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 

  • Option 1)

    \frac{2}{7}

  • Option 2)

    \frac{3}{35}

  • Option 3)

    \frac{1}{70}

  • Option 4)

    \frac{1}{7}

 

Answers (1)

As we learnt in 

Conditional Probability -

 

P\left ( \frac{A}{B} \right )= \frac{P\left ( A\cap B \right )}{P\left ( B \right )}

and

P\left ( \frac{B}{A} \right )= \frac{P\left ( A\cap B \right )}{P\left ( A \right )}

 

- wherein

where P\left ( \frac{A}{B} \right ) probability of A when B already happened.

 

 P(E\cap F )=P(E)\times P(F)

for independent events.

P(E \cup F )=0.51=0.30+P(F)-(0.3)\times P(F)

=>0.21=0.7\: P(F)

P(F)=\frac{0.21}{0.7}=\frac{21}{70}=\frac{3}{10}=0.3

P\left ( \frac{E}{F} \right )-P(\frac{F}{E})= \frac{P(E\cap F )}{P(F)}-\frac{P(E\cap F )}{P(E)}

Putting the values we get \frac{1}{70}

 


Option 1)

\frac{2}{7}

Incorrect

Option 2)

\frac{3}{35}

Incorrect

Option 3)

\frac{1}{70}

Correct

Option 4)

\frac{1}{7}

Incorrect

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