Question

If \( P(A \cup B) = 0.9 \) and} \( P(A \cap B) = 0.4 \), then \( P(A) + P(B) \) is:

• A. \( 0.3 \)
• B. \( 1 \)
• C. \( 1.3 \)
• D. \( 0.7 \)

Hint:

When calculating the probability of the union of two events, make sure to subtract the intersection to avoid double-counting the outcomes that are common to both events.

Answer 1

The Correct Option is C

We are given the following probabilities:
- \( P(A \cup B) = 0.9 \), which is the probability of the union of events A and B.
- \( P(A \cap B) = 0.4 \), which is the probability of the intersection of events A and B.
To find \( P(A) + P(B) \), we use the principle of inclusion-exclusion for two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \] This formula accounts for the fact that the probability of \( A \cup B \) (either A or B occurring) includes both the individual probabilities of A and B, but the overlap (where both A and B occur) is counted twice, so we subtract the intersection probability.
Now, substitute the known values into the formula: \[ 0.9 = P(A) + P(B) - 0.4. \] Solving for \( P(A) + P(B) \): \[ P(A) + P(B) = 0.9 + 0.4 = 1.3. \] Therefore, \( P(A) + P(B) = 1.3 \), which corresponds to option (C).