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  • Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 bla

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is \mathrm{\frac{6}{11}} , then n is equal to ________.

Option: 1

13


Option: 2

6


Option: 3

4


Option: 4

3


Answers (1)

best_answer


\mathrm{A\overset{1R,1B}{\rightarrow}\frac{^{3}C_{1}\times \, ^{1}C_{1}}{^{6}C_{2}}}
\mathrm{B\overset{1R,1B}{\rightarrow}\frac{^{2}C_{1}\times \, ^{3}C_{1}}{ ^{n+5}C_{2}}}

\mathrm{P\left ( \frac{A}{1R,1B} \right )= \frac{\frac{1}{2}\times \frac{^{3}C_{1}\times^{1}C_{1} }{^{6}C_{2}}}{\frac{1}{2}\times \frac{^{3}C_{1}\times^{1}C_{1} }{^{6}C_{2}}+\frac{1}{2}\times \frac{^{2}C_{1}\times^{3}C_{1} }{^{n+5}C_{2}}}}
\mathrm{\frac{6}{11}= \frac{\frac{3}{15}}{\frac{3}{15}+\frac{6}{\left ( n+5 \right )_{C_{2}}}}}
\mathrm{6\left [ \frac{1}{15}+\frac{2\times 2}{\left ( n+5 \right )\left ( n+4 \right )} \right ]= \frac{1}{15}\times 11}
\mathrm{\frac{24}{\left ( n+5 \right )\left ( n+4 \right )}= \frac{5}{15}}
\mathrm{n^{2}+9n+20= 72}
\mathrm{n^{2}+9n-52= 0}
\mathrm{\left ( n+13 \right )\left ( n-4 \right )= 0}
\mathrm{n= -13,n= 4}
\mathrm{\therefore \: \: n= 4}
The correct answer is option (c)

Posted by

Ajit Kumar Dubey

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