NIMCET 2008 Question Paper with solutions

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Question : 1 of 120

Marks: +1, -0

If two events A and B such that

P(A^{'})=0.3, P(B)=0.5

and

P(A ∩ B)=0.3

then

P( B /{A ∪ B^{r}})

is

${1}/{4}$
${3}/{8}$
${1}/{8}$
None of these

Solution:

P(A)=1-P(A)=1-0.3

=0.7

P(B)=1-P(B)=1-0.5

=0.5

P(A ∩ B)=P(A)-P(A ∩ B)

=0.7-0.3=0.4

⇒ P(A ∪ B)=P(A)+P(B)-P(A ∩ B)

=0.7+0.5-0.4

=0.8

Now,

P( {B}/{A ∪ B^{'}})= {P[B ∩(A ∪ B^{'})]}/{P(A ∪ B)}

= {P[(B ∩ A) ∪(B ∩ B)]}/{P(A ∪ B)}

= {P(B ∩ A)}/{P(A ∪ B)}= {0.3}/{0.8}= {3}/{8}

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