# Combined probability-flipping a coin, then rolling a die - Stuck on Part (c)

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 $$k \ge m \ge 1$$.

$$P(H)=\frac{1}{2}$$ 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 $$r \ge s \ge 1$$.

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: $$$$\sum_{n=1}^{7} \binom{n}{1} \frac{1}{2^{n+1}} \binom{8-n}{1} \frac{5^{7-n}}{6^{9-n}}$$$$