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Home  >>  CBSE XII  >>  Math  >>  Probability
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If A and B are two events such that $P(A)=\Large\frac{1}{2},\normalsize P(B)=\Large\frac{1}{3},\normalsize P(A/B)=\Large \frac{1}{4},$then $P(A'\cap B')$ equals

$\begin{array}{1 1}(A)\;\frac{1}{12}\\(B)\;\frac{3}{4}\\(C)\;\frac{1}{4}\\(D)\;\frac{3}{16}\end{array} $

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1 Answer

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Toolbox:
  • \(P(A\cap\;B)=P(\overline{A\cup\;B})\)
  • =1-P(\(A\cup\)B)
  • p(A\(\cup\)B)=p(A)+P(B)-p(\(A\cup\)B)
  • \(\large\frac{1}{4}\;=\large\;p(A\cap\;B)\;\times\;\frac{3}{1}\)
  • P(\(A\cap\)B)=\(\large\frac{1}{4}\)\(\times\)\(\large\frac{1}{3}\)=\(\large\frac{1}{12}\)
  • P(\(A\cup\)B)=\(\large\frac{1}{2}\)+\(\large\frac{1}{3}\)-\(\large\frac{1}{12}\)
  • =\(\Large\frac{3}{4}\)
  • \(1-P(\bar{A}\cup\;\bar{B}\;)=1-p(A\cup\;B)=1-\;\large\frac{3}{4}=\frac{1}{4}\)
  • 'c' option is correct

 

answered Feb 23, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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