Biostatistics/ Probability

Question

Suppose that it is known that a fraction 0.001 of the people in a town have tuberculosis (TB). A TB test is available. The sensitivity of a given test is the probability of correctly diagnosing a person who has the disease while the specificity of a test is the probability of correctly diagnosing a person who doesn’t have the disease. The sensitivity of the available TB test is 0.999 and its specificity is 0.998.

i) For one randomly selected person, what is the probability that the test will indicate that he has TB?

ii) For one randomly selected person, the test shows that he has TB, what is the probability that he really does?

iii) For one randomly selected person, the test shows that he doesn’t have TB, what is the probability that he really does not have TB? 

PS. Define all the events and detailed solution 😭 using Baye’s theorem

Answer 1

The Bayes' Theorem is this:

P(A | B) = P(B | A)· P(A)

            P(B)

Let's represent all the given with this:

P(TB) = 0.001

P( + | TB) = 0.999

P( - | nTB) = 0.998

What we are solving for are these:

(i) P ( + ) => Probability that the person is positive for TB.

(ii) P (TB | +) => Probability that he has TB given that he is tested positive.

(iii) P (nTB | - ) => Probability that he has no TB given that he is tested negative.

This is the diagram:

                 ___0.999__ ( + )

                  |

___0.001____ TB___ |

|                  |

|                  |___0.001__ ( - )

|

|                  ___0.002__ ( + )

|                  |

|___0.999_____ nTB _|

                 |

                  |___0.998__ ( - )

Therefore:

(i) P ( + ) = P ( TB ⋂ +) + P (nTB ⋂ -)

= (0.001)(0.999) + (0.999)(0.002) = 0.002997

(ii) Now using the Bayes’ Theorem:

P (TB | +) = P (+ | TB)· P(TB)

             P( + )

Substitute the values:

 P (TB | +) = (0.999) (0.001) = 0.33... or 1/3

            (0.002997)

(iii) Using again the Bayes’ Theorem:

P (nTB | -) = P (- | nTB)· P(nTB)

            P( - )

If P(TB) = 0.001, then P(nTB) =1-0.001 = 0.999.

Furthermore, using the table:

P( - ) = (0.001)(0.001) + (0.999)(0.998) =0.997003

Substitution, we’ll have:

P (nTB | -) = (0.998) (0.999) = 0.999998996994

             0.997003