3.1.2 problem 2

(a)

Since the trials are independent, the probability that the first $k-1$ trials fail is ${\left(\frac{1}{2}\right)}^{k-1}$, and the probability that the $k$-th trial is successful is $\frac{1}{2}$. Thus, for $k\ge 1$,

$$P(X=k)={\left(\frac{1}{2}\right)}^{k-1}\ast \frac{1}{2}.$$

(b)

This problem reduces to part $a$ once a trial is performed. Whatever it’s outcome, we label it failure and proceed to perform more trials until the opposite outcome is observed. Thus, for $k\ge 2$,

$$P(X=k)={\left(\frac{1}{2}\right)}^{k-2}\ast \frac{1}{2}.$$