Prob each is affected= 1/2*1/2= 1/4
prob all unaffected= (3/4)^6
prob three affected= (1/4)^3 * (3/4)^3 * ways.
now ways = 6!/3!3!
at least two will be unaffected
combination ways
uuaaaa 6!/(2!4!)= 15
uuuaaa 6!/(3!3!)=40
uuuuaa 6!/(2!4!) =15
uuuuua 6!/5!1! =6
uuuuuu 6!/6! =1
check those, I did them in my head
Pr(at least two)= sum prob*combinations
= 15((3/4)^2(1/4)^4)+40((3/4)^3(1/4)^3)+ ....
pr first child ug?
Pr= (3/4)*1/2
Phenylketonuria, a Metabolic disease in humans, is caused by a recessive allele, k. If two heterozygotes marry and plan a family of six children: a)what is the probability that all will be unaffected? b)what is the probability that three will be unaffected and three will be affected? c)what is the probability that at least two will be unaffected? d)what is the probability that the first child will be an unaffected girl?
I don't even know how to approach this problem. Much help is needed. Thanks
2 answers
I don't get b or c at alll. What are the u's and a's? What are the exclamation points?