Answer: a) No, b) 0.78, c) No.
Step-by-step explanation:
Since we have given that
P(A) = 0.7
P(B) = 0.2
P(A∩B) = 0.12
(a) Are A and B independent? Explain.
Since P(A).P(B) = [tex]0.7\times 0.2=0.14[/tex]
And P(A∩B) = 0.12
P(A∩B) ≠ P(A).P(B)
So, they are not independent.
(b) Compute P(A or B).
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\P(A\cup B)=0.7+0.2-0.12\\\\P(A\cup B)=0.9-0.12\\\\P(A\cup B)=0.78[/tex]
(c) Are A and B mutually exclusive? Explain
Since P(A∩B) ≠ 0
So, they are not mutually exclusive.
Hence, a) No, b) 0.78, c) No.
