A. 1/8

B. 1/4

C. 3/4

D. 1/3

E. 1/2

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A. 1/8

B. 1/4

C. 3/4

D. 1/3

E. 1/2

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So, the correct answer is 1/2 x 1/2 x 1/2 = 1/8 (choice A).

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Ok, let's consider it another way.

HHH (all children are healthy)

WHH (1 child is affected)

HWH ( >> )

HHW ( >> )

WWH (2 children are affected)

WHW ( >> )

HWW ( >> )

WWW (all 3 children are affected)

So, out of 8 possible outcomes, there are three among them satisfying the condition in question, i.e. the possibility that EXACTLY one of the three children is affected is 3/8.

probabilities that one of the three children is affected are:

so there are three different orders of birth

abnormal normal normal has a probab of 1/2 x 1/2 x 1/2 = 1/8

normal abnormal normal has a probab of 1/2 x1/2 x1/2 = 1/8

normal normal abnormal has a probab of 1/2 x 1/2 x 1/2 = 1/8

So the probability of 1 abnormal and 2 normal is the sum of the probabilities of these 3 possible orders

1/8 + 1/8 + 1/8 = 3/8 which is approximately 1/3

ans is 1/3 D

so lemmie get this straight,, everytime i come a across a question like this i need to think about how many children they WANT (whether it is 2 ,3 or 4), and then each of them have a 1/2 probability (for autosomal dominant), and then times it with the probability of the chances that they will have one affected child and 2 none (in this case obviously 3 times) so 3/8 approximates 1/3.. am i making sense?

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