is a high school senior who wants to go to the best college for him. There are 3 college to choose from and only 1 of the is the right one for him. His school have 3 adviser. Adviser 1 is right 80% of the times. Adviser 2 is right 10% of the times. Adviser 3 is right only when Adviser 2 and 3 agree on the same answer. Can the student get a probability to get to the right college greater than 80%?

For simplicity, I assume the right college to be A. If advisers 1 and 2 agree on the same college choice, then Adviser 3 is always going to say A, otherwise the combination is not possible.

By using this formula -> P(1 U 2 U 3) = P(1) + P(2) + P(3) - P(1 int 2) - P(1 int 3) - P( 2 int 3) + P(1 int 2 int 3) and my answer was 83.44%, however my prof. wasn't happy with the result and told me to analyse the problem better.

Please help
note: int means interception. U means union.
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May 17th, 2010, 05:53 PM   #2
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Re: Probability question about right college choice

How did you get your answer? I'm actually slightly confused, because I expected the answer to be a strategy and you gave a number.
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May 17th, 2010, 07:55 PM   #3
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Re: Probability question about right college choice

Actually,they answer have to say whether the student can do better than the 80% given by Adviser 1 (by using the other advisers recommendation as well) and if so, how much better can he do. What percentage?

In order to get the 83.44% I did the following

P(1 U 2 U 3) = P(1) + P(2) + P(3) - P(1 int 2) - P(1 int 3) - P( 2 int 3) + P(1 int 2 int 3) = .80 + .10 + .08 - .08 - .064 - .008 + .0064
= .8344 * 100
= 83.44%

Another strategy I was trying to use was to do the following
BAA for example means that Adviser 1 choose B, Adv 2 choose A, and Adv 3 choose A as well. Assume that A is the right choice.
Take all the possible outcomes:
AAA
BAA
CAA
AAB
BAB
CAB
AAC
BAC
CAC
ABA
BBA
CBA
ABB
BBB
CBB
ABC
BBC
CBC
ACA
BCA
CCA
ACB
BCB
CCB
ACC
BCC
CCC

Add all the probabilities of each case and I should get the answer. However, I don't know how to get the probability of ABC for example.
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May 17th, 2010, 08:54 PM   #4
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Re: Probability question about right college choice

Quote:
Originally Posted by locorecto
In order to get the 83.44% I did the following

P(1 U 2 U 3) = P(1) + P(2) + P(3) - P(1 int 2) - P(1 int 3) - P( 2 int 3) + P(1 int 2 int 3) = .80 + .10 + .08 - .08 - .064 - .008 + .0064
= .8344 * 100
= 83.44%
But what does that have to do with the question? You're calculating the probability that at least one of 1, 2, or 3 is correct, but that doesn't say what you choose. (Also, I think the calculation is incorrect, but that's OK since it's irrelevant regardless.)
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May 18th, 2010, 05:25 AM   #5
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Re: Probability question about right college choice

Actually, that is why I am here, because I have issues trying to solve this problem.

P(1 U 2 U 3) means that I go the all the advisors and if they all agree on the right answer, the my probability of getting the right college is about 83.44%.

Thanks for your response.
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May 18th, 2010, 05:55 AM   #6
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Re: Probability question about right college choice

What should you do if Adviser 1 and Adviser 3 agree with each other but disagree with Adviser 2? How often does this happen? What should you do if Adviser 2 and Adviser 3 agree but differ with Adviser 2? How often does that happen? What should you do if all three disagree? How often does that happen?
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