Question

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

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{5}{12}\)
• D. \(\frac{7}{12}\)

Answer 1

The Correct Option is D

To solve the problem of finding \(P\left(\frac{B}{A}\right)+P\left(\frac{A}{B}\right)\), we use the given probabilities: \(P(A)=\frac{3}{10}\), \(P(B)=\frac{2}{5}\), and \(P(A\cup B)=\frac{3}{5}\). The formula for the union of two events is:

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

Substituting the known values:

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

Solve for \(P(A\cap B)\):

\[P(A\cap B)=\frac{3}{10}+\frac{4}{10}-\frac{6}{10}=\frac{1}{10}\]

Next, determine the conditional probabilities:

\[P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{10}}{\frac{3}{10}}=\frac{1}{3}\]

\[P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{10}}{\frac{2}{5}}=\frac{1}{4}\]

Finally, add these probabilities together:

\[P\left(\frac{B}{A}\right)+P\left(\frac{A}{B}\right)=\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\]

Thus, the result is \(\frac{7}{12}\), which is the correct answer.