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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 Comment A) Need homework help? Click here. 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}$$