Question

Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn ball is drawn at random from urn $B$ and placed in urn $A$, If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is

• A. $\frac{32}{55}$
• B. $\frac{21}{55}$
• C. $\frac{19}{55}$
• D. None of these

Answer 1

The correct option is A

$\frac{32}{55}$

Explanation for correct option

Given: Let $R_1$ be the event that a red ball is drawn from urn $A$ and placed in urn $B$

$B_1$ be the event that a black ball is drawn from urn $A$ and placed in urn $B$

$R_2$ be the event that a red ball is drawn from urn $B$ and placed in urn $A$

$B_2$ be the event that a black ball is drawn from urn $B$ and placed in urn $A$

Let $R$ be the event that the red ball is drawn from urn $A$

Hence, the required probability $=P\left(R_1{R}_{2}R\right)+P\left(R_1{B}_{2}R\right)+P\left(B_1{R}_{2}R\right)+P\left(B_1{B}_{2}R\right)$

$$\begin{gathered} =\left(\frac{6}{10}\right)\left(\frac{5}{11}\right)\left(\frac{6}{10}\right)+\left(\frac{6}{10}\right)\left(\frac{6}{11}\right)\left(\frac{5}{10}\right)+\left(\frac{4}{10}\right)\left(\frac{4}{11}\right)\left(\frac{7}{10}\right)+\left(\frac{4}{10}\right)\left(\frac{7}{11}\right)\left(\frac{6}{10}\right) \\ =\frac{180}{1100}+\frac{180}{1100}+\frac{112}{1100}+\frac{168}{1100} \\ =\frac{180+180+112+168}{1100} \\ =\frac{640}{1100}\Rightarrow \frac{32}{55} \end{gathered}$$

Hence, option A is correct.