# Finding mean and variance of number of tosses needed to get exactly 2 heads

A coin with probability of getting head $$0.6$$ is tossed repeatedly till two heads appear. Let $$X$$ be the number of tosses needed to get exactly 2 heads. Describe the sample space. Find the mean and variance of $$X$$.

What I've tried so far:

Let $$P(H)=p$$ and $$P(T)=q$$ such that $$p+q=1$$. $$\displaystyle P(X=n)=\binom{n}{2}p^2q^{n-2}\tag*{}$$ Is this even correct because $$\sum\limits_{n} P(X=n)=\frac{p^2}{(1-q)^3}>1$$?

I am absolute beginner. Any help would be appreciated.

Update: Reading about negative binomial distribution, I realised the point I was missing. It's that we have to necessarily get the second head on the last toss. The correct P.M.F. would be: $$\displaystyle\boxed{ P(X=n)=\binom{n-1}{1}p^2q^{n-2}}\tag*{}$$

• Take a look at the negative binomial distribution. Feb 15, 2023 at 17:14
• @jblood94 Thank you! I've understood my mistakes. Everything now makes sense. $\displaystyle P(X=n)=\binom{n-1}{1}p^2q^{n-2}\tag*{}$ Also, $\sum\limits_{n\geqslant 2}P(X=n)=\frac{p^2}{(1-q)^2}=1\tag*{}$ Feb 15, 2023 at 19:45
• I posted my answer as you were posting your update. Feb 15, 2023 at 20:18

I see you learned that your formula was incorrect because you know the last (nth) trial must be a success (heads). That means you have $$n-1$$ trials to get $$r-1$$ successes and $$n-r$$ failures. $$P(n;r) = p\binom{n-1}{r-1}p^{r-1}q^{n-r}$$ In your case, $$r=2$$, so $$P(n;2) = p\binom{n-1}{1}pq^{n-2}$$
The First Success distribution is the number of trials $$Y$$ until the first success. $$P(Y=k) = q^{k-1}p$$ $$E(Y) = \sum_{k=1}^{\infty}kq^{k-1}p = \frac{1}{p}$$ $$Var(Y) = \frac{q}{p^2}$$ To get the variance, first get the second moment: $$E(Y^2) = \sum_{k=1}^{\infty}k^2q^{k-1}p = \frac{1+q}{p^2}$$ Then use $$Var(Y) = E(Y^2) -(EY)^2 =\frac{1+q}{p^2} - \frac{1}{p^2} = \frac{q}{p^2}$$ Consider the number of trials until success $$r$$. This is the sum of $$r$$ independent variables $$Y_1, Y_2, ..., Y_r$$ $$E(Y_1 + ... + Y_r) = E(Y_1) + ... +E(Y_r) = r \cdot \frac{1}{p} = \frac{r}{p}$$ Since they're independent, you can add their variances. $$Var(Y_1 + ... + Y_r) = r \cdot \frac{q}{p^2}$$ In your case, $$r=2, p = 0.6, q = 0.4$$. $$E(Y_1 + Y_2) = \frac{2}{0.6} = \frac{10}{3} = 3.333$$ $$Var(Y_1 + Y_2) = 2 \cdot \frac{0.4}{0.6^2} = 2 \cdot \frac{10}{9} = 2.222$$
• I had managed to get the mean & the variance using $E(X)$ and $E(X^2)$ but it was quite lengthy... This is much easier! Thank you for this interesting approach. Feb 15, 2023 at 22:29