78.1k views
4 votes
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

1 Answer

1 vote

Answer:

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.

User Lateef
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories