docarchana
Forum Guru
Topics: 70
Posts: 515
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Mother has waardenburg syndrome which is an autosomal dominant. Father is unaffected and they plan to have a family with three children. What is the probability that one of the three children will be affected?
A.1/8
B.1/4
C.3/8
D.1/3
E.7/8
This is a q posted by swapna. i guess ans is E. my calculation is
1/2x1/2x1/2 = 1/8
1-1/8=7/8
swapna is saying ans is 1/8. whats ur say. there is a similar q labelled gardners. plz help in answering these qs..
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| fox
Forum Guru
Topics: 70
Posts: 727
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How did u get the calculation? I agree that with incresing number of kids, the chances for atleast 1 to get affected should increase. But according to swapna's calculation, by simply multiplying chances, u have a decreased chance with incresing kids, which is not true. So maybe ur calculation is rite.
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| Leopard
Forum Guru
Topics: 30
Posts: 401
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Honestly speaking poor leopard is extremely weak at these biostatistical and epidemiological calculations.But
Swapna's Caculation shows the "probability that every child born in three births will be suffering /not suffering (both have equal chances) from the disease".But here the restriction "one out of three".So I do not think that it is correct calculation.
This is a tricky question needs some other modus operandi like "binomial notation" etc.which I do not know and probably not required for usmle also.
Best of luck
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| Ivil
Forum Senior
Topics: 8
Posts: 169
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doc archana is right...
probability that atleast 1 person in the group will have the disease = 1 - the propability of all the persons in the group to have the disease.
in this example, the probabilty of all 3 children having the disease is 1/2 x1/2 x 1/2 = 1/8
so 1-1/8 = 7/8 is the ans
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