Q. 3.23 - A First Course in Probability

Question

 Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is

(a) the probability that the ball selected from urn II is white?

(b) the conditional probability that the transferred ball was white given that a white ball is selected from urn II?

Step-by-Step Solution

Answer

a)The probability that the ball selected from urn II is white is $\frac{4}{9}$

b) white ball is selected from urn II is $\frac{1}{2}$

Step1:introduction

Let $W_i,i\in [1,2]$ be the event that a white ball is drawn from urn$I$ and $R_i$ be the event that a red ball is drawn from urn $i,i\in [1,2]$

Step2:The ball selected from urn II is white

$$\begin{gathered} P\left(W_2\right)=P\left(W_2\mid W_1\right)P\left(W_1\right)+P\left(W_2\mid R_1\right)P\left(R_1\right) \\ =\frac{2}{3}\times \frac{2}{6}+\frac{1}{3}\times \frac{4}{6} \\ =\frac{4}{9} \end{gathered}$$

Step3: A white ball is selected from urn I I  

$$\begin{gathered} P\left(W_1\mid W_2\right)=\frac{P\left(W_2\mid W_1\right)P\left(W_1\right)}{P\left(W_2\right)} \\ =\frac{\frac{2}{3}{\displaystyle \times }\frac{2}{6}}{\frac{4}{9}} \\ =\frac{1}{2} \end{gathered}$$