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20 July, 04:27

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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  1. 20 July, 05:52
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    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
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