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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 $A=\{e_1,e_5,e_8\},B=\{e_2,e_5,e_8,e_9\}$.Calculate $P(\bar{B})$ from P(B),also calculate $P(\bar{B})$ directly from the elementary outcomes of $\bar{B}$.

$\begin{array}{1 1}(A)\;0.68\\(B)\;0.78\\(C)\;0.48\\(D)\;0.18\end{array} $

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1 Answer

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Step 1:
Given :
$P(e_1)=P(e_2)=0.08$
$P(e_3)=P(e_4)=P(e_5)=0.1$
$P(e_6)=P(e_7)=0.2$
$P(e_8)=P(e_9)=0.07$
$A=\{e_1,e_5,e_8\}$
$B=\{e_2,e_5,e_8,e_9\}$
Addition of all probabilities will be equal to 1
Step 2:
Probability of $\bar{B}$ from P(B)
$\Rightarrow P(\bar{B})=1-P(B)$
$\Rightarrow 1-0.32$
$\Rightarrow 0.68$
Step 3:
Probability of $\bar{B}$ from the elementary outcomes
$\Rightarrow$ Probability of ($e_1,e_2.........e_9)-P(e_2,e_5,e_8,e_9)$
$\therefore P(e_1,e_3,e_3,e_6,e_1)$
$\Rightarrow 0.08+0.1+0.1+0.2+0.2$
$\Rightarrow 0.68$
Hence (A) is the correct answer.
answered Jul 8, 2014 by sreemathi.v
 

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