NIMCET 2016 Question Paper with solutions

Question : 73 of 120

Marks: +1, -0

For any two events A and B, the probability that at least one of them occur is 0.6. If A and B occurs simultaneously with a probability 0.3 then P(A’) + P(B’) is;

0.9
1.15
1.1
1.0

Validate

Solution:

Given that,

P(A∪B)=0.6=3/5

P(A∩B)=0.3=3/10

Then, by probability law
P(A ∪ B ) = P(A) + P(B) - P(A ∩ B)

⇒3/5=P(A)+P(B)-3/10

⇒P(A)+P(B)=3/5+3/10

={30+15}/50=45/50

⇒P(A)+P(B)=9/10

⇒1-P(\ov{A})+1-P(\ov{B})=9/10

[∵P(A)+P(\ov{A})=1]

⇒2-P(\ov{A})-P(\ov{B})=9/10

⇒P(\ov{A})+P(\ov{B})=2-9/10

={20-9}/10

⇒P(\ov{A})+P(\ov{B})=11/10=1.1

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