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Conditional Probability on type of coin 
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#1
Oct1812, 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? 


#2
Oct1812, 10:43 AM

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


#3
Oct1812, 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'



#4
Oct1812, 11:23 AM

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Conditional Probability on type of coin
RGV 


#5
Oct1812, 11:49 AM

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#6
Oct1812, 12:00 PM

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P: 4,959

RGV 


#7
Oct1812, 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(headcoin\, 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. 


#8
Oct1812, 12:07 PM

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#9
Oct1812, 02:08 PM

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