If A and B are two events such that \( \mathrm{P(A)}=\displaystyle\frac{3}{10},\ \mathrm{P(B)}=\displaystyle\frac{2}{5} \) and \( \mathrm{P(A\cup B)}=\displaystyle\frac{3}{5} \), then \( \mathrm{P(B|A)}+\mathrm{P(A|B)} \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{7}{12} \)
By the addition rule,
\( \displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B) \)
\( \displaystyle =\frac{3}{10}+\frac{4}{10}-\frac{6}{10} \)
\( \displaystyle =\frac{1}{10} \)
Then, by the definition of conditional probability,
\( \displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{1/10}{3/10}=\frac{1}{3} \)
\( \displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1/10}{2/5}=\frac{1}{4} \)
\( \displaystyle P(B|A)+P(A|B)=\frac{1}{3}+\frac{1}{4} \)
\( \displaystyle =\frac{7}{12} \)