Question

We have an illness, where 2% of the polulation is infected, and we have a test with sensitivity 94% and specificity 98%.

A="infected" and B="tests positive"

Question:

20 people gets tested, and all tests negative. What is the probability that atleast one of these 20 really are infected?

Answer 1

infected people =2% =0.02

non infected (or healthy) people =100-2 =98% =0.98

Sensitivity = 94% = 0.94

Specificity = 98% = 0.98

Sensitivity = (number of true positive test)/(infected people)

Specificity = (number of true negative test)/healthy people

Hence we draw the tree diagram as shown

P(infected | test negative ) =P(infected and test negative)/P(test negative)

                                             =0.06/(0.02*0.06 + 0.98*0.98)

                                             =0.06/0.9616

                                             =0.062396

So,

n=20

p=0.062396

q=1-p = 1-0.062396 =0.9676

P=nCr p^r q^(n-r)

P(atleast one infected) = P(x>=1) = 1-P(x=0)

                                                            = 1 - 20C0 (0.062396)^0 (0.9676)^20

                                                            =1-0.51750815

                                                            =0.48249