Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

Option 1)
$\frac{1}{4}$

Option 2)
$\frac{1}{3}$

Option 3)
$\frac{5}{12}$

Option 4)
$\frac{7}{12}$

Answers (1)

best_answer

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

Posted by

divya.saini

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT iteration 1 result 2025 out on bitsadmission.com; confirm allotment by July 14

anam-khan

BITSAT 2025 preference editing window open till July 5 for BE, BPharm, MSc admission

anam-khan

BITSAT cut-off for BE Computer Science slightly drops, increases for all others; session 2 result soon

Latest Question