Lisa shoots at a target. The probability of a hit in each shot is 1 /2. Given
a hit, the probability of a bull’s-eye is p. She shoots until she misses the
target. Let X be the total number of bull’s-eyes Lisa has obtained when
she has finished shooting; find its distribution.
It seems elementary, but i think this is pretty diffcult.
Best Answer
Let $q_i = P(X=i), i=0,1,2,\ldots$. Assume $i\gt 0$ and consider the outcome of the first dart. To obtain $i$ bullseyes it must either be a bullseye, in which case we need a further $i-1$ bullseyes, or a non-bullseye hit, in which case we still need $i$ bullseyes. Therefore, using the law of total probability,
\begin{eqnarray*} q_i &=& \dfrac{p}{2}q_{i-1} + \dfrac{1-p}{2}q_i \\ \therefore (p+1)q_i &=& pq_{i-1} \\ q_i &=& \dfrac{p}{p+1}q_{i-1} \\ q_i &=& \left(\dfrac{p}{p+1}\right)^iq_{0}\qquad\text{solving the simple recurrence relation.} \\ \end{eqnarray*}
We now require $q_0$. To obtain $0$ bullseyes we need $0$ or more non-bullseye hits followed by a miss. Therefore,
\begin{eqnarray*} q_0 &=& \dfrac{1}{2} \sum_{n=0}^\infty{\left(\dfrac{1-p}{2}\right)^n} \\ &=& \dfrac{1}{2}\cdot \dfrac{1}{1-\frac{1-p}{2}} \qquad\text{using geometric series formula} \\ &=& \dfrac{1}{p+1}. \end{eqnarray*}
Therefore, $$q_i = \dfrac{p^i}{\left(p+1\right)^{i+1}}, \quad i=0,1,2,\ldots.$$