0
$\begingroup$

My text defines the equation for the probability of getting exactly $k$ independent successes in an experiment of $n$ trials, with $p$ being the probability of the success event and $q$ being the probability of failure ($1-p$) is:

$b(k; n, p) = {n \choose k}(p)^k(q)^{n-k}$

The part to the right of the binomial coefficient makes sense to me intuitively. What I don't understand is why we're multiplying it all by $n \choose k$. Could someone please help clarify this for me?

$\endgroup$
1
  • 1
    $\begingroup$ Okay, my question may seem a bit stupid. I just realized that ${n \choose k}$ is the number of ways to have such an outcome, so if we know the probability of each outcome, then the probability of the outcome is # of outcomes * prob each outcome. $\endgroup$
    – AleksandrH
    Jul 14 '17 at 17:49
4
$\begingroup$

$p^k$ represents the probability of exactly $k$ consecutive successful trials

$q^{n-k}$ represents the probability of exactly $n-k$ consecutive failed trials

However, we don't care about the order. We don't need to succeed $k$ times in a row and then fail $n-k$ times in a row. Any ordering of successes and failures with $k$ successes and $n-k$ failures is valid, and there are $n \choose k$ of these orderings.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.