Suppose we have a box with $n$ red and $m$ black balls. You draw balls
sequentially one at a time without replacement. Let $X$ be the number
of red balls removed before the first black ball is chosen. Please,
compute $E(X)$
Try:
I was thinking perhaps to use indicator functions since trying to find the probability mass function is probably complicated. But, isn't it hypergeometric?
Now, if we use indicators, we can write $X_i$ to be $i$ if first $i-1$ balls selected are red and the ith ball is black and $0$ otherwise. Thus, $X = \sum X_i$, Now
$$ E(X_i) = i P(X=i) = i \cdot \frac{m {n \choose i – 1}}{{n+m \choose i}}
$$
So,
$$ E(X) = \sum_{i=1}^{\infty} E(X_i) = \sum \frac{ im {n \choose i-1 } }{ {m+n \choose i} } $$
but, this sum would be hard to evaluate. Is my approach correct?
Best Answer
Consider a particular red ball: this has a probability $\frac{1}{1+m}$ of coming before all $m$ black balls, since there are $(m+1)!$ equally likely ways of ordering these particular balls and $m!$ of those have the red ball first
So by linearity of expectation, the expected number of red balls coming before all $m$ black balls is $n$ times this, i.e. $\dfrac{n}{1+m}$