SATHEE: Probability 4 Question 20

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. $α = \frac{p}{1 - (1 - p)^{3}}, β = \frac{p (1 - p)}{1 - (1 - p)^{3}}, γ = \frac{p - 2 p^{2} + p^{3}}{1 (1 - p)^{3}}$

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, $P (W_{2}) = P (W_{1}) P (W_{2} / W_{1}) + P (B_{1}) P (W_{2} / B_{1})$

$\begin{aligned} = \frac{m}{m + n} \frac{m + k}{m + n + k} + \frac{n}{m + n} \frac{m}{m + n + k} \\ = \frac{m (m + k) + m n}{(m + n) (m + n + k)} = \frac{m (m + k + n)}{(m + n) (m + n + k)} = \frac{m}{m + n} \end{aligned}$

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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?

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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?