Given two events A and B, if the odds against Aare 2 to 1, and those in favour of A∪B are 3 to1, then
A
13≤P(B)≤12
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B
12≤P(B)≤34
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C
512≤P(B)≤34
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D
0≤P(B)≤1
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Solution
The correct option is C512≤P(B)≤34 OddsagainstA=P(Ac)/P(A)=21⇒{1−P(A)}/P(A)=2⇒1/P(A)−1=2⇒1/P(A)=3⇒P(A)=1/3AlsoOddsinfavourofA∪B=P(A∪B)P(A∪B)c=31⇒P(A∪B)1−P(A∪B)=3⇒1−P(A∪B)P(A∪B)=13⇒1P(A∪B)=1+13=43⇒P(A∪B)=34
since we know that P(A∪B)=P(A)+P(B)−P(A∩B)
Hence putting the values of results obtained above we get ⇒34=13+P(B)−P(A∩B)