Assuming a 1:1 sex ratio, what is the probability that five children produced by

Question

Assuming a 1:1 sex ratio, what is the probability that five children produced by the same parents will consist of a.) three daughters and two sons? b.) alternating sexes, starting with a son? c.) altenating sexes? d.) all daughters? e.) all the same sex? f) at least four daughters? g) a daughter as the eldest child and a son 1) as the youngest? Phenylthiocarbamide (PTC) tasting is dominant (T) to nontasting (0. If a taster woman with a nontaster father produces children with a taster man, and the man previously had a nontaster daughter, what would be the probability that a.) their first child would be a nontaster? b.) their first child would be a nontaster 2) c.) if they had six children, they would have two nontaster sons, two nontaster daughters, and two taster sons? d.) their fourth child would be a taster 3) On the average, about one child inevery ten thousand live births in the United States has phenylketonuria (PKU). What is the probability that a.) the next child bom in a Boston hospital will have PKU? b) after that child with PKU is born, the next child born will have PKU? c.) two children born in a row will have 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? 4)

Explanation / Answer

Q.1

a) In this case we have three daughters and tow sons that are not born in any particular order.

binomial theorem,where

n = the number of trials

s = the number of daughters

t = the number of boys

p = the probabality of a daughter on any given birth

q = the probabality of a son on any given birth

probabality (3D 2D) = 5! (.5^3) * (.5^2)   

(3!)* (2!)

PROB (3D ,2D) = 10/32

b)  This is given by the PRODUCT RULE, in that it is the probability of a son on the first birth times the probability of a daughter on the second birth times the probability of a son on the third birth times the probability of a daughter on the fourth birth times the probability of a son on the fifth birth.
Prob(S,D,S,D,S) = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 1/32

c)

There are two ways that we could have alternating sexes, one way starts with a son and the second way starts with a daughter.

Prob(S,D,S,D,S) = 1/32 (see above)

Prob(D,S,D,S,D) = 1/32 (product rule again)
As there is two ways to have alternating sexes, we use the sum rule to get the total probability
Prob(alternating sexes) = Prob(S,D,S,D,S) + Prob(D,S,D,S,D)
Prob(alternating sexes) = 1/32 + 1/32 = 2/32

d) We are again specifying the birth of each birth in case this daughter on each birht .The prob of all daughters is again given by the product rule, such that prob ( D,D,D,D) = 0.5 *.5*0.5*0.5*0.5 =1/32

e) There are 2 ways that we could have five children who are all the same sex,all sons or all daughter.

prob(D,D,D,D) =1/32

prob (S,S,S,S) =1/32

two ways to get all the same sex ,use the sum rule to get the total prbability.

prob=prob(D,D,D,D)+prob(S,S,S,S)

=1/32+1/32=2/32

f)the probaility of at least 4 daughters is given by the sum rule ,where

prob( at least 4 daughter ) =prob ( 4D,1S)+prob (5D)

=5/32+1/32 =6/32

g) prob of family consists of son ,daughter ,son daughter ,son

this given by the productive rule

Q.3

a) The answer is 1 in 10,000 (0.0001). The fact that the birth occurs in a Boston Hospital has no effect on the probability.

b) The answer is again 0.0001. The birth of the first child has no effect on the probability of a PKU child on the second birth. They are independent events.

c) In this case we are asking for the probability of first one PKU birth and then a second PKU birth. This is given by the PRODUCT RULE.

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