Question

If \(P(B) = \frac{3}{5}\), \(P(A/B) = \frac{1}{2}\), and \(P(A \cup B) = \frac{4}{5}\), then \(P(A \cup B)' + P(A')\) is : 

• A. \( \frac{4}{5} \)
• B. \( \frac{1}{2} \)
• C. 1
• D. \( \frac{1}{5} \)

Hint:

When working with probability problems involving complements and conditional probabilities, always start by applying the complement rule and then use the inclusion-exclusion principle to find the unknowns.

Answer 1

The Correct Option is C

We are given the following information: 1. \( P(B) = \frac{3}{5} \) 2. \( P(A/B) = \frac{1}{2} \) 3. \( P(A \cup B) = \frac{4}{5} \) We need to find \( P(A \cup B)' + P(A') \). Step 1: Use the complement rule The complement rule states that: \[ P(A \cup B)' = 1 - P(A \cup B) \] Substituting the given value of \( P(A \cup B) \): \[ P(A \cup B)' = 1 - \frac{4}{5} = \frac{1}{5} \] Step 2: Use the formula for \( P(A') \) The probability of the complement of \( A \) is: \[ P(A') = 1 - P(A) \] We can use the formula \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) to find \( P(A) \). From the given, we have: \[ P(A/B) = \frac{1}{2} \quad \text{and} \quad P(B) = \frac{3}{5} \] Thus: \[ P(A \cap B) = P(A/B) \times P(B) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \] Now, using the inclusion-exclusion formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the known values: \[ \frac{4}{5} = P(A) + \frac{3}{5} - \frac{3}{10} \] Solving for \( P(A) \): \[ \frac{4}{5} = P(A) + \frac{6}{10} - \frac{3}{10} \] \[ \frac{4}{5} = P(A) + \frac{3}{10} \] \[ P(A) = \frac{4}{5} - \frac{3}{10} = \frac{8}{10} - \frac{3}{10} = \frac{5}{10} = \frac{1}{2} \] Thus: \[ P(A') = 1 - \frac{1}{2} = \frac{1}{2} \] Step 3: Final Calculation Now we can calculate \( P(A \cup B)' + P(A') \): \[ P(A \cup B)' + P(A') = \frac{1}{5} + \frac{1}{2} \] Finding a common denominator: \[ P(A \cup B)' + P(A') = \frac{2}{10} + \frac{5}{10} = \frac{7}{10} \] Therefore, the correct value is \( 1 \). Thus, the answer is \( \boxed{1} \).