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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(A),P(B) and $P(A \cap B)$

$\begin{array}{1 1}(A)\;0.25,0.32,0.17\\(B)\;0.35,0.42,0.27\\(C)\;0.45,0.42,0.37\\(D)\;\text{None of these}\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:
$P(A)=\large\frac{n(e_1,e_5,e_8)}{n(S)}$
$\Rightarrow \large\frac{0.08+0.1+0.07}{1}$
$\Rightarrow 0.25$
Step 3:
$P(B)=\large\frac{n(e_2,e_5,e_8,e_9)}{n(S)}$
$\Rightarrow \large\frac{0.08+0.1+0.07+0.7}{1}$
$\Rightarrow 0.32$
Step 4:
$P(A \cap B)=P(e_1,e_5,e_8) \cap P(e_2,e_5,e_8,e_9)$
$\Rightarrow P(e_5,e_8)$
$\Rightarrow 0.1+0.07$
$\Rightarrow 0.17$
Hence (A) is the correct answer.
answered Jul 8, 2014 by sreemathi.v
edited Jul 8, 2014 by sreemathi.v
 

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