If $P(A)=0.3$, $P(B)=0.4$ and A, B are mutually exclusive, then $P(A\cup B)$ is:
- 0.12
- 0.1
- 1.0
- 0.7
Show answer and solution
Correct answer: D. 0.7
For mutually exclusive events, P(A∪B) = P(A)+P(B) = 0.3+0.4 = 0.7.
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Correct answer: D. 0.7
Correct answer: D. Integration by parts with $u = x$, $dv = \cos x\, dx$