# Biased coin toss problem - Understanding a problem correctly

Suppose you repeatedly toss a biased coin. Each time you toss the coin it lands $heads$ with probability $\frac13$ and $tails$ with probability $\frac23$.

Let $X$ be the random variable corresponding to the number of times you toss the coin until you’ve seen two heads. For example, when the sequence is of coin tosses is $$HT T HHT$$ . . . then X = 4.

Note that $$X = X_1 +X_2$$ where $X_1$ is the number of times you toss the coin until you see the first heads and $X_2$ is the number of additional tosses until you see the second heads.

What’s the value of:

$$P(X = 2)$$

I assume that this is a geometrical distribution. Because, there are $X_1$ and $X_2$, I would get the result of $X_1$ and add to $X_2$ However, when I plug the numbers in, I get something different from the answer which is $\frac{1}{9}$

Am I understanding this problem correctly?

It's much simpler. $X=2$ means you got 2 heads in 2 throws, i.e. you threw $HH$, which has the probability of $(1/3)^2 = 1/9$.