Throwing a biased coin $m$ times

Question

We throw a biased coin (probability for heads is $\frac{1}{n}$) $m$ times. It is known that in first two throws we got at least one tail. Calculate the probability that we got $k$ heads.

So, let's define:

$A - $ at least one tail in two first throws

$B \ - $ $k$ heads

I need to calculate $\mathbb P(B|A)=\frac{\mathbb P(B\cap A)}{\mathbb P(A)}$, so:

$\mathbb P(A)= 1-\mathbb P($first are two heads$ )=1-(\frac{1}{n})^2$

$\mathbb P(B|A)=\mathbb P(B\cap A)= \frac{1}{n}\cdot\frac{n-1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{n-1}{n}{m-2 \choose k}(\frac{1}{n})^k(1-\frac{1}{n})^{m-k-2}$

Am I correct?

Answer 1

No, you have $P(A)=P(\text{there are $k$ heads})$ and this is not true. I would argue that $P(A)=1-P(\text{first two are heads})=1-(\frac{1}{n})^2.$

So you will hve $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A\cap B)}{1-(\frac{1}{n})^2},$$ your computation of $P(A\cap B)$ seems correct to me.

Answer 2

you can use the Total probability Theorem: it results to me

$$\mathbb{P}[B]=\Bigg(\frac{1}{n}\Bigg)^k\Bigg(1-\frac{1}{n}\Bigg)^{m-k}\Bigg[\binom{m-2}{k}+2\binom{m-2}{k-1}\Bigg]$$

In fact, the number of H is always $k$ and the number of T always $(m-k)$. What changes is only the way to combine the k H in $m$ tosses.

These combinations are

• $\binom{m-2}{k}$ or $\binom{m-2}{k-1}$, it depends if you have zero or one H in the first two tosses, respectively

• The binomial coefficient $\binom{m-2}{k-1}$ has to be multiplied by 2 because we can have $TH$ or $HT$ in the first two tosses