The probabilities of events A and B are given by P(A) = 0.6 and P(B) = 0.22 respectively.?

The probabilities of events A and B are given by P(A) = 0.6 and P(B) = 0.22 respectively.


Find


(a) P(A⋃ B) , if A and B are mutually exclusive events.


(b) P(A ⋃ B) , if A and B are independent events.


(c) P(A ⋂ B ' ) , if A and B are independent events.

The probabilities of events A and B are given by P(A) = 0.6 and P(B) = 0.22 respectively.?
(a) I'll assume you're asking for P(A u B). Since A and B are mutually exclusive, then





P(A u B) = P(A) + P(B) = 0.6 + 0.22 = 0.82.





(b) OK, then it's P(A u B). Since A and B are independent, then





P(A n B) = P(A)P(B) = (0.6)(0.22) = 0.132.





So





P(A u B) = P(A) + P(B) - P(A n B) using the General Addition Rule.


= 0.6 + 0.22 - 0.132


= 0.688





Thank you for clarifying.





(c) I'll again assume P(A n B'). Since A and B are independent, then A and B' are independent. So





P(A n B') = P(A)P(B') = P(A){1-P(B)} = (0.6)(1-0.22) = 0.468.





If I assumed incorrectly, let me know, but I can't read the symbols you have above.





edit: d_c, it is against the rules to use placeholders.
Reply:Note:


P(A⋃ B) = P(A) + P(B) - P(A ⋂ B )





a. it means P(A ⋂ B ) = 0. [Answer: 0.82]


b. it means P(A ⋂ B ) = P(A) * P(B)


c. P(A ⋂ B ' ) = P(A) * P(B') = P(A) * {1-P(B)}





Finish solving this now.





d:
Reply:a) (0.6 + 0.22) - 0 = 0.82


b) (0.6 + 0.22) - (0.6 x 0.22) =


(0.82 - 0.132) = 0.688


c) (0.6 x (1-0.22)) = 0.468

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