How to find expectation of geometric distribution? On average 1 in 8 people in a particular community is left-handed and the rest are right-handed. A sample of people is chosen at random, one by one, until a left-handed person is obtained. Find the probability that the number of people in the sample is less than 12. Another sample is chosen at random, one by one, until at least one left-handed and at least one right-handed person have been obtained. The number of people in this sample is denoted by N, so that the sample consists of either : (N − 1) left-handed people followed by one right-handed person or : (N − 1) right-handed people followed by one left-handed person. Obtain P(N = r), for r ≥ 2, and show that E(N) = 57/7. Why E(N) = 8 + 8/7 - 1? Why need to -1?

You do not consider the case $r=1$. If you consider the case $r=1$ the calculation would be

$\sum _{r=1}^{\infty} r \left( \frac{1}{8}\right)\cdot \left( \frac{7}{8}\right)^{r-1}+\sum _{r=1}^{\infty } r \left( \frac{7}{8}\right)*\left( \frac{1}{8}\right) ^{r-1}=8+\frac{8}{7}$

Now you subtract in both cases the summand when $r=1$.

$E(N)=\sum _{r=1}^{\infty} \left[ r \left( \frac{1}{8}\right)\cdot \left( \frac{7}{8}\right)^{r-1} \right]-\frac{1}{8}+\sum _{r=1}^{\infty } \left[ r \left( \frac{7}{8}\right)*\left( \frac{1}{8}\right) ^{r-1} \right] -\frac{7}{8}=8+\frac{8}{7}-\left(\frac{1}{8}+\frac{7}{8}\right) =8+\frac{8}{7}-1$

The condition is, that at least one left-handed and at least one right-handed person have to be obtained. Thus r cannot be 1.