AP ICET 2017 Shift 1 Paper

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Question : 147 of 200

Marks: +1, -0

If A and B are independent events such that

P(A)=0.9

and

P(B)=0.8

then

P(A∩\ov{B})+P(\ov{A}∩B)=

$0.26$
$0.36$
$0.16$
$0.18$

Solution:

Given:

p(A)=0.9⇒p(A↖{-})=0.1

p(B)=0.8⇒P(B↖{-})=0.2

p(A∩\ov{B})+p(\ov{A}∩{B})=(0.9×0.2)+(0.1×0.8)

=0.18+0.08=0.26

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