Probability 4 Question 20
20. An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?
$(2001,5 M)$
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Answer:
Correct Answer: 20. $\alpha=\frac{p}{1-(1-p)^{3}}, \beta=\frac{p(1-p)}{1-(1-p)^{3}}, \gamma=\frac{p-2 p^{2}+p^{3}}{1(1-p)^{3}}$
Solution:
- Let $W _1=$ ball drawn in the first draw is white.
$B _1=$ ball drawn in the first draw in black.
$W _2=$ ball drawn in the second draw is white.
Then, $P\left(W _2\right)=P\left(W _1\right) P\left(W _2 / W _1\right)+P\left(B _1\right) P\left(W _2 / B _1\right)$
$$ \begin{aligned} & =\frac{m}{m+n} \frac{m+k}{m+n+k}+\frac{n}{m+n} \frac{m}{m+n+k} \\ & =\frac{m(m+k)+m n}{(m+n)(m+n+k)}=\frac{m(m+k+n)}{(m+n)(m+n+k)}=\frac{m}{m+n} \end{aligned} $$
