Question

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

Answer 1

Hint: First considered the case of that let the selected balls from \[{u_1}\]be either white or black and then on putting it into another urn look into the probability that what will be the probability if white ball is poured into the second urn and what is the probability if black ball is poured into the second urn, hence calculate the probability for both of the following cases and then add the probability if both the cases in order to obtain the exact probability of the given. Probability of any event is given as \[P(a) = \dfrac{{favourable\,outcomes}}{{total\,possible\,outcomes}}\]

Complete step by step answer:

Given:

Urn 1: \[{u_1}\]has m white and n black balls.
Urn 2: \[{u_2}\]​has p white and q black balls.
Then the probability of selecting white ball from \[{u_1}\]is \[ = \dfrac{m}{{m + n}}\]and for that of black ball is \[ = \dfrac{n}{{m + n}}\]
Similarly, the probability of selecting white ball from \[{u_2} = \dfrac{p}{{p + q}}\]and that of black ball is \[ = \dfrac{q}{{p + q}}\]

Case-I: If the transferred ball is a white ball, then in \[{u_2}\]there are total \[p + 1\]white balls and \[q\] black balls.
So, now the probability of selecting white ball from \[{u_2} = \dfrac{{p + 1}}{{p + q + 1}}\] and that of black ball is \[ = \dfrac{q}{{p + q + 1}}\]

Case-II: If the transferred ball is a black ball, then in \[{u_2}\]there are total \[p\] white balls and \[q + 1\] black balls.
So, now the probability of selecting white ball from \[{u_2} = \dfrac{p}{{p + q + 1}}\] and that of black ball is \[ = \dfrac{{q + 1}}{{p + q + 1}}\]
Hence, the probability of drawing the white ball from second urn is
\[ = \dfrac{m}{{m + n}} \times \dfrac{{p + 1}}{{P + q + 1}} + \dfrac{n}{{m + n}} \times \dfrac{p}{{P + q + 1}}\]
Now, on simplifying above, we get,
\[ = \dfrac{{m\left( {p + 1} \right) + np}}{{m + n\left( {P + q + 1} \right)}}\]
Hence, option (B) is our correct answer.

Note: Most of the students may consider only the second event case and do their calculations which is wrong. In this type of problem, we use conditional probability. Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. Conditional probability is calculated by multiplying the probability of the preceding event by the updated probability of the succeeding, or conditional, event. A conditional probability is the probability of an event, given some other event has already occurred. In the below example, there are two possible events that can occur. A ball falling could either hit the red shelf (we'll call this event A) or hit the blue shelf (we'll call this event B) or both.