Question

There are two balls in an urn whose colors are not known (each ball can be either 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

First we will assume that ${{\text{E}}_i}\left( {0 \leqslant i \leqslant 2} \right)$ denote the event that urn contains $i$ white and $2 - i$ black balls and A denotes the event that a white ball is drawn from the urn. After finding the probability, we will use the total probability rule, $P\left( A \right) = P\left( {{E_0}} \right)P\left( {A|{E_0}} \right) + P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)$ to find the required value.

Complete step by step answer:

We are given that there are two balls in an urn whose colors are not known and a white ball is put into the urn.

Let us assume that ${{\text{E}}_i}\left( {0 \leqslant i \leqslant 2} \right)$ denote the event that urn contains $i$ white and $2 - i$ black balls.
We know that the probability of any event happening is given by dividing the number of outcomes of that event divided by the total number of events, that is;
$P = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}$.

Let us also assume that A denotes the event that a white ball is drawn from the urn.

According to the question, we have

$P\left( {{E_i}} \right) = \dfrac{1}{3}$, for $i = 0,1,2$

Then we will assume $P\left( {A|{E_0}} \right)$ be the probability of drawing a white ball when there were zero white balls in the urn. So, we get
$P\left( {A|{E_0}} \right) = \dfrac{1}{3}$
Now, let$P\left( {A|{E_1}} \right)$ be the probability of drawing a white ball when there is one white ball in the urn. So, we get

$P\left( {A|{E_1}} \right) = \dfrac{2}{3}$

When we let$P\left( {A|{E_2}} \right)$ be the probability of drawing a white ball when there are two white balls in the urn, we get
$P\left( {A|{E_2}} \right) = 1$

Using the total probability rule, $P\left( A \right) = P\left( {{E_0}} \right)P\left( {A|{E_0}} \right) + P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right)$ in the above values, we get

$$\Rightarrow P\left( A \right) = \dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{2}{3} + \dfrac{1}{3} \times 1 \\\ \Rightarrow P\left( A \right) = \dfrac{1}{9} + \dfrac{2}{9} + \dfrac{1}{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} \\\$$

Hence, the option (c) is correct.

Note: In solving this question, students should note here that we cannot use Baye’s theorem as the probability of drawing a white ball is not a conditional probability. We have to consider three cases when there are no white balls, one white ball, and two white balls.SO we use the total probability rule here.