Question

An urn A contains 2 white and 3 black balls. Another urn B contains 3 white and 4 black balls. Out of these two urns, one is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that
I.It is from urn $A$ is $\dfrac{{21}}{{40}}$
II.It is from urn $B$ is $\dfrac{{20}}{{41}}$
Which of the following statements is correct
A.I only
B.II only
C.Both I and II
D.Neither I nor II

Answer 1

Hint: We will use Bayes’ theorem which formula is given as-
\[P\left( {A/E} \right) = \dfrac{{P\left( A \right)P\left( {E/A} \right)}}{{P\left( A \right)P\left( {E/A} \right) + P\left( B \right)P\left( {E/B} \right)}}\,\,\,\,\,\,\,\,\,...(1)\]
Where,
$P\left( A \right)\,{\text{and}}\,P\left( B \right)$Shows the probabilities of urns$A\& B$ respectively
$P\left( {E/A} \right)$ = Probability of blackball drawn from urn $A$
$P\left( {E/B} \right)$ = Probability of blackball drawn from urn $B$

Complete step-by-step answer:
Two Urn$A\& B$are given which contains different color balls as-
\[Urn = {\text{ }}2\] White \[ + {\text{ }}3\] Black Balls
\[Urn = {\text{ }}3\] White \[ + {\text{ }}4\] Black Balls
According to this question, we get the probabilities of urns $A\& B$ as following
$P\left( A \right) = P\left( B \right) = \dfrac{1}{2}$
Probability of blackball drawn from urn $A$
$P\left( {E/A} \right) = \dfrac{3}{5}$
Probability of blackball drawn from urn $B$
$P\left( {E/B} \right) = \dfrac{4}{7}$
Substitute all the value on equation \[\left( 1 \right)\] we get probability of blackball when it is known that it is from urn$A$
$\begin{gathered}
  P\left( {A/E} \right) = \dfrac{{P\left( A \right)P\left( {E/A} \right)}}{{P\left( A \right)P\left( {E/A} \right) + P\left( B \right)P\left( {E/B} \right)}} \\
= \dfrac{{\dfrac{1}{2} \times \dfrac{3}{5}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{4}{7}}} \\
\end{gathered} $
Further Solving we get,
$
   = \dfrac{{\dfrac{3}{{10}}}}{{\dfrac{3}{{10}} + \dfrac{2}{7}}} \\
   = \dfrac{3}{{10}} \times \dfrac{{70}}{{41}} = \dfrac{{21}}{{41}} \\
 $
Similarly, again substituting the all values in equation \[\left( 1 \right)\]we get probability of blackball when it is known that it is from urn$B$
\[
  P\left( {B/E} \right) = \dfrac{{P\left( B \right)P\left( {E/B} \right)}}{{P\left( A \right)P\left( {E/A} \right) + P\left( B \right)P\left( {E/B} \right)}} \\
   = \dfrac{{\dfrac{1}{2} \times \dfrac{4}{7}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{4}{7}}} \\
\]
Further Simplifying,
\[
   = \dfrac{{\dfrac{2}{7}}}{{\dfrac{3}{{10}} + \dfrac{2}{7}}} \\
   = \dfrac{2}{7} \times \dfrac{{70}}{{41}} = \dfrac{{20}}{{41}} \\
\]
So, the correct answer is “Option B”.

Note: Bayes’ theorem is used for calculating reverse probabilities which are asked in the given question. In the question $P\left( {E/A} \right)\,{\text{and}}\,P\left( {E/B} \right)$are already given and asked the probabilities of $P\left( {A/E} \right)\,{\text{and}}\,P\left( {B/E} \right)$ which are reverse probabilities.