# The probability of hitting a bulleye

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.

• Please edit the question to include a description of what you've tried and where you're stuck. This will help people give useful answers at an appropriate level. In general, posting verbatim homework exercises is discouraged, especially when they come across as issuing a command. – Barry Cipra Nov 28 '14 at 17:21
• @BarryCipra - I have a lot of wrong solution, they are all different and all wrong... – Victor Nov 28 '14 at 17:37
• Victor, that's great (I come up with wrong answers all the time!), please post them. If you describe what you've done, people will be able to steer you in the right direction, without giving explanations that are either too elementary or too advanced. – Barry Cipra Nov 28 '14 at 18:16

## 1 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, $$q_i = \dfrac{p^i}{\left(p+1\right)^{i+1}}, \quad i=0,1,2,\ldots.$$