P\left ( A \right ) =\frac{3}{10}, P\left ( B \right )=\frac{2}{5}\: and\ P\left ( A\cup B \right )=\frac{3}{5}\: then P\left ( \frac{B}{A} \right )+P\left ( \frac{A}{B} \right ) equals

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{3}

  • Option 3)

    \frac{5}{12}

  • Option 4)

    \frac{7}{12}

 

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(A\cup B) = P(A)+P(B)-P(A\cap B)

\frac{3}{5} = \frac{2}{5}+\frac{3}{10}-P(A\cap B)

=\frac{4+3}{10} - P(A\cap B)

=\frac{7}{10} - P(A\cap B)

P(A\cap B) = \frac{7}{10}-\frac{3}{5} = \frac{1}{10}

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

=\frac{\frac{1}{10}}{\frac{2}{5}}+\frac{\frac{1}{10}}{\frac{3}{10}}

=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}

 


Option 1)

\frac{1}{4}

Incorrect

Option 2)

\frac{1}{3}

Incorrect

Option 3)

\frac{5}{12}

Incorrect

Option 4)

\frac{7}{12}

Correct

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