Are Events A and B Independent?

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The discussion centers on the independence of two events A and B, with probabilities P(A) = 0.2 and P(B) = 0.4, and their intersection P(A ∩ B) = 0.08. The events are not independent, as the product of their probabilities (0.2 * 0.4 = 0.08) equals the intersection probability. The probability of either A or B or both occurring is calculated as P(A ∪ B) = 0.68, while the probability that neither occurs is 0.32. The probability of exactly one of the events occurring is determined to be 0.44.

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joe007
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The two events A and B have probabilities 0.2 and 0.4. Also P(A intersect B) = 0.08.

(a) Are the two events A and B independent? Explain.
well since 0.2*0.4=0.8 they are independent

(b) Find the probability that either A or B or both occur.
0.2+0.4+0.08=0.68
(c) Find the probability that neither A nor B occurs.
1-.68=0.32
(d) Find the probability that exactly one of A or B occurs.
0.2+0.4-.08=0.52

not sure could someone give feedbacks cheers
 
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joe007 said:
The two events A and B have probabilities 0.2 and 0.4. Also P(A intersect B) = 0.08.

(a) Are the two events A and B independent? Explain.
well since 0.2*0.4=0.8 they are independent

0.2*0.4=0.08

(b) Find the probability that either A or B or both occur.
0.2+0.4+0.08=0.68

Why are you adding the intersection? P(A)vP(B)= P(A)+P(B)-P(A)P(B)=0.52

(c) Find the probability that neither A nor B occurs.
1-.68=0.32

1-(P(A)+ P(B)-P(A)P(B))= 0.48

(d) Find the probability that exactly one of A or B occurs.

P(only A)= P(A)-P(A)P(B)=0.4-0.08 =0.32

P(only B)=P(B)-P(A)P(B)=0.2-0.08 = 0.12

P(only A)+P(only B)=0.44
 
Last edited:

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