Suppose that [tex]$A$[/tex] and [tex]$B$[/tex] are independent events such that [tex]$P(\bar{A})=0.30$[/tex] and [tex]$P(B)=0.10$[/tex]. Find [tex][tex]$P(A \cap B)$[/tex][/tex] and [tex]$P(A \cup B)$[/tex]. (If necessary, consult a list of formulas.)

(a) [tex]$P(A \cap B)=$[/tex] [tex]$\square$[/tex]

(b) [tex]$P(A \cup B)=$[/tex] [tex]$\square$[/tex]



Answer :

Sure, let's solve this step by step.

Given:
- [tex]\( P(\bar{A}) = 0.30 \)[/tex]
- [tex]\( P(B) = 0.10 \)[/tex]

We need to find:
- [tex]\( P(A \cap B) \)[/tex]
- [tex]\( P(A \cup B) \)[/tex]

Step 1: Calculate [tex]\( P(A) \)[/tex].

We know that [tex]\( P(\bar{A}) \)[/tex] is the probability of the complement of [tex]\( A \)[/tex]. The sum of the probabilities of an event and its complement is always 1.

[tex]\[ P(A) = 1 - P(\bar{A}) \][/tex]
[tex]\[ P(A) = 1 - 0.30 \][/tex]
[tex]\[ P(A) = 0.70 \][/tex]

Step 2: Calculate [tex]\( P(A \cap B) \)[/tex].

Since [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent events, the probability of their intersection is given by the product of their individual probabilities:

[tex]\[ P(A \cap B) = P(A) \times P(B) \][/tex]
[tex]\[ P(A \cap B) = 0.70 \times 0.10 \][/tex]
[tex]\[ P(A \cap B) = 0.07 \][/tex]

So,
[tex]\[ P(A \cap B) = 0.07 \][/tex]

Step 3: Calculate [tex]\( P(A \cup B) \)[/tex].

The probability of the union of two events is given by:

[tex]\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \][/tex]
[tex]\[ P(A \cup B) = 0.70 + 0.10 - 0.07 \][/tex]
[tex]\[ P(A \cup B) = 0.80 - 0.07 \][/tex]
[tex]\[ P(A \cup B) = 0.73 \][/tex]

So,
[tex]\[ P(A \cup B) = 0.73 \][/tex]

Final answers:

(a) [tex]\( P(A \cap B) = 0.07 \)[/tex]

(b) [tex]\( P(A \cup B) = 0.73 \)[/tex]