Question

If $P(A) = 0.4$, $P(B) = 0.8$ and $P(A | B) = 0.6$, then $P(A \cup B)$ is:

• A. 0.96
• B. 0.72
• C. 0.36
• D. 0.42

Answer 1

Answer: (B)

0.72

Answer 2

The formula for the union of two events is:

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

The conditional probability $P(A \mid B)$ is related to $P(A \cap B)$ as:

$P(A \cap B) = P(A \mid B) \cdot P(B)$.

Substitute the given values:

$P(A \cap B) = (0.6)(0.8) = 0.48$.

Now calculate $P(A \cup B)$:

$P(A \cup B) = 0.4 + 0.8 - 0.48 = 0.72$.

Thus, the probability $P(A \cup B)$ is 0.72.