If A and B are independent events such that P(A) = 0.9 and P(B) = 0.8 then $$p(A \cap \overline{B}) + P(\overline{A} \cap B) =$$
p(A) = 0.9 hence, p($$\overline{\ A}$$) = 0.1
p(B) = 0.8 hence, p($$\overline{\ B}$$) = 0.2
Now, p($$A\ intersetion\overline{\ B}$$) = p(A)*p($$\overline{\ B}$$) = 0.18
Alos, p($$\overline{\ A}\ intersetion\ B$$) =Â p($$\overline{\ A}$$)*p(B) = 0.08Â
Hence, the sum of the two terms = 0.26
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