Answer to Question #104075 in Statistics and Probability for VICTOR KASEYA

Question

A new medical drug to cure HIV/AIDS has just been developed. The probability that it will cure HIV/AIDS is 0.9 if the patient has HIV/AIDS. The probability that it will cure HIV/AIDS falls to 0.4 if the patient does not have the disease. There is a probability of 0.3 that a person chosen at random from the community has HIV/AIDS. Find the following: i. Probability that a patient is cured of HIV/AIDS [12 Marks) Probability that a person is cured of HIV/AIDS actually had HIV/AIDS. [8 Marks)

Answer 1

Let D denote presence of a HIV, A denote absence of HIV, + denote cured and - denote uncured.

i. P(+)

$P(D)=0.3, P(A)=1-P(D)= 0.7$

$P(+|D)=0.9$

Thus, $0.9=\frac{P(+,D)}{P(D)}$

Implying that $P(+,D)=0.9×P(D)=0.9×0.3=0.27$

Similarly, $P(+|A)=0.4$

Thus, $0.4=\frac{P(+,A)}{P(A)}$

Implying that $P(+,A)=0.4×P(A)=0.4×0.7=0.28$

Therefore, $P(+)=P(+,D)+P(+,A)=0.27+0.28=0.55$

ii. P(D|+)

$P(D|+)=\frac{P(+,D)}{P(+)}=\frac{0.27}{0.55}=0.49091$