$\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} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- \(E_1=A\;person\;actully\;has\;T\:B\)
- \(E_2=A\;person\;actully\;has\;T\:B\)
- \(A\; person\;actuley\;has\;T\;B\;/he\;is\;diagonised\;as\;having\;T\;B\)
- P(E\(_2\)/A)=\(\Large\frac{{P(E_2)}{P(A/E_2)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}\)

\(P(A/E_2\))=\(P(diagonised\;as\;having\;T\;B\;/actully\;has\;T\:B\))

=\(0.99\)

\(P(A/E_2\))=\(P(diagonised\;as\;having\;T\;B\;/has\;not\;have\;T\:B\))

=\(0.001\)

\(P(E_1\))=\(P(one\; in\; thousand \;having\; T\;B\))

=\(\frac{1}{1000}\)

\(p(E_2)=1-\Large\frac{1}{1000}=\frac{999}{1000}\)

P(E\(_1\)/A)

\(\Large\frac{\frac{1}{1000}\times0.99}{\frac{1}{1000}\times0,99+\frac{999}{1000}\times0.001}\)

=\(\Large\frac{.999}{1.989}\)

=\(\Large\frac{110}{221}\)

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...