There is a 2% chance that a woman has a specific chromosomal abnormality and, if

Question

There is a 2% chance that a woman has a specific chromosomal abnormality and, if she has that abnormality, there is a 10% chance that her child will have autism. Otherwise, the chance of her child having autism is only 0.1%. a) What is the probability that a randomly chosen woman has the chromosomal abnormality and her child has autism? b) What is the overall prevalence of autism in this population (i.e., what fraction of the population has autism)? c) Are the events "mom has the chromosomal abnormality" and "child has autism" independent? Explain your answer.

Explanation / Answer

let probability of having chromosomal abnormality =P(C)=0.02

and not having that =P(Cc)

probabilty of having autism =P(A) and not having =P(Ac)

a) Probabilty of having chromosomal abnormality and autism =P(C & A)=P(C)*P(A|C)=0.02*0.1=0.002

b)probabilty of autism =P(A) =P(C)*P(A|C)+P(Cc)*P(A|Cc)=0.02*0.1+0.98*0.001=0.00298

c)No as for that P(A)*P(C)=0.00298*0.02=0.0000596 which is not equal to P(A & C)

hence they are not independent

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