1
$\begingroup$

Okay so, like I said there are m cards, n of which are marked. k cards are selected at random, what is the probability of l being marked out of the k cards chosen.

So for the sample space cardinality I put n choose k.

And for the event cardinality I did m choose l {To account for the marked card combinations} multiplied by (n-m) choose (k-l) { to account for the unmarked card combinations.

So P(E) = |sample space| / |event|.

Is this correct? If not am I on the right track?

1 Answers 1

1

That is the correct track to follow.   You have a few literals juxtaposed.   You stated there were $m$ cards total, and $n$ marked among them.   You were selecting $k$ cards, and sought the probability that $l$ of them would be marked.   This is:

$$\mathsf P(E) = \dfrac{\dbinom n l \dbinom{m-n}{k-l}}{\dbinom mk}\qquad=\dfrac{{^{n}\mathrm C_{l}}~{^{m-n}\mathrm C_{k-l}}}{{^{m}\mathrm C_{k}}} \quad=\dfrac{\lvert\text{event}\rvert}{\lvert\text{sample space}\rvert}$$

That is all.

  • 0
    Thanks, I think I got things mixed up in the cardinality of the event but I meant to write what you wrote. – 2017-02-08