25% chance occurring 7/11 times.

If an event has a probability of happening 25% of the time. How do you calculate the chances of this happening 7 out of 11 times.

If A is 25% and B is 75%. What is the probability of A occurring 7 times out of a possible 11 attempts.

I have found similar threads but unable to get a particular formula / answer.

Thanks.

There are $$11 \choose 7$$ options for which $$7$$ attempts have been successful, each with $$\left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4$$ change of occurring. So the probability of $$A$$ occurring $$7$$ out of $$11$$ times is $${11 \choose 7} \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4=0.637\%$$.
If the probability of an event is $$p$$, then it happens exactly $$k$$ times out of $$n$$ with a probability of $$\binom{n}{k}p^k(1-p)^{n-k}$$
In your case, $$k=7$$, $$n=11$$, $$p=0.25$$.
The probability of event $$A$$ occurring $$7$$ times and $$B\ 4$$ times is $$p = \binom{11}{7} (0.25)^7 (0.75)^4 \approx 0.00637$$