4.3.15 problem 56

(a)

Let $I_j$ be the indicator variable for shots $j$ to $j+6$ being successful. The total number of succesful, consecutive, $7$ shots is $X={\sum }_{i=1}^{n-6}{I}_j$. Then, $$\text{E}\left(X\right)=\sum _{i=1}^{n-6}\text{E}\left(I_j\right)=\sum _{i=1}^{n-6}P(I_j=1)=\sum _{i=1}^{n-6}{p}^7=(n-6)p^7$$

(b)

Thinking of each block of $7$ shots as a single trial with probability $p^7$ of success, let $Y\sim \text{Geom}\left(p^7\right)$ be the number of failed $7$-block shots taken until the first succesful $7$-shot block. Then, $$\text{E}\left(X\right)=7(1+\text{E}(Y\left)\right)=7+\frac{7-7p^7}{p^7}=\frac{7p^7+7-7p^7}{p^7}=\frac{7}{p^7}$$

Note that it is possible that a consecutive sequence of 7 shots could happen "between" blocks - for example, this way of solving the problem does not consider the scenario where shots 2 to 8 are made. Therefore, the above calculation is a "worst case scenario" that assumes the consecutive 7 made shots must always happen in the last possible block - the actual number of blocks (and therefore shots) taken to make 7 consecutive shots is strictly less than or equal to the above calculated expectation.