My Math Forum Probability question about right college choice

 May 17th, 2010, 05:53 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
 May 17th, 2010, 07:55 PM #3 Newbie   Joined: May 2010 Posts: 3 Thanks: 0 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.
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.)

 May 18th, 2010, 05:25 AM #5 Newbie   Joined: May 2010 Posts: 3 Thanks: 0 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.
 May 18th, 2010, 05:55 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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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