The probabilities of three events {\A}, {\B} and {\C} are given by {\P}({\A})=0.6, {\P}({\B})=0.4 and {\P}({\C})=0.5 If {\P}({\A}\cup {\B})=0.8, {\P} ({\A} \cap {\C})=0.3, {\P}({\A} \cap {\B} \cap {\C})=0.2, {\P}({\B} \cap {\C})=\\beta and {\P}({\A}\cup {\B}\cup {\C})=\\alpha where 0.85 \leq \\alpha \leq 0.95, then \\beta lies in the interval: [6 Sep 2020 Shift 2]

Correct Answer is : [0.25,0.35]

Probability Part 1

Section: Mathematics

Question : 25 of 100

Marks: +1, -0

The probabilities of three events A,B and C are given by P(A)=0.6,P(B)=0.4 and P(C)=0.5 If P(A∪B)=0.8, P(A∩C)=0.3, P(A∩B∩C)=0.2, P(B∩C)=β and P(A∪B∪C)=α where 0.85≤α≤0.95, then β lies in the interval:

[6 Sep 2020 Shift 2]

[0.36,0.40]
[0.35,0.36]
[0.25,0.35]
[0.20,0.25]

Validate

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