Question

A large group of people is to be checked for two common symptoms of a certain disease. Itis thought that 20% of the people possess symptomAalone, 30% posseess symptomBalone,10% possess both symptoms, and the remainder have neither symptom. For one person chosen atrandom from this group, Ønd these probabilities:

a. (2 points) that the person has neither symptom
P(A\πB)=0:20
P(πA\B)=0:30
P(A\B)=0:10
P(πA\πB)=1°P(A[B)
P(πA\πB)=1°[P(A\πB)+P(πA\B)+P(A\B)]
=1°(0:20+0:30 + 0:10)
=1°0:60
=0:40
b. (2 points) that the person has at least one symptom
P(A[B)=P(A\πB)+P(πA\B)+P(A\B)
=0:20 + 0:30+10
=0:60
or
P(A[B)=1°P(πA\πB)
=1°0:40
=0:60
c. (2 points) that the person has both symptoms, given that he has symptom
BP(AjB)=P(A\B)/P(B)
=P(A\B)P(A\B)+P(πA\B)
=0:100:10+0:30
=0:100:40
=0:25

Answer 1

(a) and (b) are correct but (c) is not correct

Step-by-step explanation: (c) P(nA)*P(nB) + P(nA/nB)

0.20*0.30 + 0.10 = 0.060 + 0.10 = 0.160