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Conditional Probability on type of coin

by CAF123
Tags: coin, conditional, probability, type
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CAF123
#1
Oct18-12, 10:21 AM
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1. The problem statement, all variables and given/known data
Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1,2...10. When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?


2. Relevant equations
Bayes's Formula


3. The attempt at a solution
First of all, I can't make any sense out of how choosing a different coin will give a different probability of showing a head?
In solving the problem, I used Bayes's relation; [tex] P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})} [/tex]

where [tex] P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}. [/tex] Putting this together gives the wrong answer. Any ideas?
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Ray Vickson
#2
Oct18-12, 10:43 AM
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Quote Quote by CAF123 View Post
1. The problem statement, all variables and given/known data
Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1,2...10. When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?


2. Relevant equations
Bayes's Formula


3. The attempt at a solution
First of all, I can't make any sense out of how choosing a different coin will give a different probability of showing a head?
In solving the problem, I used Bayes's relation; [tex] P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})} [/tex]

where [tex] P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}. [/tex] Putting this together gives the wrong answer. Any ideas?
There is something missing from your problem statement: you say "When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?" Did you mean that one of the coins was flipped and came up heads? Did you mean 'what is the conditional probability it was coin i?" I will assume the answer is YES to both of these questions.

P{coin i|H} = P{coin i & H}/P{H}. What is the numerator equal to? How do you find the denominator?

RGV
CAF123
#3
Oct18-12, 10:58 AM
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Oh sorry, how careless of me. It should read 'When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the 5th coin'

Ray Vickson
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Oct18-12, 11:23 AM
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Conditional Probability on type of coin

Quote Quote by CAF123 View Post
Oh sorry, how careless of me. It should read 'When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the 5th coin'
OK, so my previous post is correct if we put i = 5. Now can you answer the questions I asked there? Take it one step at a time. Apply Bayes' rules, etc.

RGV
CAF123
#5
Oct18-12, 11:49 AM
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Quote Quote by Ray Vickson View Post
OK, so my previous post is correct if we put i = 5. Now can you answer the questions I asked there? Take it one step at a time. Apply Bayes' rules, etc.

RGV
I got that P(coin 5|head) = P(head|coin5)P(coin5)/ƩP(head|coin i)P(coin i). (Where the sum is from i=1 to i=5). Is this correct so far?
Ray Vickson
#6
Oct18-12, 12:00 PM
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Quote Quote by CAF123 View Post
I got that P(coin 5|head) = P(head|coin5)P(coin5)/ƩP(head|coin i)P(coin i). (Where the sum is from i=1 to i=5). Is this correct so far?
We have 10 coins, not 5.

RGV
CAF123
#7
Oct18-12, 12:00 PM
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I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10) [/itex] I now get the right result - thanks for your help!

EDIT : let i go form 1 to 10.
Ray Vickson
#8
Oct18-12, 12:07 PM
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Quote Quote by CAF123 View Post
I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) [/itex] This is still incorrect.
What's so special about the number '5'? Suppose, instead, I asked you for the conditional probability of coin 2, or coin 7 or coin 10. How would you express the conditional probabilities in those cases? Remember, I first asked you about the conditional probability of coin i, where I did not specify i.

RGV
Ray Vickson
#9
Oct18-12, 02:08 PM
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Quote Quote by CAF123 View Post
I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10) [/itex] I now get the right result - thanks for your help!

EDIT : let i go form 1 to 10.
OK now, but of course you need to say ##\sum_{i=1}^{10},## which is actually what you calculated.

RGV


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