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rkoch rkoch
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12 years ago
Albinism is an autosomal recessive condition characterized by absence of melanin pigment from the skin, eye and hair. Two carriers of albinism marry and plan to have four children. What is the probability that 3 children will be normal and one will be albino?

this one stumped me..all the answers i got weren't any of the choices
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12 years ago
Multiplication rule:

First you need to find out what the chance is of them having an albino child in general. You can use the punnet square.

Aa (mother) x Aa (father)

1/4 AA
2/4 Aa
1/4 aa

So there's a 1/4 chance of them having an albino child and 3/4 chance of them having a normal child.

Next you need to apply the multiplication rule to figure of the chance of them having just one albino child out of a total of 4 children. Since they have 3/4 chance of having a normal child, you use this value for every child that would be normal:

3/4 x 3/4 x 3/4 (all three of the normal children accounted for)

Now you have to also add in the odds of having that one albino child.

3/4 x 3/4 x 3/4 x 1/4 = 27/256  (odds of 1 albino with 3 normal)

The only problem is that you don't know which child will have it in what order. You have to account for the fact that it could be the first, the second, the third, or the fourth child. All of these events are separate possibilities and each has a 27/256 chance so you have to apply the addition rule now:

27/256 (first child has albinism) + 27/256 (second child has albinism) + 27/256 (third child has albinism) + 27/256 (fourth child has albinism) =  27/64 probability that they'll have 3 normal children and 1 albino
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