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  • An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into urn, otherwise it is replaced along with another ball of the same colour. The process i

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into urn, otherwise it is replaced along with another ball of the same colour. The process is repeated. If the probability that the third ball drawn is black is \frac{23}{\lambda}, then the value of \lambda must be

Option: 1

92


Option: 2

30


Option: 3

90


Option: 4

99


Answers (1)

best_answer

For the first two draw, the balls taken out may be Let \mathrm{E_1=} White and White
\mathrm{E_2=} White and Black
\mathrm{E_3=} Black and White
\mathrm{E_4=} Black and Black
\mathrm{ P\left(E_1\right)=P(W) \cdot P\left(\frac{W}{W}\right)=\frac{2}{4} \cdot \frac{1}{3}=\frac{1}{6} }

\mathrm{ P\left(E_2\right)=P(W) \cdot P(B / W) =\frac{2}{4} \cdot \frac{2}{3}=\frac{1}{3} }

\mathrm{ P\left(E_3\right)=P(B) \cdot P(W / B) =\frac{2}{4} \cdot \frac{2}{5}=\frac{1}{5} }

\mathrm{ \text { and } \quad P\left(E_4\right)=P(B) \cdot P(B / B) =\frac{2}{4} \cdot \frac{3}{5}=\frac{3}{10} }

\mathrm{ \therefore P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+P\left(E_4\right) =\frac{1}{6}+\frac{1}{3}+\frac{1}{5}+\frac{3}{10} }

\mathrm{ =\frac{10+20+12+18}{60} }

\mathrm{ =1 }

Then, events \mathrm{ E_1, E_2, E_3 \: and \: E_4 } are exhaustive. Obviously these events are mutually exclusive, then

\mathrm{ P\left(B / E_1\right)=\frac{2}{2}=1 ; P\left(B / E_2\right)=\frac{3}{4} }

\mathrm{ P\left(B / E_3\right)=\frac{3}{4} \text { and } P\left(B / E_4\right)=\frac{4}{6}=\frac{2}{3} }

\therefore Required probability

\mathrm{P(B)= P\left(E_1\right) \cdot P\left(\frac{B}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{B}{E_2}\right) }

                                              \mathrm{ +P\left(E_3\right) \cdot P\left(\frac{B}{E_3}\right)+P\left(E_4\right) \cdot P\left(\frac{B}{E_4}\right) }

\mathrm{= \frac{1}{6} \times 1+\frac{1}{3} \times \frac{3}{4}+\frac{1}{5} \times \frac{3}{4}+\frac{3}{10} \times \frac{2}{3}}

\mathrm{= \frac{1}{6}+\frac{1}{4}+\frac{3}{20}+\frac{1}{5} }

\mathrm{= \frac{10+15+9+12}{60} }

\mathrm{ = \frac{23}{30}=\frac{23}{\lambda} \text { (given) } }

\mathrm{\Rightarrow \lambda= 30 }

Hence option 2 is correct.









 

 

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Rishi

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