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 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 D None of these

Let the events are
$R_1$ = A red ball is drawn from urn A and placed in B
$B_1$ = A black ball is drawn from urn A and placed in B
$R_2$ = A red ball is drawn from urn A and placed in A
$B_2$ = A black ball is drawn from urn A and placed in B
R = A red ball is drawn in the second attempt from A
Then the required probability
$=P\left(R_1{R}_{2}R\right)+\left(R_1{B}_{2}R\right)+P\left(B_1{R}_{2}R\right)+P\left(B_1{B}_{2}R\right)$
$=P\left(R_1\right)P\left(R_2\right)P\left(R\right)+P\left(R_1\right)P\left(B_2\right)P\left(R\right)+P\left(B_1\right)P\left(R_2\right)P\left(R\right)+P\left(B_1\right)P\left(B_2\right)P\left(R\right)$
$=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}\times \frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}\times \frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}+\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}$
$=\frac{32}{55}$