Question

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

• A. $\frac{32}{55}$
  
• B. $\frac{30}{40}$
• C. $\frac{32}{50}$
• D. $\frac{3}{4}$

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
$P_1=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}$
Case(ii): red ball from A to B, white ball from B to A, red ball from A
$P_2=\frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}$
Case(iii): white ball from A to B, red ball from B to A, red ball from A
$P_3=\frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}$
Case(iv): white ball from A to B, white ball from B to A, red ball from A
$P_4=\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}$
Therefore required probability $=P_1+P_2+P_3+P_4=\frac{640}{1100}=\frac{32}{55}$