Probability 3 Question 32
32. 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 drawn at random from urn $A$, the probability that it is found to be red, is….
(1988, 2M)
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Solution:
- Let $R$ be drawing a red ball and $B$ for drawing a black ball, then required probability
$$ \begin{aligned} & =R R R+R B R+B R R+B B R \\ & =\frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}+\frac{6}{10} \times \frac{6}{11} \times \frac{5}{10} \\ & \quad+\frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}+\frac{4}{10} \times \frac{7}{11} \times \frac{6}{10} \\ & =\frac{640}{1100}=\frac{32}{55} \end{aligned} $$
