Question

Let A and E be any two events with positive probabilities
Statement-1: \[P\left( {\dfrac{E}{A}} \right) \geqslant P\left( {\dfrac{A}{E}} \right)P\left( E \right)\]
Statement-2: \[P\left( {\dfrac{A}{E}} \right) \geqslant P\left( {A \cap E} \right)\].
(A) Both the statements are true
(B) Both the statements are false
(C) Statement-1 is true, Statement-2 is false
(D) Statement-1 is false, Statement-2 is true

Answer 1

Hint: Here we will use the known fact that:-
\[P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\] and \[P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\]
And also that the probability of any event is greater than 0 and less than equal to 1.
And then finally find the required relations.

Complete step-by-step answer:
We know that:-
\[P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\]
Hence,
\[P\left( {\dfrac{A}{E}} \right) = \dfrac{{P\left( {A \cap E} \right)}}{{P\left( E \right)}}\]…………………….(1)
Also, we know that,
\[P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\]
Hence,
\[P\left( {\dfrac{E}{A}} \right) = \dfrac{{P\left( {A \cap E} \right)}}{{P\left( A \right)}}\]………………………………………….(2)
Now dividing equation 2 by equation 1 we get:-
\[\dfrac{{P\left( {\dfrac{E}{A}} \right)}}{{P\left( {\dfrac{A}{E}} \right)}} = \dfrac{{\dfrac{{P\left( {A \cap E} \right)}}{{P\left( A \right)}}}}{{\dfrac{{P\left( {A \cap E} \right)}}{{P\left( E \right)}}}}\]
Solving it further we get:-
\[
  \dfrac{{P\left( {\dfrac{E}{A}} \right)}}{{P\left( {\dfrac{A}{E}} \right)}} = \dfrac{{P\left( {A \cap E} \right)}}{{P\left( A \right)}} \times \dfrac{{P\left( E \right)}}{{P\left( {A \cap E} \right)}} \\
   \Rightarrow \dfrac{{P\left( {\dfrac{E}{A}} \right)}}{{P\left( {\dfrac{A}{E}} \right)}} = \dfrac{{P\left( E \right)}}{{P\left( A \right)}} \\
 \]
Now on cross-multiplying we get:-
\[P\left( {\dfrac{E}{A}} \right) = P\left( {\dfrac{A}{E}} \right) \times \dfrac{{P\left( E \right)}}{{P\left( A \right)}}\]
Now we know that the probability of any event is greater than 0 and less than equal to 1.
Hence,
\[0 < P\left( A \right) \leqslant 1\]
Therefore, this implies that:-
\[P\left( {\dfrac{E}{A}} \right) \geqslant P\left( {\dfrac{A}{E}} \right) \times P\left( E \right)\]
Therefore, statement 1 is true.
Also, in equation 1,
We know that the probability of any event is greater than 0 and less than equal to 1.
Hence,
\[0 < P\left( E \right) \leqslant 1\]
Therefore, this implies:
\[P\left( {\dfrac{A}{E}} \right) \geqslant P\left( {A \cap E} \right)\]
Therefore, statement 2 is true.
Since statement 1 as well as statement 2 are true.

So, the correct answer is “Option A”.

Note: Students might make mistakes in writing the inequalities.
Also, the probability of any event cannot be zero and is always less than 1 or equal to 1.
Hence students should not consider the probability as:
\[0 \leqslant P\left( A \right) \leqslant 1\]