Question

Two Events E and F are independent if P\left ( E \right )=0.3\: P\left ( E\cup F \right )=0.51 then P\left (\frac{ E}{F} \right )- P\left ( \frac{F}{E} \right ) is equal to

Answer 1

Answer:

$P(\frac{E}{F})-P(\frac{F}{E})=0$

Step-by-step explanation:

Concept used:

Two events A and B are said to be independent if $P(A\capB)=P(A).P(B)$

Addition theorem on probability:

Let A and B be two events.

Then

$P(A\cupB) = P(A)+P(B) - P(A\capB)$

Conditional probability of B given A:

$P(\frac{B}{A})=\frac{P(A\:\cap\:B)}{P(A)}$

Given:

$P(E)=0.3$

$P(E\cupF)=0.51$

By addition theorem

$P(E\cupF) = P(E)+P(F) - P(E\capF)$

$P(E\cupF) = P(E)+P(F) - P(E).P(F)$

$0.51=0.3+P(F)-0.3 P(F)$

$0.51-0.3=P(F)-0.3 P(F)$

$0.21=P(F)[1-0.3]$

$0.21=P(F)0.7$

$P(F)={0.21}{0.7}$

$P(F)=0.3$

Now,

$P(\frac{E}{F})-P(\frac{F}{E})$

$=\frac{P(E\capF)}{P(F)}-\frac{P(E\capF)}{P(E)}$

$=\frac{P(E).P(F)}{P(F)}-\frac{P(E).P(F)}{P(E)}$

$=P(E)-P(F)$

$=0.3-0.3$

$=0$