How to find the p.m.f when we toss $n$ coins in $5$ rounds?

Question

I'm trying to solve this problem:

My attempt:

Let $X_i$ be the number of heads from the $i$-th round, $i=1,..,4$. Then $\mathbb P(X = k) = \sum_{k_1 \le k_2 \le \ldots \le k}\mathbb P (X_1 = k_1,\ldots, X_4 = k_4, X = k)$.

We have

• $\mathbb P(X_1 = k_1) = {n \choose k_1} p^{k_1} (1-p)^{n-k_1}$.

• $\mathbb P(X_2 = k_2) = {k_1 \choose k_2} p^{k_2} (1-p)^{k_1-k_2}$.

• $\mathbb P(X_3 = k_3) = {k_2 \choose k_3} p^{k_3} (1-p)^{k_2-k_3}$.

• $\mathbb P(X_4 = k_4) = {k_3 \choose k_4} p^{k_4} (1-p)^{k_3-k_4}$.

• $\mathbb P(X = k) = {k_4 \choose k} p^{k} (1-p)^{k_4-k}$.

Because the number of heads from the $(i+1)$-th round depends on that of from the $i$-th round. There are so many cases that I can't handle. Maybe I'm in the wrong approach.

Could you please shed me some light?

Answer 1

Hint: A coin remains and shows heads in the last round if and only if it showed heads in every round. So the number of heads in the last round is binomial with $p=$ the probability that a coin shows heads all 5 rounds.

Answer 2

Let $M$ be the number of coins that are tossed at the fifth round. To reach the fifth round, you have to obtain heads for the first $4$ times.

\begin{align} Pr(X_5=k) &=\sum_{m=k}^nPr(X_5=k|M=m)Pr(M=m)\\ &=\sum_{m=k}^n \binom{m}{k}p^k(1-p)^{m-k}\binom{n}{m}(p^4)^m (1-p^4)^{n-m}\\ &=\sum_{m=k}^n\binom{n}{n-m,k,m-k} p^{k+4m}(1-p)^{m-k}(1-p^4)^{n-m} \end{align}