0
$\begingroup$

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).$

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

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

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!

$\endgroup$
11
  • $\begingroup$ Please do not cross-post across sites. $\endgroup$ Oct 22, 2019 at 17:23
  • $\begingroup$ @StubbornAtom I will take care. Have been stuck for long on this problem. Could you give me a direction please? $\endgroup$
    – MathMan
    Oct 22, 2019 at 17:24
  • 1
    $\begingroup$ You're working too hard. By considering how things change when a black ball is withdrawn, it suffices to check that the formula is correct when $b=0$ and to verify that otherwise $$e_w(b) = \frac{b}{b+w}\left(1+e_w(b-1)\right)$$ where $e_w(b) = b/(w+1).$ $\endgroup$
    – whuber
    Oct 22, 2019 at 17:37
  • $\begingroup$ @whuber I am a bit confused because of one conceptual problem. Does, the probability of finding a black ball at the $ith$ step remain the same as at every step? Shouldn't it change because at the $i-1~th$ step, a black ball could have been recovered? Thanks $\endgroup$
    – MathMan
    Oct 22, 2019 at 18:04
  • 2
    $\begingroup$ That's right, which is why "$e_w(b-1)$" appears in the recursion. $\endgroup$
    – whuber
    Oct 22, 2019 at 19:38

1 Answer 1

1
$\begingroup$

Mathematical induction. It is easy to do @whuber's proof by mathematical induction to show that with $w$ white balls and $b$ black ones in the urn, the number $X$ of black balls drawn before the first white one has $E(X) = \frac{b}{w+1}.$ [Start the induction step with $1 + e_w(b-1) = \frac{w+1}{w+1} + \frac{b-1}{w+1}.]$

Simple case with seven balls in the urn. Also, in the specific case where $w=2$ and $b = 5,$ simple combinatorial arguments show that $P(X = k) = \frac{6-k}{21},$ for $k = 0, 1, \dots, 5.$ [For example, $P(X = 1) = \frac{2 \cdot 5}{7\cdot 6} = \frac{5}{21}.$]

Then you can use a calculator to find $$E(X) = \sum_{k=0}^5 k\frac{6-k}{21} = \frac{b}{w+1} = \frac{5}{3} = 1.6667.$$

k = 0:5; sum(k*(6-k)/21)
[1] 1.666667

Simulation of specific case. A simulation in R of a million such experiments (drawing without replacement and counting the draws before getting a white ball) approximates the distribution of $X.$ [The R function match finds the draw with the first white ball (1). The sample function draws all the balls in sequence without replacement.]

set.seed(1022)
b = 5;  w = 2
balls = c(rep(0,b),rep(1,w))  # `0` for black, `1` for white
x = replicate(10^6, match(1,sample(balls))-1)
mean(x)
[1] 1.66686
b/(w+1)
[1] 1.666667
table(x)/10^6
x
       0        1        2        3        4        5 
0.286038 0.238137 0.189911 0.142529 0.095611 0.047774 

These simulated probabilities are accurate to a couple of decimal places as shown in the histogram below. Histogram bars show simulated probabilities and dots show exact ones.

enter image description here

Note; If balls were drawn with replacement, then $X$ would have a geometric distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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