Question 340354
 Two local surgeons demonstrated that, when surgeons perform leg or arm amputations, it is conceivable that they might operate on the wrong leg or arm. Suppose that two surgeons performing on a leg are able to choose the correct leg with a probability of 0.80 and that their choice in one operation is independent of their choice in any other. Suppose that they perform three amputations. What is the probability that they will amputate the wrong leg in two of the operations?
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Binomial Problem with p = 0.2 and n=3
P(x = 2) = 3C2(0.2)^2(0.8) = 3*0.04*0.8 = 0.096
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Cheers,
Stan H
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a 0.48
b 0.032
c 0.384
d 0.096