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Probability Part 2

Section: Mathematics

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Question : 90 of 100

Marks: +1, -0

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

\frac{6}{11}

n

[24-Jun-2022-Shift-1]

13
6
4
3

Solution: 👈: Video Solution

P(1 R

and

1 B)=P(A) \cdot P\left(\frac{1\text{ R }1\text{ B}}{A}\right)+P(B) \cdot P\left(\frac{1\text{ R }1\text{ B}}{B}\right)

=\frac{1}{2} \cdot \frac{{}^{3}C_{1} \cdot {}^{1}C_{1}}{{}^{6}C_{2}} + \frac{1}{2} \cdot \frac{{}^{2}C_{1} \cdot {}^{3}C_{1}}{{}^{n+5}C_{2}}

P\left(\frac{1\text{ R }1\text{ B}}{A}\right)=\frac{\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \frac{3}{15} + \frac{1}{2} \cdot \frac{6 \cdot 2}{(n+5)(n+4)}}=\frac{6}{11}

\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10} + \frac{6}{(n+5)(n+4)}} = \frac{6}{11}

\Rightarrow \frac{11}{10} = \frac{6}{10} + \frac{36}{(n+5)(n+4)}

\Rightarrow \frac{5}{10 \times 36} = \frac{1}{(n+5)(n+4)}

\Rightarrow n^2 + 9n - 52 = 0

\Rightarrow n = 4

is only possible value

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