Let A and E be any two events with positive probabilities Statement I: PE/A≥PA/EPE. Statement II: PA/E≥PA∩E.

Correct answer: (B) Both the statements are true. We know that PE/A=(PE∩A)/PA ⇒PE∩A=PE/A·PA ...i Also, PA/E=(PE∩A)/PE ⇒PE∩A=PA/E·PE ii From the equations i and ii, we get PE/A·PA=PA/E·PE Now, since 0<PA≤1 ⇒PE/A≥PA/E·PE Again, we have PA/E=(PE∩A)/PE ⇒PA/E·PE=PE∩A Now, since 0<PE≤1 PA/E≥PA∩E ⇒Both the statements are true.

Q48 / 90 · JEE Main 2014 (19 Apr Online) · Mathematics

Let $A$ and $E$ be any two events with positive probabilities

Statement I: $P (E / A) \geq P (A / E) P (E) .$

Statement II: $P (A / E) \geq P (A \cap E) .$

A
Both the statements are false
Correct answer
B
Both the statements are true
C
Statement - I is false, Statement - II is true
D
Statement - I is true, Statement - II is false

Solution

We know that $P (E / A) = \frac{P (E \cap A)}{P (A)}$

$\Rightarrow P (E \cap A) = P (E / A) \cdot P (A) . . . (i)$

Also, $P (A / E) = \frac{P (E \cap A)}{P (E)}$

$\Rightarrow P (E \cap A) = P (A / E) \cdot P (E) (i i)$

From the equations $(i)$ and $(i i),$ we get

$P (E / A) \cdot P (A) = P (A / E) \cdot P (E)$

Now, since $0 < P (A) \leq 1$

$\Rightarrow P (E / A) \geq P (A / E) \cdot P (E)$

Again, we have $P (A / E) = \frac{P (E \cap A)}{P (E)}$

$\Rightarrow P (A / E) \cdot P (E) = P (E \cap A)$

Now, since $0 < P (E) \leq 1$

$P (A / E) \geq P (A \cap E)$

$\Rightarrow$ Both the statements are true.

Hence, the correct answer is option: 'B'.

Let A and E be any two events with positive probabilities Statement I: PE/A≥PA/EPE.Statement II: PA/E≥PA∩E.

Solution

Relevant Questions

Let $A$ and $E$ be any two events with positive probabilities

Statement I: $P (E / A) \geq P (A / E) P (E) .$

Statement II: $P (A / E) \geq P (A \cap E) .$