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#1




Autosomal Dominant, Planning to have 3 children
A mother has XYZ syndrome{AD} 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. 3/8 D. 1/3 E. 7/8 
#2




What is the XYZ syndrome ? I dont think I've heard of it lol
And if its autosomal dominant, we would assume the mother is heterozygous for the dx & since the father is not affected he is homozygous for normal alleles. Therefore, the probability for every child to be affected is the same isnt it ? In case of AD its 50% (1/2) chance of producing an affected offspring. Is that right? 
The above post was thanked by:  
Mashee (07202011) 
#3




Does anyone know a good site on odd calculation? I am tired of guessing all the time

#4




Ya exactly what I thought its supposed to be 50% chance for each child doesn't matter how many children...

#5




although the chance of each child affected is 1/2, but the question is asking the chance of disease in 1 of the three children, that is to say what is the chances of one child has disease and the other two not

#6




c
I think is C, since the probability of one child is 1/2, then one of the three children is 3/8

#7




Ok,
mother Aa Father aa Probability affected/unaffected = 1/2  ALL 3 affected = 1/2^3 = 1/8  ALL 3 unaffected = 1/2^3 = 1/8 Anything in between (2 affected + 1 unaffected // 1 affected + 2 unaffected) = 3/8 Then, the sum of all probabilities is: 1/8 + 3/8 + 3/8 + 1/8 = 1 So, the answer should be C) 
#8




that was a great explanation, thank you so much bebix........

#9




why mother Aa? She is AD affectid AA

#10




She has a AD disease
If she is AA...then a 100% of the children will have the disease 
#11




thank you bebix

#12




correct answer CC
Thank you Bebix,
I want to make it simple, and easier to understand: probabilities that one of the three children is affected are: so there are three different orders 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 
#13




it is the probability of one kid is affected + probability of the other 2 kids are unaffected
so 1/2 + 1/2 *1/2 = 1/2 +1/4 =3/8 to me i think that way is easier to calculate and understand 
The above post was thanked by:  
usluipek (07212011) 
#14




yup...same thing (unions and intersections)

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dr.muhamad (07212011) 
#15




yup sure bebix they r the same,,, i meant just to follow the rules of adding and multiplying

#16




Dr. Muhamad: great explanations! Thank you!
Can you pls recommend a book or site on probability calculations? 
The above post was thanked by:  
dr.muhamad (07212011) 
#17




Quote:
okay i guess u have to use tow things kaplan vids and lange as it is almost just 100 pages and it is good but u need to hear the vid first and then read the chapter and u will be great with biostat and probabilities 
The above post was thanked by:  
usluipek (07222011) 
#18




Dr. muhamad, your calculation is not quite right, have a closer look

#19




Quote:
i think bebix way is the right one ill take it off 
#20




Quote:

#21




Because the affected child could be the first, second or third one...

#22




A
A. 1/8
Because the chance in one kid is 1/2 if there are three kids then, 1/2x1/2x1/2=1/8 
#23




Quote:
So, we have 3 different scenarios: 1. first child affected, second unaffected, third unaffected = the probability of this event is 1/2^3 = 1/8 2. first child unaffected, second affected, third unaffected = the probability of this event is 1/2^3 = 1/8 3. first child unaffected, second affected, third affected = the probability of this event is 1/2^3 = 1/8 What is the probability that one of the three children will be affected? since this 3 different scenarios are possible, the final probability is equal to 1/8*3 = 3/8 Last edited by bebix; 07252011 at 07:28 AM. 
#24




Quote:
Besides, if she were to have 2 kids, we'd just multiply 1/2 by 1/2 and get 1/4. Last edited by tigriki; 07252011 at 07:50 AM. 
#25




Quote:
is NOT the same as Probability of first event AND Probability of second event AND Probability of third event 
#26




"Besides, if she were to have 2 kids, we'd just multiply 1/2 by 1/2 and get 1/4."
nop... because you could have first child affected + second unaffected OR first child unaffected + second affected This TWO probabilities are different (even if the number are the same) 1/4 + 1/4... 
#27




Quote:

The above post was thanked by:  
bebix (07252011) 
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