Question

If $A$ and $B$ are two events such that $P ( A )=\frac{1}{3}, P ( B )=\frac{1}{5}$ and $P ( A \cup B )=\frac{1}{2}$, then $P \left( A \mid B ^{\prime}\right)+ P \left( B \mid A ^{\prime}\right)$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{5}{8}$
• C. $\frac{5}{4}$
• D. $\frac{7}{8}$

Answer 1

The Correct Option is B

We are given that \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{5} \), and \( P(A \cup B) = \frac{1}{2} \). We need to find \( P(A \mid B^{\prime}) + P(B \mid A^{\prime}) \).

First, we use the formula for the probability of the union of two events:

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

Substitute the given values:

\(\frac{1}{2} = \frac{1}{3} + \frac{1}{5} - P(A \cap B)\)

Simplify the right-hand side:

\(\frac{1}{2} = \frac{5}{15} + \frac{3}{15} - P(A \cap B)\) \(\frac{1}{2} = \frac{8}{15} - P(A \cap B)\)

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

\(P(A \cap B) = \frac{8}{15} - \frac{1}{2}\) \(P(A \cap B) = \frac{8}{15} - \frac{15}{30}\) \(P(A \cap B) = \frac{16}{30} - \frac{15}{30}\) \(P(A \cap B) = \frac{1}{30}\)

Next, we calculate \( P(A \mid B^{\prime}) \):

The probability \( P(A \mid B^{\prime}) \) is defined as:

\(P(A \mid B^{\prime}) = \frac{P(A \cap B^{\prime})}{P(B^{\prime})}\)

We know that \( P(B^{\prime}) = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5} \).

Also, \( P(A \cap B^{\prime}) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{30} = \frac{10}{30} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10} \).

Therefore,

\(P(A \mid B^{\prime}) = \frac{\frac{3}{10}}{\frac{4}{5}} = \frac{3}{10} \times \frac{5}{4} = \frac{3}{8}\)

Now, calculate \( P(B \mid A^{\prime}) \):

The probability \( P(B \mid A^{\prime}) \) is defined as:

\(P(B \mid A^{\prime}) = \frac{P(B \cap A^{\prime})}{P(A^{\prime})}\)

We know \( P(A^{\prime}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \).

Also, \( P(B \cap A^{\prime}) = P(B) - P(A \cap B) = \frac{1}{5} - \frac{1}{30} = \frac{6}{30} - \frac{1}{30} = \frac{5}{30} = \frac{1}{6} \).

Therefore,

\(P(B \mid A^{\prime}) = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}\)

Finally, adding the two probabilities:

\(P(A \mid B^{\prime}) + P(B \mid A^{\prime}) = \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8}\)

Thus, the answer is \( \frac{5}{8} \).