If `A and B` are two events, the probability that at most one of these events occurs isA. `1-P(A cap B)`B. `P(overline(A))+P(overline(B))-P(overline(A) cap overline(B))`C. `P(overline(A))+P(overline(B))+P(A cup B)-1`D. all the above

If `A and B` are two events, the probability that at most one of these

1.	If `A and B` are two events, the probability that at most one of these events occurs isA. `1-P(A cap B)`B. `P(overline(A))+P(overline(B))-P(overline(A) cap overline(B))`C. `P(overline(A))+P(overline(B))+P(A cup B)-1`D. all the above
Answer» Correct Answer - D We have, Required probability `=P(overline(A) cup overline(B))=P(overline(A cap B))=1-P(A cap B)` So, alternative (a) is correct. Again, `P(overline(A) cup overline(B))=P(overline(A))+P(overline(B))-P(overline(A) cap overline(B))` [ By add. Theorem] So, alternative (b) is correct. Again, `P(overline(A) cup overline(B))` `=P(overline(A))+P(overline(B))-P(overline(A) cap overline(B))` `=P(overline(A))+P(overline(B))-P(overline(A cup B))` `=(P(overline(A))+P(overline(B))-{1-P(A cup B)}` `=P(overline(A))+P(overline(B))+P(A cup B)-1`. So, alternative (c ) is also correct.

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If `A and B` are two events, the probability that at most one of these events occurs isA. `1-P(A cap B)`B. `P(overline(A))+P(overline(B))-P(overline(A) cap overline(B))`C. `P(overline(A))+P(overline(B))+P(A cup B)-1`D. all the above

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