# how many times will get exactly two heads if I toss 2 not-fair coins (chances of getting a head is 0.68) for 100 times?

I want to calculate how many times I will get exactly two heads if I toss 2 not-fair coins (chances of getting a head is 0.68) for 100 times?

Can somebody gives me a formula that I can generalize this question? (n not-fair coins where is p the chance of head for m times)

This is just a binomial distribution where “success” is getting two heads. Let’s calculate the probability of such an event.

$$P(HH) = P(H)P(H)=0.4624$$

So this is our “p” in the binomial distribution. The other parameter in the binomial distribution is $$n$$, the number of attempts, which is $$100$$. Now calculate the probability of getting exactly two “successes”, which is exactly what a binomial tells you.

$$f(x)=\binom{100}{x} 0.4624^x (1-0.4624)^{100-x}$$

The trick to generalizing this is recognizing the “p” parameter in the binomial distribution. It’s always the probability of getting the number of flips.

EDIT

Binomial PMF in general:

$$f(x\vert n,p)=\binom{n}{x}p^x(1-p)^{n-x}$$