A) \[\frac{32}{55}\]
B) \[\frac{21}{55}\]
C) \[\frac{19}{55}\]
D) None of these
Correct Answer: A
Solution :
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 \[B\] and placed in \[A\] \[{{B}_{2}}=A\] black ball is drawn from urn B and placed in A \[R=A\] red ball is drawn in the second attempt from \[A\] Then the required probability \[=P({{R}_{1}}{{R}_{2}}R)+({{R}_{1}}{{B}_{2}}R)+P({{B}_{1}}{{R}_{2}}R)+P({{B}_{1}}{{B}_{2}}R)\] \[=P({{R}_{1}})P({{R}_{2}})P(R)+P({{R}_{1}})P({{B}_{2}})P(R)+P({{B}_{1}})P({{R}_{2}})P(R)+\] \[P({{B}_{1}})P({{B}_{2}})P(R)\] \[=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}+\frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}+\frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}+\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}\] \[=\frac{32}{55}\].
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