Part (a). Using the negative binomial distribution, write an expression for the probability that exactly $k$ flips of a coin are needed to get $m$ heads, where $k \ge m \ge 1$.
My answer:
$P(H)=\frac{1}{2}$ and $P(T)=1-P(H)=\frac{1}{2}$
$\therefore P(X=m)=\binom{k-1}{m-1} (\frac{1}{2})^m (\frac{1}{2})^{k-m}$
$(\frac{1}{2})^m (\frac{1}{2})^{k-m}=(\frac{1}{2})^{m+k-m}=(\frac{1}{2})^k$
$\therefore P(X=m)=\binom{k-1}{m-1} (\frac{1}{2^k})$
**Part (b).**Find an expression for the probability that $r$ rolls of a die are needed to get s sixes, where $r \ge s \ge 1$.
My answer:
Probability of success, $P(6)= \frac{1}{6}$, and probability of failure$=1-\frac{1}{6}= \frac{5}{6}$
$ \therefore P(Y=s)=\binom{r-1}{s-1} (\frac{1}{6})^s (\frac{5}{6})^{r-s}$
$ (\frac{1}{6})^s (\frac{5}{6})^{r-s}=\frac{5^{r-s}}{6^{s+r-s}}=\frac{5^{r-s}}{6^r}$
$\therefore P(Y=s)=\binom{r-1}{s-1}\frac{5^{r-s}}{6^r}$
This is the part I'm stuck on:
**Part (c).**The coin is flipped until 2 heads are obtained, and then the die rolled until 2 sixes are obtained. Counting each flip as a trial, and each roll as a trial, show that the probability that exactly 10 trials are needed to obtain 2 heads and 2 sixes is: \begin{equation} \sum_{n=1}^{7} \binom{n}{1} \frac{1}{2^{n+1}} \binom{8-n}{1} \frac{5^{7-n}}{6^{9-n}} \end{equation}
My Answer below.
