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Home  >>  CBSE XI  >>  Math  >>  Probability
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If A and B are mutually exclusive events,$P(A)=0.35$ and $P(B)=0.45$ find $P(A \cup B)$

$\begin{array}{1 1}(A)\;0.70\\(B)\;0.80\\(C)\;0.50\\(D)\;0.60\end{array} $

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1 Answer

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  • $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
A and B are mutually exclusive events
$\therefore P(A \cap B)=\phi$=0
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.35+0.45-0$
$\Rightarrow 0.80$
Hence (B) is the correct answer.
answered Jul 7, 2014 by sreemathi.v
 
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