= Solution
For events with $P(B)>0$, the <conditional probability> is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives <Bayes' theorem>:
$$
P(B\mid A)=\frac{P(A\cap B)}{P(A)}=P(A\mid B)\frac{P(B)}{P(A)}.
$$
Conditionally on $N=n$, independent fair tosses give a <binomial distribution> for the head count, so $P(H=1\mid N=n)=n2^{-n}$. The joint <probability mass function> is consequently
$$
P(N=n,H=1)=n4^{-n}.
$$
Use the <law of total probability> and the differentiated <geometric series>, $\sum_{n\geq1}nr^n=r/(1-r)^2$, to obtain $P(H=1)=4/9$. The <posterior coin count after exactly one head> is therefore
$$
\boxed{P(N=n\mid H=1)=\frac{9n}{4^{n+1}},\qquad n\geq1}.
$$
These <probabilities> sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.
Back to article page
