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  #1  
Old 07-20-2011
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Genetics 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
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  #2  
Old 07-20-2011
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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?
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  #3  
Old 07-20-2011
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Does anyone know a good site on odd calculation? I am tired of guessing all the time
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  #4  
Old 07-20-2011
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Ya exactly what I thought its supposed to be 50% chance for each child doesn't matter how many children...
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  #5  
Old 07-20-2011
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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
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  #6  
Old 07-20-2011
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Default c

I think is C, since the probability of one child is 1/2, then one of the three children is 3/8
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Old 07-20-2011
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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)

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  #8  
Old 07-20-2011
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that was a great explanation, thank you so much bebix........
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  #9  
Old 07-20-2011
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why mother Aa? She is AD affectid AA
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  #10  
Old 07-20-2011
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Quote:
Originally Posted by dr.Irina View Post
why mother Aa? She is AD affectid AA
She has a AD disease

If she is AA...then a 100% of the children will have the disease
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Old 07-20-2011
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thank you bebix
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Old 07-21-2011
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Default 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
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  #13  
Old 07-21-2011
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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
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  #14  
Old 07-21-2011
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Quote:
Originally Posted by dr.muhamad View Post
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
yup...same thing (unions and intersections)
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  #15  
Old 07-21-2011
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yup sure bebix they r the same,,, i meant just to follow the rules of adding and multiplying
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Old 07-21-2011
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Dr. Muhamad: great explanations! Thank you!

Can you pls recommend a book or site on probability calculations?
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  #17  
Old 07-21-2011
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Quote:
Originally Posted by usluipek View Post
Dr. Muhamad: great explanations! Thank you!

Can you pls recommend a book or site on probability calculations?

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
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  #18  
Old 07-24-2011
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Dr. muhamad, your calculation is not quite right, have a closer look
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Old 07-24-2011
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Quote:
Originally Posted by usluipek View Post
Dr. muhamad, your calculation is not quite right, have a closer look
Yea you are right thank you
i think bebix way is the right one
ill take it off
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  #20  
Old 07-25-2011
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Quote:
Originally Posted by laurier View Post
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
Should not it b 1/8.?? Why we are adding it 3 times? Im unable to understand!!
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  #21  
Old 07-25-2011
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Quote:
Originally Posted by dr.aysa View Post
Should not it b 1/8.?? Why we are adding it 3 times? Im unable to understand!!
Because the affected child could be the first, second or third one...
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  #22  
Old 07-25-2011
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Post 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
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  #23  
Old 07-25-2011
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Quote:
Originally Posted by plasticsurgeon View Post
A. 1/8

Because the chance in one kid is 1/2

if there are three kids then, 1/2x1/2x1/2=1/8
Nop..the answer is 3/8

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; 07-25-2011 at 07:28 AM.
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  #24  
Old 07-25-2011
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Quote:
Originally Posted by bebix View Post
Nop..the answer is 3/8

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 we have 3 different scenarios are possible, the final probability is equal to 1/8*3 = 3/8

It doesn't seem right to me. Each birth will be an independent event, so since she is going to have 3 kids and the chance in each birth to have an affected child is 1/2, I'd just multiply it: 1/2*1/2*1/2=1/8. I still don't get why you then multiply it by 3, probably because the number of scenarios doesn't really matter. In some cases you will not even be able to count the number of possible scenarios.

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; 07-25-2011 at 07:50 AM.
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  #25  
Old 07-25-2011
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Quote:
Originally Posted by tigriki View Post
It doesn't seem right to me. Each birth will be an independent event, so if the chance in each birth to have an affected child is 1/2, I'd just multiply it 1/2*1/2*1/2=1/8. I still don't get why you then multiply it 3 times...
Probability of first event OR Probability of second event OR Probability of third event

is NOT the same as

Probability of first event AND Probability of second event AND Probability of third event

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  #26  
Old 07-25-2011
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"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...
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Old 07-25-2011
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Quote:
Originally Posted by bebix View Post
Nop..the answer is 3/8

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

I get it! Thanx so much!
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