Probability Ques 72
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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Answer:
Correct Answer: 72.$(\frac{32}{55})$
Solution:
Formula:
- 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} $
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