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Balls are taken one by one out of an urn containing $w$ white and $b$ black balls until the first white ball is drawn. Prove that the expectation of the number of black balls preceding the first white ball is $\frac {b}{w+1}$

Attempt: Let $X_i$ be the random variable that denotes the number of black balls that are drawn at the $i_{th}$ step before a white ball is drawn.

Then, the total number of such balls $ X= X_1 + \cdots+X_n \implies E(X)=\sum E(X_i).$

$E(X_i)= 1 \cdot \dfrac {^bC_r}{^{b+w}Cr}\cdot \dfrac {^wC_1}{^{b+w-r}C_1}$

Thus, $\sum E(X_i) = \sum_{i=1}^{b} ~ 1 \cdot \dfrac {^bC_r}{^{b+w}Cr}\cdot \dfrac {^wC_1}{^{b+w-r}C_1}$

Could someone please tell me if I attempted this correctly? Because I get a very complicated answer in the end after evaluating the above.

Thanks a lot!

MathMan
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1 Answers1

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Let $X$ be the number of black balls preceding the first white ball. Then clearly we have $\mathbb P(X=0) = \frac w{w+b}$. For $1\leqslant n\leqslant b$ we can show by induction that $$ \mathbb P(X=n) = \frac w{w+b}\cdot\frac{\prod_{i=0}^{n-1}b-i}{\prod_{i=1}^n w+b-i}. $$ Hence, \begin{align} \mathbb E[X] &= \sum_{n=0}^b n\cdot\mathbb P(X=n)\\ &=\frac w{w+b}\cdot \sum_{n=1}^b\frac{\prod_{i=0}^{n-1}b-i}{\prod_{i=1}^n w+b-i}\\ &=\frac b{w+1}. \end{align}

Math1000
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  • Thank you for answering. There is one thing which is confusing me. Why is the probability of finding a black ball at any step always = $\frac {w}{w+b}$. If a black ball had been recovered in previous steps, the probability will definitely change? – MathMan Oct 22 '19 at 18:34
  • The probability of finding a black ball at any step is not always $\frac w{w+b}$, but it is a common factor that can be pulled out. – Math1000 Oct 22 '19 at 18:37