A and B, P(A and B) = 0.08 and P(A) = 0.2. If events A and B are independent, which of the following represents P(B)?

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### Question 18:

**Given two events, A and B, P(A and B) = 0.08 and P(A) = 0.2. If events A and B are independent, which of the following represents P(B)?**

#### Options:
- A) 0.016
- B) 0.12
- C) 0.28
- D) 0.4

### Explanation:
To solve this, we use the formula for the probability of the intersection of two independent events:
\[ P(A \cap B) = P(A) \times P(B) \]

Given:
\[ P(A \cap B) = 0.08 \]
\[ P(A) = 0.2 \]

We need to find \( P(B) \).

So, we can rearrange the formula to solve for \( P(B) \):
\[ P(B) = \frac{P(A \cap B)}{P(A)} \]

Plugging in the given values:
\[ P(B) = \frac{0.08}{0.2} = 0.4 \]

Hence, the correct answer is:
- **D) 0.4**

---

This question tests your understanding of the probabilities of independent events. Independent events' probabilities multiply together to give the joint probability of both events occurring together. By rearranging the formula, you can find the unknown probability when given specific conditions.
Transcribed Image Text:### Question 18: **Given two events, A and B, P(A and B) = 0.08 and P(A) = 0.2. If events A and B are independent, which of the following represents P(B)?** #### Options: - A) 0.016 - B) 0.12 - C) 0.28 - D) 0.4 ### Explanation: To solve this, we use the formula for the probability of the intersection of two independent events: \[ P(A \cap B) = P(A) \times P(B) \] Given: \[ P(A \cap B) = 0.08 \] \[ P(A) = 0.2 \] We need to find \( P(B) \). So, we can rearrange the formula to solve for \( P(B) \): \[ P(B) = \frac{P(A \cap B)}{P(A)} \] Plugging in the given values: \[ P(B) = \frac{0.08}{0.2} = 0.4 \] Hence, the correct answer is: - **D) 0.4** --- This question tests your understanding of the probabilities of independent events. Independent events' probabilities multiply together to give the joint probability of both events occurring together. By rearranging the formula, you can find the unknown probability when given specific conditions.
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