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Home  >>  CBSE XI  >>  Math  >>  Probability
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A sample space consists of a elementary outcomes $e_1,e_2........e_9$ whose probabilities are $P(e_1)=P(e_2)=0.08,P(e_3)=P(e_4)=P(e_5)=.1,P(e_6)=P(e_7)=.2,P(e_8)=p(e_9)=0.07$.Suppose $A=\{e_1,e_5,e_8\},B=\{e_2,e_5,e_8,e_9\}$.List the composition of the event $A\cup B$,and calculate $P(A \cup B)$ by adding the probabilities of the elementary outcomes.

$\begin{array}{1 1}(A)\;0.40\\(B)\;0.35\\(C)\;0.45\\(D)\;\text{None of these}\end{array} $

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1 Answer

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Step 1:
$P(e_1)=P(e_2)=0.08$
$P(e_3)=P(e_4)=P(e_5)=0.1$
$P(e_6)=P(e_7)=0.2$
$P(e_8)=P(e_9)=0.07$
$A=\{e_1,e_5,e_8\}$
$B=\{e_2,e_5,e_8,e_9\}$
Addition of all probabilities will be equal to 1
Step 2:
Considering the composition of the event
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow P(e_1,e_5,e_8)+P(e_2,e_5,e_8,e_9)-P(e_5,e_8)$
$\Rightarrow 0.08+0.1+0.07+0.08+0.1+0.07+0.07-0.1-0.07$
$\Rightarrow 0.40$
Hence (B) is the correct answer.
answered Jul 8, 2014 by sreemathi.v
 

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