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

Hint: First of all, find the sample space for the given question. Now use the sample space to find the probability of drawing white balls from the urn containing two balls which are either white or black. Then use conditional probability and total probability theorem to find the probability to draw white ball.

Complete step by step answer :
Here, we have to find the probability of drawing a white ball when we already have two balls in an urn. The color of both the balls is unknown but it can be either white or black.

The sample space of urn with two balls that are either white of black can be \[S = \left\{ {WW,WB,BB} \right\}\],where \[W\] denotes white ball and \[B\] denotes black ball. Let \[{E_0},{E_1},{E_2}\] be the events of urn containing 0, 1, 2 white balls respectively.
The probability of \[{E_0},{E_1},{E_2}\] can be computed by dividing the favorable number of cases by the total number of cases in the sample space. So, we have \[P\left( {{E_0}} \right) = \dfrac{1}{3}\], \[P\left( {{E_1}} \right) = \dfrac{1}{3}\] and \[P\left( {{E_2}} \right) = \dfrac{1}{3}\].
Let \[A\] be the event of drawing a white ball from the urn where the colors of two balls are either black or white but the color of one ball is white for sure.
The probability \[P\left( {A|{E_0}} \right)\] is the probability of drawing a white ball when there was zero white ball in the urn. So, \[P\left( {A|{E_0}} \right) = \dfrac{1}{3}\] because now the urn has 3 balls out of which only one ball is white.
The probability \[P\left( {A|{E_1}} \right)\] is the probability of drawing a white ball when there is one white ball in the urn already. So, \[P\left( {A|{E_0}} \right) = \dfrac{2}{3}\] because now the urn has 3 balls out of which 2 balls are white.
The probability \[P\left( {A|{E_2}} \right)\] is the probability of drawing a white ball when there are two white balls in the urn already. So, \[P\left( {A|{E_2}} \right) = \dfrac{3}{3}\] because now the urn has 3 balls and all are white.

According to total probability theorem, if \[S = {E_0} \cup {E_1} \cup \ldots \cup {E_n}\] for \[\left\{ {{E_0},{E_1}, \ldots ,{E_n}} \right\}\] be partitions of sample space \[S\] and \[A\] is any event associated with sample space \[S\], then \[P\left( A \right) = \sum\limits_{j = 0}^n {P\left( {{E_i}} \right)P\left( {A|{E_i}} \right)} \]. So, we can use this theorem for our question, therefore, probability of event \[A\] is given by,
 \[
  P\left( A \right) = P\left( {{E_0}} \right)P\left( {A|{E_0}} \right) + P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right) \\
  \Rightarrow P\left( A \right) = \dfrac{1}{3} \cdot \dfrac{1}{3} + \dfrac{1}{3} \cdot \dfrac{2}{3} + \dfrac{1}{3} \cdot \dfrac{3}{3} \\
 \Rightarrow P\left( A \right) = \dfrac{{1 + 2 + 3}}{9} \\
  \Rightarrow P\left( A \right)= \dfrac{6}{9} \\
  \Rightarrow P\left( A \right)= \dfrac{2}{3} \\
\]

Therefore, the correct option is C.

Note: In this question, note that we cannot use Baye's theorem as the probability of drawing a white ball is not a conditional probability. We have to consider two cases in this question, first is when there are only two balls in the urn. The second case when there is a fixed white ball also. This case will lead us to use the total probability rule.