# Conditional Probability on type of coin

by CAF123
Tags: coin, conditional, probability, type
 PF Gold P: 2,305 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; $$P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})}$$ where $$P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}.$$ Putting this together gives the wrong answer. Any ideas?
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P: 5,016
 Quote by CAF123 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; $$P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})}$$ where $$P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}.$$ 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
 PF Gold P: 2,305 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'
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P: 5,016
Conditional Probability on type of coin

 Quote by CAF123 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
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P: 2,305
 Quote by Ray Vickson 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?
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P: 5,016
 Quote by CAF123 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
 PF Gold P: 2,305 I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said $\sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10)$ I now get the right result - thanks for your help! EDIT : let i go form 1 to 10.
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P: 5,016
 Quote by CAF123 I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said $\sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10)$ 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
 Quote by CAF123 I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said $\sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10)$ I now get the right result - thanks for your help! EDIT : let i go form 1 to 10.