Two fair coins are tossed. Given that at least one of them lands heads, what is the conditional probability that the first coin lands heads?

We give two proofs. The first one uses Bays’ theorem and the second one simply uses the definition of the conditional probability.

Solution 1.

Let $E$ be the event that the first coin lands heads. Let $F$ be the event that at least one of two coins lands heads.
By Bayes’ rule, the required probability can be calculated by the formula:
\[P(E \mid F) = \frac{P(E) \cdot P(F \mid E)}{P(F)}.\] We know $P(E)=1/2$. When the first coin lands heads, then of course at least one of two coins lands heads. So, we have $P(F \mid E) = 1$.

Since $F = \{\text{hh}, \text{ht}, \text{th}\}$, we see that $P(F) = 3/4$.
Plugging these values into the formula, we obtain
\[P(E \mid F) = \frac{\frac{1}{2}\cdot 1}{\frac{3}{4}} = \frac{2}{3}.\]

Solution 2.

Let $E$ be the event that the first coin lands heads. Let $F$ be the event that at least one of two coins lands heads.
Then we have $F = \{\text{hh}, \text{ht}, \text{th}\}$.
Also, we have
\[E \cap F = \{\text{hh}, \text{ht}\}.\] Thus, the required probability is given by
\begin{align*}
P(E \mid F) &= \frac{|E \cap F|}{|F|}\\
&= \frac{2}{3}.
\end{align*}

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