# Probability Question involving three coins.

Suppose we have three coins. One has probability 0.25 of landing heads, one has probability 0.5 of landing heads, and one has probability 0.6 of landing heads. Suppose we choose one of the coins at random and toss it.

(i) Find the probability that the coin toss shows heads. (ii) Given that the coin toss showed heads, fi nd the probabilities that each of the three coins was the one chosen.

Do a probability tree:

First branch, choosing (randomly, of course), a coin

Second branch, the probability of each coin to show heads.

Thus, for the first one we get

$$\;P(H)=\frac13\left(\frac14+\frac12+\frac35\right)$$

For the second one: supppose you want to find out the prob. the first coin was chosen, then

$$P(C_1/H)=\frac{P(C_1\cap H)}{P(H)}$$

and

$$P(C_1\cap H)=\frac13\cdot\frac14$$

$$P(H|C_1)= \frac{1}{4}, P(H|C_2)= \frac{1}{2}, P(H|C_3)= \frac{3}{5}$$

$$P(C_i) = \frac{1}{3}$$

See other answer for $P(H)$

then given

$$P(H|C_i) = \frac{P(C_i\cap H)}{P(C_i)}$$

and

$$P(C_i|H) = \frac{P(C_i\cap H)}{P(H)}$$

so

$$P(C_i|H) = \frac{P(H|C_i)P(C_i)}{P(H)}$$