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fodog414 fodog414
wrote...
Posts: 279
11 years ago Edited: 11 years ago, fodog414
http://session.masteringgenetics.com/problemAsset/1511921/6/Sand.ch2.p15.jpg
Part C
The female I-1 and her mate, male I-2, had four children, one of whom has albinism. What is the probability that they could have had a total of four children with any other outcome except one child with albinism and three with normal pigmentation?
Express your answer using three decimal places.

Part D is shown after C is completed only have one attempt remaing on
Post Merge: 11 years ago

exclude part C figured out using binomial probability.. For Part D What is the probability that female I-3 is a heterozygous carrier of the allele for albinism? Express your answer using three decimal places. answer cannot be 1/2 or 3/4
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wrote...
Staff Member
11 years ago
What were your findings?
- Master of Science in Biology
- Bachelor of Science
fodog414 Author
wrote...
11 years ago
Using pascals triangle we can conclude that their are n number of events which is 4 being 4 children... Now you have to ignore the outcome of one child with albinism and three with normal pigmentation.

the binomial coefficients are 1 4 6 4 1

you have a 3/4 of chance being normal
1/4 having albinism

Now:
1(3/4)^4=.316
exclude 4(3/4)^3(1/4)^1
6(3/4)^2(1/4)^2=.211
4(3/4)^1(1/4)^3=.047
1(1/4)^4=.0039

=.578

Got part D wrong and she withheld the answer so not sure on that.
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