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An urn contains 3 coins; 2 biased with $P(H) = p$, and the other is a two-headed coin. All the coins are tossed at once. If a coin is selected from those that came up heads, what is the probability that it was the two-headed coin?

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Hint: For each $n\in\{0,1,2,3\}$,

  • calculate the probability $P_n$ that there are exactly $n$ heads.

  • calculate the probability $Q_n$ that, if $n$ heads have appeared, then the two-headed coin is chosen

Then the probability that the two-headed coin is chosen overall is $$P_0Q_0+P_1Q_1+P_2Q_2+P_3Q_3.$$ (Do you understand why?)

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very helpful thank you – user71501 Apr 8 '13 at 2:22

HINT: The possible results are three heads, with probability $p^2$; two heads and a tail, with probability $2p(1-p)$; and one head and two tails, with probability $(1-p)^2$. That gives half of the information embodied in the following table:

$$\begin{array}{rccc} &3\text{ heads}&2\text{ heads}&1\text{ head}\\ \text{Probability}:&p^2&2p(1-p)&(1-p)^2\\ \text{Probability of choosing a head}:&?&?&? \end{array}$$

You should have no trouble completing the table. How can you use these six numbers to calculate the desired probability?

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