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Lets assume that a phone number lies in one of three pages with equal probability (1/3).

The probability of selecting the pages follows a Bell Curve with probabilities as :

.2    .6    .2 

Now the user selects a page. What is the probability that the phone number is on that page

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Bad news: the probabilities of selecting the pages do not sum to 1. Good news: they are not needed to solve your problem. Hint for the solution: imagine the user selects the page first, and then the phone number chooses the page it wants to appear on. –  Did Aug 4 '11 at 9:14
    
ops that was a typo ... now they sum to 1 :) –  basarat Aug 4 '11 at 9:20
    
@Didier: Ah, I hadn't seen your hint :-). I thought the analogy might be even more convincing when built in two stages... –  joriki Aug 4 '11 at 9:22
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@Basarat: If the hints aren't helping you, it would help us help you if you give us a hint which part you don't understand. There's an equivalence between A) and B) in my answer and an equivalence between B) and your problem. Which of these two equivalences do you not see? Or do you not see the answer to A)? –  joriki Aug 4 '11 at 9:34
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@Basarat, I see you saw it is not (variable), so everything is fine. –  Did Aug 4 '11 at 16:30
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1 Answer 1

up vote 2 down vote accepted

Hint: It might help to think about these two similar problems first:

A)

Let's assume that one of three pages is selected from a phone book with probabilities

.2    .6    .2

The user selects a page. Now we throw a die and write a phone number on one of the three pages with equal probability (1/3).

What is the probability that we write the phone number on the selected page?

When you feel sure about the answer to that one, consider

B)

Let's assume that we throw a die and write a phone number on one of three pages of a phone book with equal probability (1/3).

Now the user selects one of the three pages from the phone book with probabilities

.2    .6    .2

What is the probability that the page that has the phone number written on it is selected?

What difference do you think it could make whether we throw the die before or after the selection?

Then, when you feel sure about the answer to that one, think about what difference you think it could make whether we throw a die before the selection or the number is already there with the same probabilities as given by the die.

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So I guess it will have a fixed value only if I know that the number is on the page n (where n = 1, 2, 3). e.g. for n = 3 the probability will be .2 . Therefore correct me if I am wrong : "One cannot tell the probability without knowing on which page the number is?" –  basarat Aug 4 '11 at 14:53
    
@Basarat: No. Think about problem A) again. One of three pages is selected. Then we throw a die and write a phone number on one of the three pages with equal probability (1/3). What is the probability that we write the phone number on the selected page? If you don't see the answer, try it out with a real die :-) Select page 1 if the die shows 1 or 2, page 2 if it shows 3 or 4, and page 3 if it shows 5 or 6, and record how often you write the phone number on the selected page. –  joriki Aug 4 '11 at 15:08
    
Thanks for the explanation :D sorry for taking so long to understand what you were saying. I feel soo stupid 1/3 was staring me in the face all along! –  basarat Aug 4 '11 at 15:51
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