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?

Question

Leo Burton

Mathematics

28 February, 01:00

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?

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Answers (1)

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Penelope Warren · Answer 1

0.4007

Step-by-step explanation:

Let's define the following events:

A: method A is used

B: method B is used

NR: the eye has not recovered in a month

R: the eye is recovered in a month

The probability that the eye has not recovered in a month is 0.002 if method A is used, i. e., P (NR|A) = 0.002, so P (R|A) = 0.998.

When method B is used, the probability that the eye has not recovered in a month is 0.005, i. e., P (NR|B) = 0.005, so P (R|B) = 0.995.

40% of eye surgeries are done with method A, i. e., P (A) = 0.4

60% of eye surgeries are done with method B, i. e., P (B) = 0.6

If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed? We are looking for P (A|R), then, by Bayes' Formula

P (A|R) = P (R|A) P (A) / (P (R|A) P (A) + P (R|B) P (B)) = 0.998*0.4 / (0.998*0.4 + 0.995*0.6) = 0.4007