Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A population has $G$ good and $B$ bad elements, $G+B=N$. Elements are drawn one by one at random without replacement. Suppose the first good element appears on draw number $X$. Find a simple formula, not involving any summation from $1$ to $N$, for $E(X)$. Hint: Write $X-1$ as sum of $B$ indicators.

Ok, so we know in the first $X-1$ draws we only get bad elements. Let $I_j$ be $1$ if $j$th draw gives bad ball and $0$ otherwise. $X-1=I_1+I_2+ \cdots +I_{x-1}$, all of which have value $1$. $E(I_1)= \cdots=E(I_n)=B/N=(N-G)/N$. Now we can write $E(X)=E((X-1)+1)$ $=E(X-1)+E(1)=E(X-1)+1.$ I'm not exactly sure on the formula for $E(X-1)$. Hopefully, I'm on the right track. Thanks!

share|improve this question
@MikeSpivey +1 Of course! Serves me right for thinking through the problem too quickly! I'll remove it to avoid confusion. –  Mike Wierzbicki Oct 20 '11 at 5:12

1 Answer 1

up vote 4 down vote accepted

Your current attempt doesn't express $X-1$ as the sum of $B$ indicators. Instead, number the bad balls $1$ through $B$. Try $I_j = 1$ if bad ball $j$ is chosen before any of the good balls and $0$ otherwise. Then you have $X-1 = \sum_{j=1}^B I_j$. Now, can you finish off the problem by finding $E[I_j]$?

(Added, for completeness): We have $E[I_j] = \frac{1}{G+1}$, the probability that bad ball $j$ is chosen before any of the good balls. Thus $$E[X] = \frac{B}{G+1} + 1 = \frac{B + G + 1}{G+1} = \frac{N+1}{G+1}.$$

share|improve this answer
@caligirl11: No, $E(I_j)$ is the probability that bad ball $j$ is chosen before any of the good balls. This is not $B/N$. Instead, you have bad ball $j$ vs $G$ good balls; what's the probability that, out of those $G+1$ balls, the bad one is chosen first? –  Mike Spivey Oct 20 '11 at 5:30
Ok thank you! I got it now E($I_j$)=1/(G+1). –  caligirl11 Oct 20 '11 at 5:35
@caligirl11: Yep, nice job! –  Mike Spivey Oct 20 '11 at 5:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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