Question

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

• A. $\frac{32}{55}$
• B. $\frac{30}{45}$
• C. $\frac{32}{50}$
• D. $\frac{1}{50}$

Answer 1

The correct option is A $\frac{32}{55}$

Case I: Red ball from A to B, red ball from B to A, then red ball from $A\to BB\to A$,Red from A
$\left(R\right)\left(R\right)$
$P_1=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}$
Case II : $A\to BB\to A$,R from A
$\left(R\right)\left(W\right)$
$P_2=\frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}$
Case III : $A\to BB\to A$, R from A
$\left(W\right)\left(R\right)$
$P_3=\frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}$
Case Iv : $A\to BB\to A$, R from A
$\left(W\right)\left(W\right)$
$P_4=\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}$
$\therefore$ Required Probaility =$\frac{180+180+112+168}{1100}$
$=\frac{640}{1100}=\frac{32}{55}$