A television quiz game operates as follows. In the first part of the game, a contestant is asked a series of difficult questions; the probability of answering any question correctly is $p$, independently of all other questions. This first part of the game ends as soon as the contestant answers a question incorrectly. Denote by $N$ the total number of questions answered correctly in this first part of the game (excluding the final incorrect one). In the second part of the game, the contestant is asked a series of easy questions; here, the probability of answering any question correctly is $r$, independently of all other questions. The total number of questions asked in this second part of the game is $N$ (i.e. if $N = n$ in the first part of the game, the contestant answers $n$ questions in the second part). The contestant's final score, $y$ say, is equal to the number of correct answers in the second part of the game.
(a) Find the probability mass function of $N$. Also state the conditional distribution of $Y$ was given $N = n$, including the value(s) of its parameter(s).
(b) Use the Law of Total Probability to find the probability mass function of $Y$. Show that this pmf has the same form as that of $N$, but with $p$ replaced by another quantity which you should define.
My answer:
(a) $N$ is obviously a geometric distribution, and $Y$ binomial.
$P(N=n)=p^n(1-p)$, $P(Y=k|N=n)={n \choose k}r^k(1-r)^{n-k}$
(b) Use Law of total probability,
$P(Y=k)=\sum_{n=k}^\infty {n \choose k}r^k(1-r)^{n-k}p^n(1-p)$
Could you please check if my answer is correct. And then help me with the last step. I have no idea how to simplify this series into a geometric distribution-like form. It gives me a hint to use binomial expansion, but I don't know how to use it with $p^n$ presents.
You are completely right. The final proof is the following
$$\mathbb{P}[Y=k]=(pr)^k(1-p)\sum_{n-k=0}^{\infty}\binom{n}{n-k}[p(1-r)]^{n-k}=$$
$$=(pr)^k(1-p)\sum_{n-k=0}^{\infty}\binom{(n-k)+k}{n-k}[p(1-r)]^{n-k}=$$
$$=(pr)^k(1-p)\underbrace{\sum_{i=0}^{\infty}\binom{i+k}{i}[p(1-r)]^i}_{\text{It's a known series}}=$$
$$=(pr)^k(1-p)\frac{1}{[1-p(1-r)]^{k+1}}=$$
$$=\left[\frac{pr}{1-p(1-r)}\right]^k\cdot\frac{1-p}{1-p(1-r)}$$
Thus
$$Y\sim Geo\left[\frac{1-p}{1-p(1-r)}\right]$$
For the list of math series see here; the one I used is in the list of Binomial Coefficients [3]