Question

Suppose that in an adult population the proportion of people who are both overweight and suffer hypertension is 0.03; the proportion of people who are not overweight but suffer hypertension is 0.22; the proportion of people who are overweight but do not suffer hypertension is 0.05; and the proportion of people who are neither overweight nor suffer hypertension is 0.82. An adult is randomly selected from this population.

Find the probability that the person selected suffers hypertension given that he is overweight.

Find the probability that the selected person suffers hypertension given that he is not overweight.

Answer 1

let probability of being overweight and suffer hypertension is P(O) and P(H)

therefore P(O n H)=0.03

P(O n H^c)=0.05

P(O^c n H^c) =0.82

hence

P(O)=P(O n H^c)+P(O n H) =0.03+0.05 =0.08

also as P(O u H) =1-P(O^c n H^c) =1-0.82 =0.18

P(H) =P(O u H)+P(O n H)-P(O)=0.18+0.03-0.08=0.13

a)probability that the person selected suffers hypertension given that he is overweight =P(H|O)=P(O n H)/P(O)

=0.03/0.08=0.3750

b)

probability that the selected person suffers hypertension given that he is not overweight =P(H|O^c)=P(H n O^c)/P(O^c)

=P(H n O^c)/P(O^c)=(0.13-0.03)/(1-0.08)=0.1087