Q. 3.4 - A First Course in Probability

Question

Urn $A$ contains $2$ white balls and $1$ black ball, whereas urn  $B$ contains$1$ white ball and $5$ black balls. A ball is drawn at random from urn $A$ and placed in urn $B$. A ball is then drawn from urn $B$. It happens to be white. What is the probability that the ball transferred was white?

Step-by-Step Solution

Answer

By Bayer's formula,the colour was trasfered to the ball at $P\left(W_A\mid W_B\right)=\frac{4}{5}.$

Step: 1 Events:

The white ball drawn from Urn $A$ - $W_A$

The white ball drawn from Urn $B$ - $W_B$

The number of balls from $A$ is

$$\begin{gathered} P\left(W_A\right)=\frac{2}{3} \\ P\left(W_A^c\right)=\frac{1}{3} \end{gathered}$$

The ball transfers from $A$ to $B$,the conditional distribution of ball $B$ is

$\begin{array}{r}P\left(W_B\mid W_A\right)=\frac{2}{7}\\ P\left(W_B\mid W_A^c\right)=\frac{1}{7}\end{array}$

Step: 2 Finding values:

By Bayes formula,

$$\begin{gathered} P\left(W_A\mid W_B\right)=\frac{P\left(W_B\mid W_A\right)P\left(W_A\right)}{P\left(W_B\mid W_A\right)P\left(W_A\right)+P\left(W_B\mid W_A^c\right)P\left(W_A^c\right)} \\ P\left(W_A\mid W_B\right)=\frac{\left(\frac{2}{7}\times \frac{2}{3}\right)}{{\displaystyle \left(\frac{2}{7}\times \frac{2}{3}\right)}+\left(\frac{1}{7}\times \frac{1}{3}\right)} \\ P\left(W_A\mid W_B\right)=\frac{4}{5}. \end{gathered}$$