Find probability of certain event, total probability theorem

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Homework Statement


Suppose you're at a college campus. 3/4 of the people on the campus are students or professors from that college, and the rest 1/4 aren't. When asked a question, students and professors from that college will give you a correct answer every time, and those that aren't from the college will give you a correct answer 3/8th of the time.

a) You stop a random person and ask for directions to place A belonging on campus and he gives you an answer. What is the probability that the answer is true?

b) You ask the same person the same question again and he gives you the exact same answer. What is the probability that answer is correct now?

Homework Equations


Total probability theorem: P(A) = P(A|H1)P(H1) + P(A|H2)P(H2) + ... + P(A|Hn)P(Hn)

The Attempt at a Solution


Solution a):

Using the total probability theorem, I make two hypotheses:

H1:The person is from the college
H2:The person is not from the college

And the event:

A:The answer is correct
Then:

P(H1) = 3/4, P(H2) = 1/4, P(A|H1) = 1, P(A|H2) = 3/8

The probability for A would be:

P(A) = P(H1)P(A|H1) + P(H2)P(A|H2) = 3/4 + 1/4 * 3/8 = 0.84375

However, when it comes to part b), I am stuck. How do I proceed?
 

Answers and Replies

  • #2
Delta2
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Question (b) is a bit weird because it assumes that a person has a truly random response, that is a specific person when asked the same question many times will have random response each time (though we know in real life this is not the case, usually the same person will reply to the same question with the same answer).

I am not sure (probably I am wrong in what I am saying but maybe I give you some inspiration towards the right answer), but I think what b) does is changing the probability that the asked person is from campus or outside. The probability distribution now will not be 3/4 (for campus members) and (1/4 for not campus members) but y for campus members, 1-y for not campus members where y>0. You have to calculate the y, while intuitively you can understand that y>3/4.
 
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  • #3
Ray Vickson
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Homework Statement


Suppose you're at a college campus. 3/4 of the people on the campus are students or professors from that college, and the rest 1/4 aren't. When asked a question, students and professors from that college will give you a correct answer every time, and those that aren't from the college will give you a correct answer 3/8th of the time.

a) You stop a random person and ask for directions to place A belonging on campus and he gives you an answer. What is the probability that the answer is true?

b) You ask the same person the same question again and he gives you the exact same answer. What is the probability that answer is correct now?

Homework Equations


Total probability theorem: P(A) = P(A|H1)P(H1) + P(A|H2)P(H2) + ... + P(A|Hn)P(Hn)

The Attempt at a Solution


Solution a):

Using the total probability theorem, I make two hypotheses:

H1:The person is from the college
H2:The person is not from the college

And the event:

A:The answer is correct
Then:

P(H1) = 3/4, P(H2) = 1/4, P(A|H1) = 1, P(A|H2) = 3/8

The probability for A would be:

P(A) = P(H1)P(A|H1) + P(H2)P(A|H2) = 3/4 + 1/4 * 3/8 = 0.84375

However, when it comes to part b), I am stuck. How do I proceed?
Like Delta2 I, too, find the second question a bit strange, or at least highly ambiguous. There are two possible interpretations: (1) 3/8 of the non-college people will give correct answers (but the same person asked the same question twice of more will give the same answer---just because it is the same person). (2) A non-college person gives random answers, correct 3/8 of the time; in this case the same such person can give a different answer to exactly the same question asked twice or more times.

However, I do not agree that the 3/4-1/4 probabilities will be changed by knowing the first answer unless you can verify the truth of the same answer before asking the question again. In case (1), asking the question any number of times (without checking the truth of the answer before additional askings) will have no effect. In case (2), the situation is akin to tossing a coin twice and asking for the probability of getting Heads twice in a row (again, under the assumption that the truth of the first answer is not checked before asking the second question).
 
  • #4
haruspex
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Like Delta2 I, too, find the second question a bit strange, or at least highly ambiguous. There are two possible interpretations: (1) 3/8 of the non-college people will give correct answers (but the same person asked the same question twice of more will give the same answer---just because it is the same person). (2) A non-college person gives random answers, correct 3/8 of the time; in this case the same such person can give a different answer to exactly the same question asked twice or more times.
I'm almost sure (2) is intended, but reading the question quite literally suggests a third interpretation: that the non-collegiate folk count, giving exactly 3 correct answers to each sequence of 8 questions. Without knowing the pattern, this would imply that TF and FT each have a 15/56 chance, TT a 6/56 chance, and FF a 20/56 chance.
 
  • #5
Delta2
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I think we need to define an event ##B_n##: The answer remains the same the ##n##th time (we ask the same question to the same person).
##A## is as defined in the OP.

The problem asks for ##P(A/B_2)## though ##P(A/B_n)## might also be interesting to calculate.
 
  • #6
Ray Vickson
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I think we need to define an event ##B_n##: The answer remains the same the ##n##th time (we ask the same question to the same person).
##A## is as defined in the OP.

The problem asks for ##P(A/B_2)## though ##P(A/B_n)## might also be interesting to calculate.
Sorry: I mis-interpreted what you wrote in #2. Indeed, given the answer to two questions, that will change the 1/4-3/4 probabilities. (However, without checking the veracity of the first answer, the 1/4-3/4 ratio would remain unchanged from the first to the second question.)
 
  • #7
Delta2
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Sorry: I mis-interpreted what you wrote in #2. Indeed, given the answer to two questions, that will change the 1/4-3/4 probabilities. (However, without checking the veracity of the first answer, the 1/4-3/4 ratio would remain unchanged from the first to the second question.)
I wasn't sure when I was writing that, what I was actually referring too is the probability ##P(H_i/B_n)## , purely intuitively we expect ##\lim_{n \to \infty}P(H_1/B_n)=1##
 

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