Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

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} $

Can you answer this question?

1 Answer

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)=\(\Large\frac{{P(E_2)}{P(A/E_2)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}\)
\(P(E_1\))=\(P(one\; in\; thousand \;having\; T\;B\))


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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App