fox wrote:
Every child has a 50% (1/2) chance of inheriting the defective copy of the Allele. If you ask the chance for one to be affected i think it should be 50/150 (sum of all chances) = 1/3. i.e. D. Its just my assumption. Criticism welcomed!
I was under the impression that you multiply your odds of getting an affected child from each pregnancy.
ALTHOUGH i do know that there are some scenarios where we do have to ADD the probabilities. Anyone here taken stats?
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| yoga
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d
isnt that probability 4 each preganancy?? i mean wether u r taking about 1in 3 or i in 10 it doesnt really matter right??
this is old post i hope though 1 can carify
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| yoga
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this is the trick genetics guy killed him self explaining
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| perfectcube
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Without having reviewed genetics I would suppose that the answer is E.
1/2*1+1/2*(1-1/2)+1/2*[1/2*(1-1/2)]=7/8
I think that intuitively it's obvious that if the prob. of sth happening is 50% if you do the experiment 1 time the prob. is always 50% but if you repeat the experiment let's say 10 times then the prob. that at least one time this certain thing will happpen is greater than 50%.
If the question was what is the prob. that all three will be affected the answer would be 1/8.
If the question was what is the prob. that the second one for example would be affected the probability would be 1/2. (The same goes with the first or the third one)
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| yoga
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