Question

For two events $A$ and $B$, if $P\left( A \right) = P\left( {A/B} \right) = \dfrac{1}{4}$ and $P\left( {B/A} \right) = \dfrac{1}{2}$, then which of the following option is correct?
A. $A$ and $B$ are independent.
B. $P\left( {A'/B} \right) = \dfrac{3}{4}$
C. $P\left( {B'/A'} \right) = \dfrac{1}{2}$
D. All of the above.

Answer 1

Hint: Check the options one by one.
For option A, Use the fact that says two events $A$ and $B$ are independent if and only if $P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$. The probabilities $P\left( A \right)$ is given. You need to find out the probabilities $P\left( B \right)$ and $P\left( {A \cap B} \right)$ using the definition of conditional probability $P\left( {A/B} \right)$ and $P\left( {B/A} \right)$.
For option B and C, use the fact that says if two events $A$ and $B$ are independent then the events $A'$ and $B$, $A'$ and $B'$ are also independent and for any event $A$, $P\left( A \right) + P\left( {A'} \right) = 1$

Formula Used:
$P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$
$P\left( {A/B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$
$P\left( {B/A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$
For any event $A$, $P\left( A \right) + P\left( {A'} \right) = 1$

Complete step by step solution:
Option A:-
Given that $P\left( A \right) = P\left( {A/B} \right) = \dfrac{1}{4}$ and $P\left( {B/A} \right) = \dfrac{1}{2}$
We have $P\left( {B/A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{{P\left( {A \cap B} \right)}}{{\left( {\dfrac{1}{4}} \right)}}$
$ \Rightarrow P\left( {A \cap B} \right) = \dfrac{1}{2} \times \dfrac{1}{4}$
$ \Rightarrow P\left( {A \cap B} \right) = \dfrac{1}{8}$
Similarly, we have $P\left( {A/B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$
$ \Rightarrow \dfrac{1}{4} = \dfrac{{\left( {\dfrac{1}{8}} \right)}}{{P\left( B \right)}}$
$ \Rightarrow P\left( B \right) = \dfrac{1}{8} \times 4 = \dfrac{1}{2}$
Now, $P\left( A \right)P\left( B \right) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8} = P\left( {A \cap B} \right)$
$\therefore P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$
Thus, the events $A$ and $B$ are independent.
Hence option A is correct.
Option B:-
If two events $A$ and $B$ are independent then the events $A'$ and $B$ are also independent.
So, $P\left( {A'/B} \right) = \dfrac{{P\left( {A' \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{P\left( {A'} \right)P\left( B \right)}}{{P\left( B \right)}} = P\left( {A'} \right)$
Now, we have for any event $A$, $P\left( A \right) + P\left( {A'} \right) = 1$
Here $P\left( A \right) = \dfrac{1}{4}$
So, $P\left( {A'} \right) = 1 - P\left( A \right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
$\therefore P\left( {A'/B} \right) = \dfrac{3}{4}$
Hence option B is correct.
Option C:-
If two events $A$ and $B$ are independent then the events $A'$ and $B'$ are also independent.
So, $P\left( {B'/A'} \right) = \dfrac{{P\left( {A' \cap B'} \right)}}{{P\left( {A'} \right)}} = \dfrac{{P\left( {A'} \right)P\left( {B'} \right)}}{{P\left( {A'} \right)}} = P\left( {B'} \right)$
Now, we have for any event $B$, $P\left( B \right) + P\left( {B'} \right) = 1$
Here $P\left( B \right) = \dfrac{1}{2}$
So, $P\left( {B'} \right) = 1 - P\left( B \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
$\therefore P\left( {B'/A'} \right) = \dfrac{1}{2}$
Hence option C is correct.
Thus all the options A, B and C are correct.

Option ‘D’ is correct

Note: Here you need to check all the options here and memorize all the formulas of probability Also note that if P(A)≠P(B) then the events will not be equally likely and if P(A)=P(B) then the events are equally likely.