The probability of a duck exhibiting botulism (caused by C. botulinum) is .3 and

Question

The probability of a duck exhibiting botulism (caused by C. botulinum) is .3 and the probability that the duck actually has botulism is .35. The probability that a duck shows symptoms that are ascribed to botulism and is actually infected with the bacteria is .24.

Q: A duck selected at random does not show botulism symptoms, what is the probability that this duck is infected with the C. botulinum.

The answer is P(Inf.|No symptoms) = P(Inf| No symptoms) / P(No symptoms) = .11/.7=.843.

How do you calculate P(Inf|No symptoms)? I used P(A & B)/ P(B), but wouldn't the P(B) just cancel out? Please show work.

Explanation / Answer

probabilty of symptoms =P(S)=0.3

probability of actuually having =P(A)=0.35

probabilty of symptoms and actually having=P(S&A)=0.24

probabilty that duck does not show symptoms =P(Sc)=1-0.3 =0.7

a) probability that this duck is infected with the C. botulinum, given does not show symptoms

=P(A|Sc )=P(A&Sc)/P(Sc) =(P(A)-P(A&S))/P(Sc)=(0.35-0.24)/0.7 =0.157

please revert

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