Coin chosen is two headed coin in this probability question

I have a probability question that reads:

Question:

A box has three coins. One has two heads, another two tails and the last is a fair coin. A coin is chosen at random, and comes up head. What is the probability that the coin chosen is a two headed coin.

My attempt:

P(two heads coin| given head) = P(two heads coin * given head)/P(given head)
= 1/3/2/3 = 1/2


Not sure whether this is correct?

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Hint: No, it isn't. Let $H \equiv$ obtaining heads, $A \equiv$ picking a two-headed coin, $B \equiv$ picking a two-tailed coin, and $C \equiv$ picking a fair coin. Then observe that the probability of obtaining a head is: \begin{align*} P(H) &= P(A)\cdot P(H \mid A) + P(B)\cdot P(H \mid B) + P(C)\cdot P(H \mid C) \\ &= \dfrac{1}{3} \cdot \dfrac{2}{2} + \dfrac{1}{3}\cdot \dfrac{0}{2} + \dfrac{1}{3}\cdot \dfrac{1}{2} \end{align*}

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so my numerator is correct and my denominator is wrong. am i right? –  lakesh Aug 9 '13 at 10:34
No, the numerator is wrong as well. Using my notation, you have calculated $P(A)$ instead of $P(A \text{ and } H)$. –  Adriano Aug 9 '13 at 10:35
I don't think you're answering the question being asked. You correctly find $P(H)=1/2$, but the question was about $P(A\mid H)$ which your hint seems to say nothing about at all. –  Henning Makholm Aug 9 '13 at 12:17
@HenningMakholm I was hoping that the OP knew the formula: $$P(A \mid H) = \dfrac{P(A) \cdot P(H \mid A)}{P(H)}$$ so I helped him out with the harder part, which was calculating the denominator. –  Adriano Aug 9 '13 at 16:32

For such a small number of options its easy to count them

The possible outcomes are:

tails or tails using the double tail coin
heads or tails using the fair coin

All these outcomes are equally likely. How many of these are heads and of those how many use the double headed coin?

$$Answer = \frac{2}{3}$$

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