Question

Urn I has 2 white and 3 black balls, urn II has 4 white and 1 black ball and urn III has 3 white and 4 black balls. An urn is selected at random and a ball is drawn at random
(i) What is the probability of drawing a white ball?
(ii) If the ball is white, then what is the probability that urn I was selected?

Answer 1

Hint: Here we will first assume the events of selecting the urn I, urn II, urn III as \[{E_1},{E_2},{E_3}\]respectively and A as the event of drawing a white ball. We will then find the probability of each event and then at the last we will use Bayes’ Theorem to find the probability that urn I was selected.
Bayes’ Theorem states that
\[P\left( {\dfrac{A}{B}} \right){\text{ }} = \;\dfrac{{P\left( A \right){\text{ }}P\left( {\dfrac{B}{A}} \right)}}{{P(B)}}\]

Complete step-by-step answer:
Let us assume the events of selecting the urn I, urn II, urn III as \[{E_1},{E_2},{E_3}\]respectively
Now since the probability of selecting each urn is equally likely
Hence,
\[P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = P\left( {{E_3}} \right) = \dfrac{1}{3}\]
(i) We have to find the probability of drawing a white ball.
We will find the probability of drawing a white ball when urn 1 is already selected.
Hence, we have to find \[P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right)\]
Probability of selecting a white ball from urn 1 is:-
Since total number of balls in urn 1 is 5
No of white balls in urn 1 is 2
And we know that Probability is given by:-
\[{\text{Probabilty}} = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]
Therefore, Probability of selecting a white ball from urn 1 is:-
Probability of selecting a white ball from urn 1 is:-
\[P\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{2}{5}\]
And we know that:-
\[P\left( {{E_1}} \right) = \dfrac{1}{3}\]
Hence,
\[P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{1}{3} \times \dfrac{2}{5}\]………………………..(1)
Now we will find the probability of drawing a white ball when urn 2 is already selected.
Hence, we have to find \[P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right)\]
Probability of selecting a white ball from urn 2 is:-
Since total number of balls in urn 2 is 5
No of white balls in urn 2 is 4
And we know that Probability is given by:-
\[{\text{Probabilty}} = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]
Therefore, Probability of selecting a white ball from urn 2 is:-
Probability of selecting a white ball from urn 2 is:-
\[P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{4}{5}\]
And we know that:-
\[P\left( {{E_2}} \right) = \dfrac{1}{3}\]
Hence,
\[P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{1}{3} \times \dfrac{4}{5}\]………………………..(2)
Now we will find the probability of drawing a white ball when urn 3 is already selected.
Hence, we have to find \[P\left( {{E_3}} \right) \times P\left( {\dfrac{A}{{{E_3}}}} \right)\]
Probability of selecting a white ball from urn 3 is:-
Since total number of balls in urn 3 is 5
No of white balls in urn 3 is 3
And we know that Probability is given by:-
\[{\text{Probabilty}} = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]
Therefore, Probability of selecting a white ball from urn 3 is:-
Probability of selecting a white ball from urn 3 is:-
\[P\left( {\dfrac{A}{{{E_3}}}} \right) = \dfrac{3}{5}\]
And we know that:-
\[P\left( {{E_3}} \right) = \dfrac{1}{3}\]
Hence,
\[P\left( {{E_3}} \right) \times P\left( {\dfrac{A}{{{E_3}}}} \right) = \dfrac{1}{3} \times \dfrac{3}{5}\]………………………..(3)
Now for total probability of drawing white ball we have to add (1), (2) and (3)
\[P\left( A \right) = P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right) + P\left( {{E_3}} \right) \times P\left( {\dfrac{A}{{{E_3}}}} \right)\]
Putting in the values we get:-
\[P\left( A \right) = \dfrac{1}{3} \times \dfrac{3}{5} + \dfrac{1}{3} \times \dfrac{4}{5} + \dfrac{1}{3} \times \dfrac{2}{5}\]
Solving it further we get:-
\[P\left( A \right) = \dfrac{3}{{15}} + \dfrac{4}{{15}} + \dfrac{2}{{15}}\]
Taking LCM we get:-
\[P\left( A \right) = \dfrac{{3 + 4 + 2}}{{15}}\]
\[ \Rightarrow P\left( A \right) = \dfrac{9}{{15}}\]
\[ \Rightarrow P\left( A \right) = \dfrac{3}{5}\]
(ii) Here we have to find the probability that urn I was selected if white ball turns out.
Therefore, we have to find \[P\left( {\dfrac{{{E_1}}}{W}} \right)\]
Hence, we will use Bayes’ theorem to find the required probability.
According to Bayes’ Theorem,
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( A \right)}}\]
We know that:-
 \[P\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{2}{5}\]
\[P\left( A \right) = \dfrac{3}{5}\]
Hence putting the values we get:-
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{\dfrac{1}{3} \times \dfrac{2}{5}}}{{\dfrac{3}{5}}}\]
Solving it further we get:-
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{2}{{3 \times 3}}\]
\[ \Rightarrow P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{2}{9}\]

Hence the probability that urn I was selected if white ball turns out is \[\dfrac{2}{9}\].

Note: Students might make mistakes in finding the probability of drawing a white ball, so they should note that we have multiplied the probability of white ball in each urn by probability of selecting each urn to get the correct answer.
Also, students should note that the probability of any event is always greater than 0 and less than equal to 1.