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Home  >>  CBSE XII  >>  Math  >>  Probability
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If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find \[ \begin{array} ((i) P(A \cap B) \quad & (ii) P(A|B) \quad & (iii) P(A ∪ B) \end{array} \]

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  • \(p(B/A)\;=\;\large \frac{p(A\;\cap\;B)}{p(A)}\) \(\Rightarrow\;p(A\;\cap\;B)=p(B/A)\;p(A)\)
  • \(\;p(A\;\cup\;B)\;=\;p(A)\;+\;p(B)\;-\;p(A\;\cap\;B)\;\)
  • \(p(A/B)\;=\;\large \frac{p(A\;\cap\;B)}{p(B)}\)
Given \(p(A)=0.8\;p(B)=0.5\;p(B/A)=0.4\)
$\Rightarrow$ \(p(A\;\cap\;B)=0.4\;\times\;0.8\) = \(0.32\)
\(p(A/B)=\large \frac{p(A\;\cap\;B)}{p(B)}\) $\rightarrow$ \(=\large\frac{0.32}{0.5}=\frac{32}{50}\)\(=0.64\)
\(\;p(A\;\cup\;B)\;=\;p(A)\;+\;p(B)\;-\;p(A\;\cap\;B)\;\) $\rightarrow$ \(p(A\;\cup\;B)=0.8\;+\;0.5\;-\;0.32\) =\(0.98\)
answered Mar 20, 2013 by poojasapani_1
edited Jun 18, 2013 by balaji.thirumalai
 

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