Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A ∪ B)=\frac{3}{5}, $ then $P(B|A)+P(A|B)$ is equal to :

Options:

$\frac{5}{12}$

$\frac{7}{12}$

$\frac{11}{12}$

$\frac{1}{3}$

Correct Answer:

$\frac{7}{12}$

Explanation:

The correct answer is Option (2) → $\frac{7}{12}$

$P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A ∪ B)=\frac{3}{5}$

$P(A∩B)=-P(A ∪ B)+P(A)+P(B)=\frac{1}{10}$

so $P(B|A)+P(A|B)=P(A∩B)(\frac{1}{P(A)}+\frac{1}{P(B)})=\frac{7}{12}$