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JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-II

  • question_answer
    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. Then, the probability that the person actually has TB is

    A) \[\frac{110}{221}\]

    B) \[\frac{2}{223}\]

    C) \[\frac{110}{223}\]       

    D) \[\frac{1}{221}\]

    Correct Answer: A

    Solution :

    [a] Let A denote the event that the person has TB Let B denote the event that the person has not TB. Let E denote the event that the person is diagnosed to have TB. \[\therefore P(A)=\frac{1}{1000},P(B)=\frac{999}{1000}\] \[P\left( \frac{E}{A} \right)=0.99,P\left( \frac{E}{B} \right)=0.001\] The required probability is given by \[P\left( \frac{A}{E} \right)=\frac{P(A)\times \left( \frac{E}{A} \right)}{P(A)\times P\left( \frac{E}{A} \right)+P(B)\times P\left( \frac{E}{B} \right)}\] \[=\frac{\frac{1}{1000}\times 0.99}{\frac{1}{100}\times 0.99+\frac{999}{1000}\times 0.01}=\frac{110}{221}\]


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