Question

There are two urns. There are m white and n black balls in the first urn and p white and q black balls in the second urn. One ball taken from the first turn is placed into the second. Now the probability of drawing a white ball from second urn is
A) $\dfrac{{pm + \left( {p + 1} \right)n}}{{\left( {m + n} \right)\left( {p + q + 1} \right)}}$
B) $\dfrac{{\left( {p + 1} \right)m + pn}}{{\left( {m + n} \right)\left( {p + q + 1} \right)}}$
C) $\dfrac{{qm + \left( {q + 1} \right)n}}{{\left( {m + n} \right)\left( {p + q + 1} \right)}}$
D) $\dfrac{{\left( {q + 1} \right)m + qn}}{{\left( {m + n} \right)\left( {p + q + 1} \right)}}$

Answer 1

First find the probabilities of white and black balls in both the Urns. Then we will replace a ball from Urn1 into Urn2. Again find the probabilities. Finally we will get the probability of drawing a white ball.

Complete step by step solution:
There are two urns given. Urn1 contains m white and n black balls whereas Urn2 contains p white and q black balls.
Now let’s find the probabilities of the balls in urns separately.
Probability of balls in Urn1:
P (white balls) = $\dfrac{m}{{m + n}}$
P (black balls) = $\dfrac{n}{{m + n}}$
Probability of balls in Urn2:
P (white balls) = $\dfrac{p}{{p + q}}$
P (black balls) = $\dfrac{q}{{p + q}}$
Now we have taken out a ball from Urn1 and placed it in Urn2. But since we have no idea whether it is a black or white ball we will consider the probabilities of both.
Probabilities of balls in Urn2 after replacement:
Case 1: If the replaced ball is white then the number of white balls in Urn2 will change only.
P (white balls) = $\dfrac{{p + 1}}{{p + q + 1}}$ P (black balls) = $\dfrac{q}{{p + q}}$
Case 2: If the replaced ball is black then the number of black balls in Urn2 will change only.

P (white balls) = $\dfrac{p}{{p + q}}$ P (black balls) = $\dfrac{{q + 1}}{{p + q + 1}}$
But we are asked to find the probability of drawing a white ball only.
So probability of drawing a white ball from Urn2 is
$\Rightarrow \dfrac{m}{{m + n}} \times \dfrac{{p + 1}}{{p + q + 1}} + \dfrac{n}{{m + n}} \times \dfrac{p}{{p + q + 1}}$
Since denominators are same add the numerators
This is the probability of drawing a white ball.

Hence option B is correct.

Note:
Remember we have no idea which ball is replaced from Urn1 to Urn2. So we will find the probability of both white and black balls. But when calculating we know that we have to find the probability of white ball only. So don’t go for black.