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. Assume a 1:1 sex ratio.What is the probability that at least 2 children will be normal?
Gentoype of affected individual = aa
Genotype of normal individual = AA, Aa (carrier)
Aa × Aa
| Gametes | A | a |
| A | AA (normal) | Aa (normal, carrier) |
| a | Aa (normal, carrier) | aa (affected) |
Probability of normal individual = 3/4
Probability of affected individual = 1/4
Probability that atleast 2 children will be normal = 2 children will be normal + 3 children will be normal + 4 children will be normal
Using combinations
4C2 (3/4)^2 (1/4)^2 + 4C3 (3/4)^3 (1/4)^1 + 4C4 (3/4)^4 (1/4)^0
= (4!/2!2!) (3^2)/(4^4) + (4!/3!1!) (3^3)/(4^4) + (4!/4!0!) (3^4)/(4^4)
= (4×3/2)(9)/(256) + (4)(27)/(256) + (1)(81)/(256)
= (63)/256 + (108)/256 + (81)/256
= 252/256
= 0.984
Please rate.
albinism is an autosomal recessive condition characterized by absence of melanin pigment from the skin, eye...
Albinism is an autosomal recessive condition characterized by absence of melanin pigment from the skin, eye and hair. Two heterozygote carriers of albinism marry and plan a family of five. What is the probability that 4 children will be normal and one will be albino? What is the probability that at least 3 children will be normal? I know the answers are 0.396 & 0.896... please show the math and explain why
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