# Conditional Probability on type of coin

1. Oct 18, 2012

### 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?

2. Oct 18, 2012

### Ray Vickson

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

3. Oct 18, 2012

### 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'

4. Oct 18, 2012

### 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

5. Oct 18, 2012

### 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?

6. Oct 18, 2012

### Ray Vickson

We have 10 coins, not 5.

RGV

7. Oct 18, 2012

### 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.

Last edited: Oct 18, 2012
8. Oct 18, 2012

### Ray Vickson

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

9. Oct 18, 2012

### Ray Vickson

OK now, but of course you need to say $\sum_{i=1}^{10},$ which is actually what you calculated.

RGV