# Conditional probability and partitions

1. May 26, 2010

### Kate2010

1. The problem statement, all variables and given/known data

I'm currently trying to revise for exams and really struggling on this problem:

Suppose you have 3 coins that look identical (ie don't know which is which) with probabilites of 1/4, 1/2 and 3/4 of showing a head.

1. If you pick a coin at random and flip it, what is the probability of a head?
2. If you flip all 3 coins 1 after the other what is the probability of getting 3 heads? Are the three events Aj that the jth coin shows a head for j=1,2,3 independent?
3. You pick a coin at random and flip it twice and get a head both times. What is the conditional probability that you picked the coin with a probability of 3/4 of getting a head?
4. You pick one of the other 2 coins at random and flip it. What is the probability of a head?

2. Relevant equations

Partition Theorem P(A) = $$\sum$$ P(A | Bi)P(Bi)

P(B|A) = P(A|B)P(B)/P(A)

3. The attempt at a solution

1. I think this is 1/2, by adding up the probabilities and dividing by 3.

2. Would I get the probability of all heads by doing 1/4 x 1/2 x 3/4 = 3/32? As I assume that getting a head on one coin does not affect the other coins?

But then I'm not sure about the next part of (2). Is P(A1) = 1/2 by (1). The I tried to use the partition theorem for P(A2)

P(A2) = P(A2|A1)P(A1) + P(A2|not A1) P(not A1)
P(A1) = 1/2 = P(not A1) but apart from this I don't know.

3. P(3/4 coin | 2 heads) = P (2 heads| 3/4 coin)P(3/4 coin) / P(2 heads) = 3/4 x 3/4 x 1/3 / P(2 heads), but I don't know P(2 heads). Would drawing a probability tree work to find this?

4. I have not yet attempted this as I wanted to be clear on the rest of the question 1st.

Thanks :)

2. May 26, 2010

### hgfalling

1) Yes, fine.

2) Yes, the probability of getting all heads if you flip all three coins is 3/32. For that part, flipping the coins one after the others is equivalent to flipping them all at once, because you're not going to stop or distinguish one flip from the others.

Now for this second part.
Suppose that the Ajs were independent. Then the probability of getting three heads would be (1/2)3 = 1/8. But that doesn't match your answer above. So clearly they aren't independent.

To calculate p(A2 | A1), the probability that if flip 1 were a head, flip 2 will then be a head, try using the following questions:

What is the probability (given that the first coin was a head) that the first coin's inherent heads probability is 1/4? 1/2? 3/4? What do those probabilities sum to? If the first coin was the 1/4 coin, what is the probability of A2? Likewise for the other coins. If you have all these facts, then you should be able to figure out p(A2 | A1).

3) A tree would help you out in finding P(2 heads), but you can probably do it in your head. What's the probability that you get 2 heads in a row if your coin is 1/4? 1/2? 3/4? What are the a priori probabilities that you have these coins (ignoring your given outcomes)?

4) Just an extension of the last problem.

3. May 27, 2010

### Kate2010

Thanks, that's really helpful.

For question 2, the first part of your answer seems to make it unnecessary to calculate P(A2 | A1) but as I'm still finding it hard to do I shall persevere. Or do I still need to do it for the question?

I get P(1/4 | A1) = P(A1 | 1/4) P(1/4)/P(A1) = (1/4)(1/3)/(1/2) = 1/6. P(1/2 | A1) = 1/3, P(3/4 | A1) = 1/2. These sum to 1.

The next bit I'm more unsure of:
P(A2| 1/4 coin 1st) = (1/2 + 3/4) / 2 = 5/8
P(A2 | 1/2 coin 1st) = (1/4 + 3/4) / 2 = 1/2
P(A2 | 3/4 coin 1st) = (1/4 + 1/2) / 2 = 3/8
These don't sum to 1?

Then, again kind of guessing, P(A2 | A1) = P(A1|A2)P(A2)/P(A1) = P(A2) as P(A1 | A2) = P(A1)?

P(A2) = (1/6)(5/8) + (1/2)(3/8) + (1/2)(1/3) = 11/24

And for 3:

P(2 heads | 1/4 coin) = 1/16
P(2 heads | 1/2 coin) = 1/4
P(2 heads | 3/4 coin) = 9/16

We choose each coin with probability 1/3 as they look identical, so using the partition theorem we get P(2 heads) = 1/3 (1/16 + 1/4 + 9/16) = 1/3

Then my final answer for 3 is 9/16

Last edited: May 27, 2010
4. May 27, 2010

### hgfalling

Yup, good so far.

Yup, they don't sum to one, because they are conditional on all different things. Imagine there were 1000 coins which all impacted A2 by just a little bit away from 1/2. Those 1000 numbers wouldn't sum to 1 either. :>

OK, off track now.

Don't use Bayes' Theorem for this one.

Try P(A2 | A1) = P (A2 $\cap$ A1) / P(A1). You'll need to figure out how to calculate P (A2 $\cap$ A1), but you've already got all the pieces you need for that.

1/3(1/16 + 1/4 + 9/16) is not 1/3.

5. May 29, 2010

### Kate2010

Thank you very much for clearing some things up :)
I still don't totally understand but I think I'll just wait for my class.

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