Let A and E be any two events with positive probabilities -Statement - 1 - P E-A - P A-E PEStatement - 2 - P A-E - PA- E

The correct option is A Both the statement are true P(E|A)= P(E∩ A) / P(A) P(A|E)= P(E∩ A) / P(E) 0≤ P(E)≤ 1 P(A|E)≥ #160;P(E∩ A) #16 ...

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Byju's Answer

Standard XII

Mathematics

Conditional Probability

Let A and E b...

Question

Let A and E be any two events with positive probabilities :
Statement - 1 : $P (\frac{E}{A}) \geq P (\frac{A}{E}) P (E)$
Statement - 2 : $P (\frac{A}{E}) \geq P (A \cap E)$

Both the statement are true

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Both the statement are false

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Statement - 1 is true, Statement - 2 is false

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Statement - 1 is false, Statement - 2 is true.

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Solution

The correct option is A Both the statement are true
$P (E | A) = \frac{P (E \cap A)}{P (A)}$

$P (A | E) = \frac{P (E \cap A)}{P (E)}$

$0 \leq P (E) \leq 1$

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

$P (E | A) = \frac{P (A | E) P (E)}{P (A)}$ .

$0 \leq P (A) \leq 1$
So, $P (E | A) \geq P (A | E) P (E)$

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Let A and E be any two events with positive probabilities :Statement - 1 : P(EA)≥P(AE)P(E)Statement - 2 : P(AE)≥P(A∩E)

The correct option is A Both the statement are trueP(E|A)=P(E∩A)P(A)P(A|E)=P(E∩A)P(E)0≤P(E)≤1P(A|E)≥P(E∩A)P(E|A)=P(A|E)P(E)P(A) .0≤P(A)≤1So,P(E|A)≥P(A|E)P(E)

Let A and E be any two events with positive probabilities :
Statement - 1 : $P (\frac{E}{A}) \geq P (\frac{A}{E}) P (E)$
Statement - 2 : $P (\frac{A}{E}) \geq P (A \cap E)$

The correct option is A Both the statement are true
$P (E | A) = \frac{P (E \cap A)}{P (A)}$

$P (A | E) = \frac{P (E \cap A)}{P (E)}$

$0 \leq P (E) \leq 1$

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

$P (E | A) = \frac{P (A | E) P (E)}{P (A)}$ .

$0 \leq P (A) \leq 1$
So, $P (E | A) \geq P (A | E) P (E)$