Question

There are two balls in an urn whose colours are not known (each ball can either be white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is
A) $\dfrac{1}{4}$
B) $\dfrac{1}{3}$
C) $\dfrac{2}{3}$
D) $\dfrac{1}{6}$

Answer 1

Hint:
Firstly, let ${X_n}$ be the event that the urn contains n white balls and 2 – n black balls, where $0 \leqslant n \leqslant 2$ and A be the event that a white ball is selected
Finally, after finding the probability of selecting one white ball, when there are 0, 1 and 2 white balls individually and use the formula of total probability i.e. $P(A) = P({X_0})P\left( {\dfrac{A}{{{X_0}}}} \right) + P({X_1})P\left( {\dfrac{A}{{{X_1}}}} \right) + P({X_2})P\left( {\dfrac{A}{{{X_2}}}} \right)$.

Complete step by step solution:
Here, it is given that there are two balls in an urn whose colours are not known and another white ball is put into the urn.
Let, ${X_n}$ be the event that the urn contains n white balls and 2 – n black balls, where $0 \leqslant n \leqslant 2$ .
As per the information given in question, $P({X_n}) = \dfrac{1}{3}$ , for every n = 0, 1, 2.
Let A be the event that a white ball is drawn from the urn.
Also, let $P\left( {\dfrac{A}{{{X_0}}}} \right)$ be the probability of drawing a white ball from the urn when there are 0 white balls in the urn. So, we get $P\left( {\dfrac{A}{{{X_0}}}} \right) = \dfrac{1}{3}$ .
Now, let $P\left( {\dfrac{A}{{{X_1}}}} \right)$ be the probability of drawing a white ball from the urn when there are 1 white ball in the urn. So, we get $P\left( {\dfrac{A}{{{X_1}}}} \right) = \dfrac{2}{3}$ .
And, let $P\left( {\dfrac{A}{{{X_2}}}} \right)$ be the probability of drawing a white ball from the urn when there are 2 white balls in the urn. So, we get $P\left( {\dfrac{A}{{{X_2}}}} \right) = \dfrac{3}{3} = 1$ .
Now, to get the probability of selecting a white ball, we use the equation $P(A) = P({X_0})P\left( {\dfrac{A}{{{X_0}}}} \right) + P({X_1})P\left( {\dfrac{A}{{{X_1}}}} \right) + P({X_2})P\left( {\dfrac{A}{{{X_2}}}} \right)$
Thus, putting the respective values in the above equation, we get
 $
  P(A) = P({X_0})P\left( {\dfrac{A}{{{X_0}}}} \right) + P({X_1})P\left( {\dfrac{A}{{{X_0}}}} \right) + P({X_2})P\left( {\dfrac{A}{{{X_2}}}} \right) \\
   = \left( {\dfrac{1}{3} \times \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} \times \dfrac{2}{3}} \right) + \left( {\dfrac{1}{3} \times 1} \right) \\
   = \dfrac{1}{9} + \dfrac{2}{9} + \dfrac{1}{3} \\
   = \dfrac{{1 + 2 + 3}}{9} \\
   = \dfrac{6}{9} \\
   = \dfrac{2}{3} \\
 $

Thus, the probability of selecting a white ball from the urn is $\dfrac{2}{3}$.
So, option (C) is correct.

Note:
Here, we cannot apply the Bayes theorem as the probability of selecting white ball is not a conditional probability.
Also, we have to take three cases when there are 0, 1 and 2 white balls in the urn. So, here we use the formula for total probability.