Let probability of a child having genotype AB: "p=1\/2"
Then probability of a child having genotype BB: "1-p=1-1\/2=1\/2"
Let random variable "X=" number of children having genotype AB: "X\\sim B(n, p)"
Given that "n=5, p=1\/2"
Find the probability that two of the children have genotype AB and three others have genotype BB .
Comments
Leave a comment