Certainly! Let's work through this problem step by step to find [tex]\( P(A \cap B) \)[/tex] and [tex]\( P(A \cup B) \)[/tex] given that [tex]\( P(A) = 0.60 \)[/tex] and [tex]\( P(\bar{B}) = 0.20 \)[/tex].
### Step 1: Determine [tex]\( P(B) \)[/tex]
First, we know that [tex]\( P(\bar{B}) \)[/tex] is the probability that event [tex]\( B \)[/tex] does not occur. The complement rule states that the probability of [tex]\( B \)[/tex] occurring, [tex]\( P(B) \)[/tex], is given by:
[tex]\[ P(B) = 1 - P(\bar{B}) \][/tex]
Given [tex]\( P(\bar{B}) = 0.20 \)[/tex]:
[tex]\[ P(B) = 1 - 0.20 = 0.80 \][/tex]
### Step 2: Calculate [tex]\( P(A \cap B) \)[/tex]
Since events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent, the probability of both events occurring, [tex]\( P(A \cap B) \)[/tex], is the product of their individual probabilities:
[tex]\[ P(A \cap B) = P(A) \times P(B) \][/tex]
Given [tex]\( P(A) = 0.60 \)[/tex] and [tex]\( P(B) = 0.80 \)[/tex]:
[tex]\[ P(A \cap B) = 0.60 \times 0.80 = 0.48 \][/tex]
Hence, [tex]\( P(A \cap B) = 0.48 \)[/tex].
### Step 3: Calculate [tex]\( P(A \cup B) \)[/tex]
The probability of either event [tex]\( A \)[/tex] or [tex]\( B \)[/tex] occurring, [tex]\( P(A \cup B) \)[/tex], for independent events, is given by:
[tex]\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \][/tex]
Using the values we have:
[tex]\[ P(A \cup B) = 0.60 + 0.80 - 0.48 \][/tex]
Simplifying:
[tex]\[ P(A \cup B) = 1.40 - 0.48 = 0.92 \][/tex]
Hence, [tex]\( P(A \cup B) = 0.92 \)[/tex].
### Summary
(a) [tex]\( P(A \cap B) = 0.48 \)[/tex]
(b) [tex]\( P(A \cup B) = 0.92 \)[/tex]
These are the probabilities for the given events [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
