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 November 9th, 2012, 07:01 AM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 probability problem Q. A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from (i) LONDON? I'm supposed to solve this problem using Bayes' theorem, but I don't know how to do this. Please help. And please explain the solution in detail.
 November 9th, 2012, 10:36 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications Re: probability problem These derivations could be wrong depending upon the way the problem is interpreted. Plus, this is a lunchtime post and I don't really have time to perform checks as I like to (I usually like to write a Ruby program to check). And I could be wrong anyway. Besides, I would like to get others' comments and interpretations. The way that I am interpreting the problem is that a random mechanism resulted in only two consecutive letters being visible. This mechanism is equally likely to occur at London and Clifton. Bayes' Theorem states: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$ where P(A|B) is the probability of A given B and P(B|A) is the probability B given A. Let's let: A = the letter came from London B = the two adjacent letters, ON, are the ones visible. Some words about B: LONDON has 5 pair of adjacent letters, 2 of which are ON; CLIFTON has 6 pair of adjacent letters, 1 of which is ON. 1) -------------------------------------------- Let's assume that the probability of such letters coming from London and Clifton are equal. This is probably what is intended. Bayes' Theorem calculation: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{\frac{2}{5} \cdot \frac{1}{2}}{\frac{2}{5} \cdot \frac{1}{2}+\frac{1}{6} \cdot \frac{1}{2}}=\frac{12}{17}$ One may argue that the denominator should be 3/11; after all, there are 11 pairs of adjacent letters, 3 of which are ON. However, consider the following: 300 such letters arrive, and since the probabilities for the letters being from London and Clifton are equal, 150 letters arrive from each. 2/5 of the letters from London arrive with the letters ON visible and (2/5)*150 = 60. 1/6 of the 150 letters from Clifton arrive with the letters ON visible, and (1/6)*150 = 25. So (60+25)/300 = 85/300 which equals a probability of 17/60, which is the denominator that is used in the calculation. I think that the difference is that there are different numbers of letter pair combinations between the two cities. 2) -------------------------------------------- Let's assume that two of three such letters come from London and one of 3 such letters comes from Clifton (due to population differences let's say). This is probably not the desired problem, but I present it anyway. Bayes' Theorem calculation: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{\frac{2}{5} \cdot \frac{2}{3}}{\frac{2}{5} \cdot \frac{2}{3}+\frac{1}{6} \cdot \frac{1}{3}}=\frac{24}{29}$ I will try to write a program to verify later.
November 10th, 2012, 04:27 AM   #3
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Joined: Sep 2012

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Re: probability problem

Quote:
 Originally Posted by jks These derivations could be wrong depending upon the way the problem is interpreted. Plus, this is a lunchtime post and I don't really have time to perform checks as I like to (I usually like to write a Ruby program to check). And I could be wrong anyway. Besides, I would like to get others' comments and interpretations. The way that I am interpreting the problem is that a random mechanism resulted in only two consecutive letters being visible. This mechanism is equally likely to occur at London and Clifton. Bayes' Theorem states: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$ where P(A|B) is the probability of A given B and P(B|A) is the probability B given A. Let's let: A = the letter came from London B = the two adjacent letters, ON, are the ones visible. Some words about B: LONDON has 5 pair of adjacent letters, 2 of which are ON; CLIFTON has 6 pair of adjacent letters, 1 of which is ON. 1) -------------------------------------------- Let's assume that the probability of such letters coming from London and Clifton are equal. This is probably what is intended. Bayes' Theorem calculation: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{\frac{2}{5} \cdot \frac{1}{2}}{\frac{2}{5} \cdot \frac{1}{2}+\frac{1}{6} \cdot \frac{1}{2}}=\frac{12}{17}$ One may argue that the denominator should be 3/11; after all, there are 11 pairs of adjacent letters, 3 of which are ON. However, consider the following: 300 such letters arrive, and since the probabilities for the letters being from London and Clifton are equal, 150 letters arrive from each. 2/5 of the letters from London arrive with the letters ON visible and (2/5)*150 = 60. 1/6 of the 150 letters from Clifton arrive with the letters ON visible, and (1/6)*150 = 25. So (60+25)/300 = 85/300 which equals a probability of 17/60, which is the denominator that is used in the calculation. I think that the difference is that there are different numbers of letter pair combinations between the two cities. 2) -------------------------------------------- Let's assume that two of three such letters come from London and one of 3 such letters comes from Clifton (due to population differences let's say). This is probably not the desired problem, but I present it anyway. Bayes' Theorem calculation: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{\frac{2}{5} \cdot \frac{2}{3}}{\frac{2}{5} \cdot \frac{2}{3}+\frac{1}{6} \cdot \frac{1}{3}}=\frac{24}{29}$ I will try to write a program to verify later.
After going through your answer I could solve the problem. Your explanation helped me a lot. You're a genius. Thank you so much...

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# LONDON probability letter comes

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