An urn initially contains one red and one blue ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let X denote the selection number of the first chosen ball that is blue. For instance, if the first selection is red and the second blue, then X is equal to $2$.

• (a) Find $P\{X>i\},i\ge 1$
• (b) Show that with probabilityrole="math" $1$, a blue ball is eventually chosen. (That is, show that$P\{X<\mathrm{\infty }\}=1$.)
• (c) Find$E\left[X\right].$

Given in the question that $P\{X>i\},i\ge 1$

An urn initially contains one red and one blue ball.

Let $X$be the selected number of the first chosen ball is blue.

Find $P(X>i),i>1$

For $X=1$, it means to draw a blue ball on the first draw. The probability of drawing a blue ball on the first draw is simply $\frac{1}{2}$

localid="1648022111169" $P(X>i)=1-P(X=1)-P(X<i)$

$=1-\frac{1}{2}$

$=\frac{1}{2}.$

For $X=2$this means the first draw is red (which occurs with probability $1/2$) and the second is blue. Then the probability of drawing a blue ball is role="math" $1/3.$

So $P(X=2)=\frac{1}{2}\left(\frac{1}{3}\right)$

$P(X>2)=1-P(X=2)-\underset{P(X=1)}{\underbrace{P(X<2)}}$

$=1-\frac{1}{2}\left(\frac{1}{3}\right)-\frac{1}{2}$

$=1-\frac{2}{3}$

We get,

$=\frac{1}{3}.$

For $X=3$is the case that the first blue ball drawn is chosen on the third round. This means the first two draws are red and the third draw is the blue ball, which occurs with probability $1/4$.

Thus,

$P(X=3)=\frac{1}{2}\left(\frac{2}{3}\right)\left(\frac{1}{4}\right)$

$P(X>3)=1-P(X=3)-\underset{-P(X=1)+P(X=2)}{\underbrace{P(X<3)}}$

$=1-\frac{1}{2}\left(\frac{2}{3}\right)\left(\frac{1}{4}\right)-\frac{1}{2}-\frac{1}{2}\left(\frac{1}{3}\right)$

$=1-\frac{3}{4}$

We get,

$=\frac{1}{4}$

So on,

For$X>i$, $P(X>i)=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)\dots \left(\frac{i-1}{i}\right)\left(\frac{i}{i+1}\right)$

$=\frac{1}{i+1}.$

Given in the question that$P\{X<\mathrm{\infty }\}=1$

From the result of part (a), take the limit as$i\to \mathrm{\infty }$

$P(X>\mathrm{\infty })={\mathrm{Lim}}_{i\to \mathrm{\infty }}\frac{1}{i+1}$

$={\mathrm{Lim}}_{i\to \mathrm{\infty }}\frac{\frac{1}{i}}{1+\frac{1}{i}}$

$=0.$

From the properties of probability theory,

$P(X>\mathrm{\infty })=1-P(X<\mathrm{\infty })$

$=1-0$

$=1.$

Given in the question that the expected value$E\left(X\right)$

The expected value of the random variable $X$be defined can be defined as,

$E\left(X\right)=\sum _{i=1}^{\mathrm{\infty }} iP(X=i)$

Now find,$P(X=i)$

For $X=i,$this means that the first $i-1$draws were all red and the $i^{\text{th}}$draw is blue.

Thus,

$P(X=i)=\underset{i-1\text{red balls drawn}}{\underbrace{\frac{1}{2}\left(\frac{2}{3}\right)\cdots \cdot \left(\frac{i-1}{i}\right)}}\underset{\text{blue ball drawn}}{\underbrace{\left(\frac{1}{i+1}\right)}}$

$=\frac{1}{i(i+1)}$

The expected value of the random variable $X$can be be defined as,

$E\left(X\right)=\sum _{i=1}^{\mathrm{\infty }} iP(X=i)$

$=\sum _{i=1}^{\mathrm{\infty }} i\frac{1}{i(i+1)}$

We get,

$=\sum _{i=1}^{\mathrm{\infty }} \frac{1}{(i+1)}$

Since $\sum _{i=1}^{\mathrm{\infty }} \frac{1}{i}$diverges, the above expression diverges by the basic comparison test.

Therefore$E\left(X\right)=\mathrm{\infty }.$