Let A and B be events in a sample space S such that P(A) = .4 and P(BA) = 2. Find: P(ANB). OP(AB)-0.99 O PAB)-0.97 O PAB)-0.08 OPAB)-0.25 Finder 2 NOV 11

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Chapter1: Combinatorial Analysis
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**Question 2** (1 point)

Let \( A \) and \( B \) be events in a sample space \( S \) such that \( P(A) = 0.4 \) and \( P(B \mid A) = 0.2 \). Find \( P(A \cap B) \).

Choices:
- \( P(A \cap B) = 0.99 \)
- \( P(A \cap B) = 0.97 \)
- \( P(A \cap B) = 0.08 \)
- \( P(A \cap B) = 0.25 \)

**Explanation:**
To solve for \( P(A \cap B) \), you can use the formula for conditional probability:

\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \]

Given:
\( P(A) = 0.4 \)
\( P(B \mid A) = 0.2 \)

Substitute the values into the formula:

\[ 0.2 = \frac{P(A \cap B)}{0.4} \]

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

\[ P(A \cap B) = 0.2 \times 0.4 = 0.08 \]

Correct choice: \( P(A \cap B) = 0.08 \)
Transcribed Image Text:**Question 2** (1 point) Let \( A \) and \( B \) be events in a sample space \( S \) such that \( P(A) = 0.4 \) and \( P(B \mid A) = 0.2 \). Find \( P(A \cap B) \). Choices: - \( P(A \cap B) = 0.99 \) - \( P(A \cap B) = 0.97 \) - \( P(A \cap B) = 0.08 \) - \( P(A \cap B) = 0.25 \) **Explanation:** To solve for \( P(A \cap B) \), you can use the formula for conditional probability: \[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \] Given: \( P(A) = 0.4 \) \( P(B \mid A) = 0.2 \) Substitute the values into the formula: \[ 0.2 = \frac{P(A \cap B)}{0.4} \] Solving for \( P(A \cap B) \): \[ P(A \cap B) = 0.2 \times 0.4 = 0.08 \] Correct choice: \( P(A \cap B) = 0.08 \)
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