daughters, and two taster sons? d.) their fourth child would be a taster daughte
ID: 143436 • Letter: D
Question
daughters, and two taster sons? d.) their fourth child would be a taster daughter? On the average, about one child in every ten thousand live births in the United States has phenylketonuria (PKU). What is the probability that a) the next child born in a Boston hospital 3) 8) will have PKU? b.) after that child with PKU is born, the 9) next chiid bom will have PKU? c.) two children born in a row will have PKU? 4) PKU and albinism are two autosomal recessive disorders, unlinked in human beings. If two people, each heterozygous for both traits, produce a child, what is the chance of them having a child with a.) PKU? b.) either PKU or albinism? c.) both traits?Explanation / Answer
3)a) The probability that the next child born in Boston will jave PKU is 1/10,000
b) The probability of the next child being born with PKU is also 1/10,000 as it is not related to the previous birth.
c) The probability of two consecutive children being born with PKU is 1/10,000 X 1/10,000 = 1 X 10-8
4)a)The probability of them having a child with PKU (autosomal recessive disorder) is 1/4 or 25%.
b) In this question basic probability is applied. Since having either of the diseases is an independent event and the probability of each event is 1/4, then the probability of either having PKU or albinism is 1/4 + 1/4 - 1/16 = 7/16
In this case 1/16 is the probability of having both the diseases. The formula used is probability of event A(having PKU) + probability of event B (having albinism) - probability of both events.
c) The probability of the child having both the traits is 1/4 X 1/4 = 1/16 or 6.25%
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