An urn has $m$ black balls. At each stage, a black ball is removed and a new ball that is black with probability $p$ and white with probability $1-p$ is put in its place. Find the expected number of stages needed until there are no more black balls in the urn. note: The preceding has possible applications to understanding the AIDS disease. Part of the body’s immune system consists of a certain class of cells known as T-cells. There are$2$ types of T-cells, called CD4 and CD8. Now, while the total number of T-cells in AIDS sufferers is (at least in the early stages of the disease) the same as that in healthy individuals, it has recently been discovered that the mix of CD4 and CD8 T-cells is different. Roughly 60 percent of the T-cells of a healthy person are of the CD4 type, whereas the percentage of the T-cells that are of CD4 type appears to decrease continually in AIDS sufferers. A recent model proposes that the HIV virus (the virus that causes AIDS) attacks CD4 cells and that the body’s mechanism for replacing killed T-cells does not differentiate between whether the killed T-cell was CD4 or CD8. Instead, it just produces a new T-cell that is CD4 with probability .$6$ and CD8 with probability .$4$. However, although this would seem to be a very efficient way of replacing killed T-cells when each one killed is equally likely to be any of the body’s T-cells (and thus has probability .$6$ of being CD4), it has dangerous consequences when facing a virus that targets only the CD4 T-cells

Step by step solution

01

Given Information

Let $X_1$define as the number of the stage until the first black ball is replaced with the white ball.

$X_2$define as the number of additional stages until the second black is replaced with the white ball.

02

Explanation

At each stage, a black ball is removed and a new ball,

black with probability $p$and white with probability $1-p$is put in its place.

Let us consider the variable $X$indicating the number

of stages needed until there are no more black balls in the urn.

$X=X_1+X_2+\dots +X_m$

Let $X_1$define as the number of stage until the first black ball is replaced with the white ball,

$X_2$define as the number of additional stages until the second black is replaced with white ball and this continuous.

$X_3,X_4,X_5,\dots ,X_m$.

03

Explanation

$E\left[X\right]=E\left[X_1+X_2+\dots +X_m\right]$

So, $E\left[X_1\right]=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1-p$}\right.fori=1,2,3,..,m.$

$=\frac{1}{1-p}+\frac{1}{1-p}+\dots +\frac{1}{1-p}$

$=\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$1-p$}\right.$

04

Final Answer

The expected number is$E\left(X\right)=\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$1-p$}\right.$.

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