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Help needed regarding probability question

  • #1
utkarshakash
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Homework Statement


A letter is known to have come either from London or Clifton; on the postmark only the two consecutive letters ON are legible; what is the chance that it came from London?

Homework Equations



The Attempt at a Solution


Let us assume that probability that it came from London or Clifton is 1/2. Now the probability that letters ON is selected from London is 2/4 and for Clifton it is 1/6. So the total probability is (1/2 * 2/4 + 1/2 * 1/6). Now the chance that it is from London is (1/2 * 2/4) / (1/2 * 2/4 + 1/2 * 1/6). I can't figure out what's wrong with my solution.
 

Answers and Replies

  • #2
HallsofIvy
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I guess you are arguing that since there are two "on"s in "London" and only one in "Clifton", the fact that "on" is legible means it is twice as likely to have come from London as from Clifton. I'm not sure I agree with that reasoning, but assuming it is correct, it still makes no sense to say "let us assume that probability it came from London or Clifton is 1/2" when the problem is precisely to find that probability. If we take the probability the letter came form Clifton to be "p", the probability it came from London is "2p" and, if those are the only two possibilities, we must have p+ 2p= 3p= 1 so that p= 1/3 and 2p= 2/3.
 
  • #3
Ray Vickson
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Homework Statement


A letter is known to have come either from London or Clifton; on the postmark only the two consecutive letters ON are legible; what is the chance that it came from London?

Homework Equations



The Attempt at a Solution


Let us assume that probability that it came from London or Clifton is 1/2. Now the probability that letters ON is selected from London is 2/4 and for Clifton it is 1/6. So the total probability is (1/2 * 2/4 + 1/2 * 1/6). Now the chance that it is from London is (1/2 * 2/4) / (1/2 * 2/4 + 1/2 * 1/6). I can't figure out what's wrong with my solution.
You could look at the problem differently: 'Clifton' has 6 adjacent letter pairs ('cl', 'li', etc.) of which exactly one is 'on', while 'London' has 5 adjacent letter pairs ('lo', 'on',etc.) of which two are 'on'. If you start with an a priori probability of London or Clifton and assume a probability distribution over the letter pairs, you can find posterior (conditional) probabilities for London or Clifton, given 'on'.
 
  • #4
utkarshakash
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I guess you are arguing that since there are two "on"s in "London" and only one in "Clifton", the fact that "on" is legible means it is twice as likely to have come from London as from Clifton. I'm not sure I agree with that reasoning, but assuming it is correct, it still makes no sense to say "let us assume that probability it came from London or Clifton is 1/2" when the problem is precisely to find that probability. If we take the probability the letter came form Clifton to be "p", the probability it came from London is "2p" and, if those are the only two possibilities, we must have p+ 2p= 3p= 1 so that p= 1/3 and 2p= 2/3.
But this answer is not correct.
 
  • #5
haruspex
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You could look at the problem differently: 'Clifton' has 6 adjacent letter pairs ('cl', 'li', etc.) of which exactly one is 'on', while 'London' has 5 adjacent letter pairs ('lo', 'on',etc.) of which two are 'on'. If you start with an a priori probability of London or Clifton and assume a probability distribution over the letter pairs, you can find posterior (conditional) probabilities for London or Clifton, given 'on'.
I think that's pretty much what utkarshakash meant (the original 1/2 being the a priori probability), but went wrong in calculating 2/4 instead of 2/5 for two consecutives in London being "on".
 
  • #6
utkarshakash
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You could look at the problem differently: 'Clifton' has 6 adjacent letter pairs ('cl', 'li', etc.) of which exactly one is 'on', while 'London' has 5 adjacent letter pairs ('lo', 'on',etc.) of which two are 'on'. If you start with an a priori probability of London or Clifton and assume a probability distribution over the letter pairs, you can find posterior (conditional) probabilities for London or Clifton, given 'on'.
This was a better method. Thanks
 
  • #7
utkarshakash
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I think that's pretty much what utkarshakash meant (the original 1/2 being the a priori probability), but went wrong in calculating 2/4 instead of 2/5 for two consecutives in London being "on".
Yes. I started with this method but instead of taking consecutives I assumed ON to be one letter. So this gave me a total of 4 letters in London out of which 2 were ON. And that's how everything messed up.
 

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