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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.

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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 ${11 \choose 7} \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4=0.637\%$.

In general, you use the binomial distribution formula:

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, $k=7$, $n=11$, $p=0.25$.

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You can think of this as a binomial distribution.

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 $

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