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Given P(A cup B)=0.6 , P(A capB)=0.2 , ...

Given ` P(A cup B)=0.6 , P(A capB)=0.2` , then probability of exactly one of the event occurs is

A

`0.4`

B

`0.2`

C

`0.6`

D

`0.8`

Text Solution

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The correct Answer is:
A
Given , `P( A cup B)=0.6, P(A cap B)=0.2`
Probability of exactly one of the event occurs is
`P(bar(A) cap B)+P(A cap barB)`
`=P(B)-P(A cap B)+P(A)-P(A cap B)`
`=P(A cup B) + P(A cap B)-2P(A cap B)" "[ :' P( A cupB)=P(A)+P(B)-P(A cap B)]`
`= P( A cup B) -P( A cap B)`
`=0.6-0.2`
`=0.4 `
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