# Thread: Conditional probability problem

1. ## Conditional probability problem

A box contains 4 fair coins, and one two headed coin. A man selects a coin from the box, tosses it twice, and gets two heads. What is the probability that he selected the two headed coin?

Possible answers: 1/5, 1/3, 1/2, 4/5, 2/3.

My answer was 1/2. Can anyone disprove this?

2. Originally Posted by smallfry A box contains 4 fair coins, and one two headed coin. A man selects a coin from the box, tosses it twice, and gets two heads. What is the probability that he selected the two headed coin?

Possible answers: 1/5, 1/3, 1/2, 4/5, 2/3.

My answer was 1/2. Can anyone disprove this?
I assume that you know Bayes' theorem

Let A be the event that you draw the two headed coin and
Let B be the event that you get two heads then

$\displaystyle \displaystyle P(A|B)=\frac{P(B|A)P(A)}{P(B)}=...$

3. Yes, I know Bayes' theorem. So the answer IS 1/2 right?

4. did you plug in the numbers? Please show your work!

5. I got 1/2

6. Ok I did it, and the answer is 1/2. Thanks guys I figured it out with a probability tree.

7. It's

$\displaystyle {1/5\over (1)(1/5)+(1/4)(4/5)}$

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