Q.7.20 - A First Course in Probability

Question

 In an urn containing n balls, the ith ball has weight W(i),i = $1$,...,n. The balls are removed without replacement, one at a time, according to the following rule: At each selection, the probability that a given ball in the urn is chosen is equal to its weight divided by the sum of the weights remaining in the urn. For instance, if at some time i$1$,...,ir is the set of balls remaining in the urn, then the next selection will be ij with probability $W\left(i_j\right)/\sum _{k=1}^{r}W\left(i_k\right)$, j = 1,...,r  Compute the expected number of balls that are withdrawn before the ball number $1$ is removed.

Step-by-Step Solution

Answer

$\text{The required mean number is}\sum _{j=2}^n\frac{W\left(j\right)}{W\left(j\right)+W\left(1\right)}\text{.}$

Step 1: Given Information

Given the probability that, 

$W\left(i_j\right)/\sum _{k=1}^{r}W\left(i_k\right)$ j = 1,...,r 

Step 2: Explanation

Define indicator random variables $I_{j, }$j$=2,...,$n that marks if we have chosen the ball number j out from the jar before we have chosen ball number $1$ . The weight of the ball j is W(j) and the weight of ball $1$ is W($1$). Observe that the probability that we have chosen ball j before ball number $1$ is exactly

$P(I_j=1)=\frac{W\left(j\right)}{W\left(j\right)+W\left(1\right)}$

So, the expected number of balls that have been drawn before ball number $1$ is

$\sum _{j=2}^{n}P(I_j=1)=\sum _{j=2}^n\frac{W\left(j\right)}{W\left(j\right)+W\left(1\right)}$

Step 3: FInal Answer

$\text{The required mean number is}\sum _{j=2}^n\frac{W\left(j\right)}{W\left(j\right)+W\left(1\right)}\text{.}$