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Home/Practice questions/General Aptitude Test
General Aptitude TestQuantitative ReasoningProbability

If A and B are mutually exclusive events, then following probability…

If A and B are mutually exclusive events, then following probability of the events are given below:

(A) \(P(A)=0.4\), \(P(B)=0.4\) and \(P(A\cup B)=0.8\)

(B) \(P(A)=0.3\), \(P(B)=0.6\) and \(P(A\cup B)=0.9\)

(C) \(P(A)=0.3\), \(P(B)=0.3\) and \(P(A\cup B)=0.6\)

(D) \(P(A)=0.1\), \(P(B)=0.1\) and \(P(A\cup B)=0.2\)

which of the following options is correct?

Answer & explanation+

Correct answer: option 3

The correct answer is Option (3) → (A), (B), (C) and (D)

Mutually exclusive events cannot happen together, so \(P(A\cap B)=0\) and the addition rule becomes \(P(A\cup B)=P(A)+P(B)\). Each set of values is tested against this rule (every value also lies between 0 and 1).

(A) \(P(A)=0.4\), \(P(B)=0.4\) and \(P(A\cup B)=0.8\) Correct. \(0.4+0.4=0.8\), as given.

(B) \(P(A)=0.3\), \(P(B)=0.6\) and \(P(A\cup B)=0.9\) Correct. \(0.3+0.6=0.9\), as given.

(C) \(P(A)=0.3\), \(P(B)=0.3\) and \(P(A\cup B)=0.6\) Correct. \(0.3+0.3=0.6\), as given.

(D) \(P(A)=0.1\), \(P(B)=0.1\) and \(P(A\cup B)=0.2\) Correct. \(0.1+0.1=0.2\), as given.

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