# Coin tossing conditional probability with 5 coins

A box contains 5 coins and each has a different probability of showing heads. Let $p_1, p_2, p_3, p_4, p_5$ denote the probability of heads on each coin.

Suppose that $p_1 = 0, p_2 = 1/4, p_3 = 1/2, p_4 = 3/4$ and $p_5 = 1$. Select a coin at random and toss it. Suppose a head is obtained. Toss the coin again. What is the probability of another head. In other words, find $P(H_2 | H_1)$, where $H_j$ denotes "heads on toss j".

The progress I have been able to make is to compute the probability $P(H_1)$ using total probability theorem, but I don't know how to compute $P(H_2 \cap H_1)$.

• Let $A_1,A_2,\dots,A_5$ denote the probability of having selected each respective coin. $H_2\cap H_1=H_2\cap H_1 \cap (A_1\cup A_2\cup\dots\cup A_5) = (H_2\cap H_1\cap A_1)\cup (H_2\cap H_1\cap A_2)\cup\dots$. Do you see how to continue at this point? – JMoravitz Dec 5 '17 at 16:02

Let $E_i$ denote the event that coin $i$ is selected.
Then: $$P(H_2\cap H_1)=\sum_{i=1}^5P(H_2\cap H_1\mid E_i)P(E_i)$$