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May 17th, 2010, 02:59 PM  #1 
Newbie Joined: May 2010 Posts: 3 Thanks: 0  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. 
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  
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 Quote:
 
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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