I don't know how to compute the probability in the problem below.
$10$ people take an exam, $6$ male and $4$ female.
The probability for a male to pass is $0.5$, while the probability for a female to pass is $0.4$.
What is the probability that at least one female passes, and what is the probability that exactly $1$ male and $1$ female passes?
I think for the first part you use a binomial distribution with $p = 0.5, q =0.1$.
And result is $$1-\sum\limits_{k = 1}^5 {\left( \begin{array}{l} 5\\ k \end{array} \right){p^k}{q^{5 - k}}} $$
But I'm not sure if this is right or not.
Which the second part, I really doesn't know how to compute.
Firstly, in a binomial distribution, you only have the parameters $n$ and $p$, and $q$ is defined as $1-p$. So $p=0,5$, $q=0.1$ makes no sense.
The probability that at least 1 female passes is 1 minus the probability that no woman passes which equals 1 minus the probability that all woman do not pass.
The probability that one woman does not pass is $1 - 0.4 = 0.6$, so.... (apply independence).
The second probability is the product (because of indepence) of the probability that exactly one male passes and the probability that exactly one woman passes.
Both of these probabilities are indeed binomially distributed. The former with $p=0.5$ (and what is $n$?) the latter with $p = 0.4$ (and $n = $?). Now apply your binomial formula and multiply these two probabilities.