Answer :
Certainly! Let's solve this step by step.
Given:
- \( P(A) = 0.4 \)
- \( P(B \mid A) = 0.8 \)
We need to find \( P(A \cap B) \), which is the probability that both events \( A \) and \( B \) occur simultaneously.
From the definition of conditional probability, we know that:
[tex]\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \][/tex]
We can rearrange this formula to solve for \( P(A \cap B) \):
[tex]\[ P(A \cap B) = P(B \mid A) \times P(A) \][/tex]
Now, substitute the given values:
[tex]\[ P(A \cap B) = 0.8 \times 0.4 \][/tex]
Multiplying the values, we get:
[tex]\[ P(A \cap B) = 0.32 \][/tex]
Thus, the probability \( P(A \cap B) \) is \( 0.32 \).
The correct answer is:
B. 0.32
Given:
- \( P(A) = 0.4 \)
- \( P(B \mid A) = 0.8 \)
We need to find \( P(A \cap B) \), which is the probability that both events \( A \) and \( B \) occur simultaneously.
From the definition of conditional probability, we know that:
[tex]\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \][/tex]
We can rearrange this formula to solve for \( P(A \cap B) \):
[tex]\[ P(A \cap B) = P(B \mid A) \times P(A) \][/tex]
Now, substitute the given values:
[tex]\[ P(A \cap B) = 0.8 \times 0.4 \][/tex]
Multiplying the values, we get:
[tex]\[ P(A \cap B) = 0.32 \][/tex]
Thus, the probability \( P(A \cap B) \) is \( 0.32 \).
The correct answer is:
B. 0.32
