0
$\begingroup$

An urn contains $N$ balls, $K$ balls of which are red. You toss a coin that has a probability $p$ of landing heads. If the coin lands heads, you pick a ball from the urn at random, without replacement. If the coin lands tails, you do nothing. You repeat the coin toss for $n$ times in total.

Let $x$ be the number of red balls that you picked after $n$ coin tosses. What is the expected value and probability distribution of $x$?

Without the coin toss condition (or $p=1$), $x$ follows a hypergeometric distribution. I'm having trouble with incorporating the coin toss however.

$\endgroup$

1 Answer 1

0
$\begingroup$

We can rework the scenario so all the coin-tosses happen before the drawing from the Urn. For the coin landing heads $k$ times the probability is just

$$P_{1,n}(k)={n \choose k}p^k(p-1)^{n-k}$$ (Binomial distribution). So if the probability to get $a$ red Balls after picking $b$ Balls from the Urn (without the coin) is $P_{2,b}(a)$, the probability (with cointoss) of picking $x$ red balls is

$$P(x)=\sum_{i=x}^n P_{1,n}(i) \cdot P_{2,i}(x) $$

. You have the possibility for each amount of heads-coin-tosses greater or equal to $x$ to pick $x$ red balls, so you just summ over all the possible amounts of coin tosses. That would be explicit (with the probability of a event in a hypergeometric distribution from wikipedia)

$$P(x) = \sum_{i=x}^n p^i (1-p)^{(n-i)} {n \choose i} \frac{{K \choose x} {N-K \choose i-x}} {N \choose i}$$

which does not seem to simplify further (at least at first glance).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .