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By examining the chest X ray,the probability that TB is detected when a person is actually suffering is 0.99.The probability of an healthy person diagnosed to have TB Is 0.001.In a certain city,1 in 1000 people suffers from TB.A person is selected at random and is diagnosed to have TB.What is the probability that he actually has TB?

$\begin{array}{1 1}(A)\;\large\frac{110}{221}\\(B)\;\large\frac{120}{221}\\(C)\;\large\frac{130}{221}\\(D)\;\large\frac{112}{221}\end{array} $

1 Answer

Toolbox:
  • \(E_1=A\;person\;actully\;has\;T\:B\)
  • \(E_2=A\;person\;actully\;has\;T\:B\)
  • \(A\; person\;actuley\;has\;T\;B\;/he\;is\;diagonised\;as\;having\;T\;B\)
  • P(E\(_2\)/A)=\(\Large\frac{{P(E_2)}{P(A/E_2)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}\)
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\(P(A/E_2\))=\(P(diagonised\;as\;having\;T\;B\;/actully\;has\;T\:B\))
=\(0.99\)
\(P(A/E_2\))=\(P(diagonised\;as\;having\;T\;B\;/has\;not\;have\;T\:B\))
=\(0.001\)
\(P(E_1\))=\(P(one\; in\; thousand \;having\; T\;B\))
=\(\frac{1}{1000}\)
\(p(E_2)=1-\Large\frac{1}{1000}=\frac{999}{1000}\)
P(E\(_1\)/A)
\(\Large\frac{\frac{1}{1000}\times0.99}{\frac{1}{1000}\times0,99+\frac{999}{1000}\times0.001}\)
=\(\Large\frac{.999}{1.989}\)
=\(\Large\frac{110}{221}\)

 

answered Mar 2, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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