# Which kind of probability should use in here?

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.

• I have tried to edit your question to make it easier to understand what you are asking. I had to make some guesses as to what the question was meant to be, please let me know if I made any mistakes. Jun 1 '14 at 10:47
• Thank you very much. But i want to add one request. Jun 1 '14 at 10:57

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.

• No, here have 10 persons, and you don't know who pass, who fail. 0,4 is the probability of 1 pass. Jun 1 '14 at 11:45
• @phuong There is no way to describe this as 1 binomial distribution, because for those the chance of success should always be equal for all subexperiments. Which is not the case here, so you should treat it as two separate, independent binomial distributions with different success probabilities (and different $n$ in this case). Jun 1 '14 at 11:49
• Ok. May be you right in first question. I apply binomial is not correct. But in the second part, how you can apply binomial? Please give me some clearly advice? Jun 1 '14 at 11:59
• I think I spelled it out pretty clearly. I'll let others give more hints, if they like... Maybe check up on first principles like independence and the basic applications of binomial distributions? Jun 1 '14 at 12:01
• As you wrote, I think result is 1-(0.6)^5. is it right? But in the second, as you write may be the result is sum nCk (p_1)^k.(1-p_1)^k . sum nCk(p_2)^k.(1-p_2)^k. Is it right? But I think the right answer must be sum nc2(p_1).(p_2).(1-p_1-p_2)^(n-2). where p_1=0.5 and p_2=0.4. Pls help me again. Jun 1 '14 at 12:16