There are $2^4=16$ total possible outcomes of which only one outcome gives rise to all heads.

Thus the probability that all coins land heads is $1/16$.

Solution (2)

Consider the event that the first coin is heads.

In this case, there are total $2^3=8$ possible outcomes for the rest of coins (2nd, 3rd, and 4th).

Hence, the probability that all coins land heads given that the first coin is heads is $1/8$.

Solution (3)

Let $H$ be the event that all coins land heads. Let $F$ be the event that at least one coin lands heads. Then the required conditional probability is given by
\begin{align*}
P(H \mid F) &= \frac{P(H \cap F)}{P(F)}.
\end{align*}
The complement $F^c$ of $F$ is the event that all lands tails whose probability $P(F^c)$ is $1/16$ just like part (a). Hence
\[P(F) = 1 – P(F^c) = 1 – \frac{1}{16} = \frac{15}{16}.\]

It follows that
\begin{align*}
P(H \mid F) &= \frac{P(H \cap F)}{P(F)}\\[6pt]
&= \frac{P(H)}{P(F)}\\[6pt]
&= \frac{1/16}{15/16}\\[6pt]
&= \frac{1}{15}.
\end{align*}

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