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Author  27 Posts  
avir 
Waardenburg syndrome is an autosomal dominant condition that accounts for 1.4% of cases of congenital deafness. In addition to deafness, patients with this condition have an atypical facial appearance, including lateral displacement of the inner canthi, hypertelorism ,white forelock and white patches of skin on the ventral midline . A mother has Waardenburg syndrome, her husband 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. 3/4 D. 1/3 E. 1/2  
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OlgaG
Forum Junior Topics: 20 Posts: 806 
The probability for each of 3 children will be the same. Let's suppose that mother is heterozygose Aa, father is normal aa, so 50% children have chance to be normal aa and 50%  sick with genotype Aa, the correct answer is E.  
avir 
no, is not E  
GoodGirl
. Topics: 154 Posts: 2516 
A ?  
OlgaG
Forum Junior Topics: 20 Posts: 806 
I am still thinking this is E  
avir 
is not E, E is only if they plan to have 1 kid and they ask probability that the child will be affected probability that all 3 kids will be affected is 1/8, so is not A  
OlgaG
Forum Junior Topics: 20 Posts: 806 
In a question they asked "What is the probability that one of the three children will be affected? "  
madoo
Forum Newbie Topics: 6 Posts: 143 
my first answer was 1/2 but because u said it is wrong i gave it more shoot so if u mean only one child is affected it would be 3/8 then option D. 1/3 would be correct !!!!!! only if i am correct read this explaination becuse it is autosomal dominat and the risk in one child is 50% treat is as coin flip then u can do it in two ways >first long way is calculate the risk of only two children by 2X2 table and u will get four option (25% each) then add the risk of another child by 4x2 table and you will get eight results (12.5% each) only three of the eight results will have the content of one affected child and two normal others > second one imagine if the asked about the risk of the first child is affected and the next two is normal u can calculate this as 50% x 25% = 12.5% (50% is the chance the first child is affected 25% is the chance that the 2nd child and 3rd child are normal ) then back to the real question : they asked about if the 1st or 2nd or 3rd and because these risks are exclusive in big single event u should use " +" for the word " or" so all the risks would be 12.5% + 12.5% + 12.5% = 3/8 or approx 1/3 Edited by madoo on Sep 13, 2009  9:52 PM  
avir 
thank you madoo, yes answer 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  
Passerby 
But 3/8 is not one of the options in the answers?  
OlgaG
Forum Junior Topics: 20 Posts: 806 
Yes! Avir , where did you get this test and did it come with an answer?  
Melu
Forum Junior Topics: 16 Posts: 1510 
I am thinking of A as the answer still. What is the correct answer?  
usmdee
Forum Guru Topics: 0 Posts: 500 
answer is 3/8 i also did this question but they asked probability of having 3 girls or 3 boys , so basically is the same concept (as avir and madoo said, are mutually exclusive events so you have to ADD probabilities) (so probab of having 3 girls or 3 boys is 2/8)  
madoo
Forum Newbie Topics: 6 Posts: 143 
usmdee wrote: answer is 3/8 i also did this question but they asked probability of having 3 girls or 3 boys , so basically is the same concept (as avir and madoo said, are mutually exclusive events so you have to ADD probabilities) (so probab of having 3 girls or 3 boys is 2/8) yup having 3 girls probability is 1/8 probability of "this or that" is calculated by adding if they are exclusive nice question  
avir 
thanks guys Olga, is mcq from free online step1 Qbank, nothing fancy  
OlgaG
Forum Junior Topics: 20 Posts: 806 
Now I understand, I think the right answer is 1/8. If they plan to have 3 children  each of them has 1/2 chance to be sick or healthy so probability to have one of three children affected is 1/2*1/2*1/2=1/8 Interesting question.  
madoo
Forum Newbie Topics: 6 Posts: 143 
OlgaG wrote: Now I understand, I think the right answer is 1/8. If they plan to have 3 children  each of them has 1/2 chance to be sick or healthy so probability to have one of three children affected is 1/2*1/2*1/2=1/8 Interesting question. nop answer is 3/8 1/8 is risk of 3 affected children  
avir 
sorry guys, you made me smile... so Olga, you cannot understand the concept you have the explanation, the maths, the correct answer, is all above ok, i will type again correct answer is 3/8 now please answer (to yourself) what is the probability for two of three children to be affected, this way you will understand  
OlgaG
Forum Junior Topics: 20 Posts: 806 
I understand your math and concept but! I am sorry there is no answer 3/8, and 1/3 is not really equal 3/8 1/3=0.33 3/8=0.375 They make math in this tests not so complicated and we don't need to try answer 3/8 make equal 1/3 What i understood is that your tests didn't come with a real explanation and here we trying to find it Right?  
usmdee
Forum Guru Topics: 0 Posts: 500 
Avir never posted a Qs without explanation, and he explained very well is free online Qbank for step1, so you can find the Qs and the explanation it doesn't matter what the options are, u have to pick the one who approximates better 3/8, in this case is option D  
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