# Conditional probability problem

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• March 6th 2011, 08:46 AM
smallfry
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?
• March 6th 2011, 08:55 AM
TheEmptySet
Quote:

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

$\displaystyle P(A|B)=\frac{P(B|A)P(A)}{P(B)}=...$
• March 6th 2011, 07:35 PM
smallfry
Yes, I know Bayes' theorem. So the answer IS 1/2 right?
• March 6th 2011, 07:46 PM
TheEmptySet
did you plug in the numbers? Please show your work!
• March 6th 2011, 09:20 PM
matheagle
I got 1/2
• March 7th 2011, 08:48 AM
smallfry
Ok I did it, and the answer is 1/2. Thanks guys :) I figured it out with a probability tree.
• March 7th 2011, 11:44 AM
matheagle
It's

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