Question

Suppose two different methods are available for eye surgery. The probability that the eye has not recovered in a month is 0.002 if method A is used. When method B is used, the probability that the eye has not recovered in a month is 0.005. Assume that 40% of eye surgeries are done with method A and 60% are done with method B in a certain hospital. If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed (write it up to second decimal place)? a) 0.5 b) 0.6 c) 0.8 d) 0.4 e) 0.7

Answer 1

d) 0.4

Explanation:

P(Ra) = 0.998

P(Rb) = 0.995

P(A) = 0.4

P(B) = 0.6

The probability that method A was performed, given that the eye was recovered within a month of the surgery, is determined by the probability that the eye recovered within a month with method A (P(A) x P(Ra)),divided by the probability that the eye recovered within a month with either method (P(A) x P(Ra) + P(B) x P(Rb)):

`P(A|R)=(P(A) *P(R_A))/(P(A) *P(R_A)+P(B) *P(R_B))\\P(A|R) = (0.4*0.998)/(0.4*0.998+0.6*0.995) \\P(A|R) = 0.40`

The probability that method A was performed is 0.40.