0
$\begingroup$

Embarrassingly simple question here and apologies if this is all over the web, I was lacking some of the vocab I needed to Google it.

Let's say I have a set of samples with varying probabilities for the same event (we'll use a baseball analogy here, as that seems to be common). I have nine batters who have batting averages of: [0.250, 0.340, 0.210, 0.220, 0.300, 0.230, 0.290, 0.180, 0.110].

What I want to know is if all of these batters go up once, what is the probability of 9 hits, what is the probability of 8 hits, 7 hits, etc...

How do I calculate this more generally? In my case, I have a mean and I have a bunch of probabilities, but I can't figure out how to create some kind of distribution from that.

Essentially what I'm wondering is how to calculate a binomial distribution when the different trials have different probabilities of success.

$\endgroup$
0
$\begingroup$

Unfortunately, there's no good way of calculating this explicitly. You'll need to sum over all possible ways to get $H$ hits. To find $P(H)$ abstractly, you are solving for the coefficient of $x^H$ in $\prod_{i=1}^n[p_ix+(1-p_i)]$, or equivalently if $[n]:=\{1,2,\cdots,n\}$,

$$P(H)=\sum_{S\subset [n], |S|=H}\left[\prod_{p\in S} p\right]\left[\prod_{q\in S^c}(1-q)\right]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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