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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