A human gene carries a certain disease from the mother to the child with a probability rate of 35%. That is, there is a 35% chance that the child becoems infected. Suppose a female carrier of the gene has five children. Assume that the infections of the 5 children are independent of one another. Find the probability that at least one of the children get the disease from their mother.
2007-02-28 07:20:56 · 3 answers · asked by Student12345 2 in Education & Reference ➔ Homework Help
The first answer is wrong. 1- 0.65^5 only included the case which 1 of the children doesn't have the disease, but not 2, 3, 4 or 5 children having the disease. The second answer is incorrect also because the word independent doesn't not imply same probability.
This question, in fact, is a binomial distribution. For Binomial distribution, the probability of a specific number of success can be found by formula:
P(n = x) = (n chose x) * (p)^x * (1-p)^(n-x)
which:
p: probability of success of each trial
x: number of success
(n chose x) = n!/[(x!) * (1- x)!]
The probability can be found by using the
P(n>=1) = 1 - P(n=0)
= 1 - (5 chose 0) * (0.35)^0 * (1-0.35)^5
= 1 - 0.0406
= 0.9594
2007-02-28 07:47:20 · answer #1 · answered by Ben 3 · 0⤊ 0⤋
The probablity for each child remains the same if they are independent of each other. They all have the same probablity it would be 3.5 out of ten children and 1.75 out of five children.
2007-02-28 07:27:08 · answer #2 · answered by fancyname 6 · 0⤊ 0⤋
1-.65^5=.88
2007-02-28 07:24:07 · answer #3 · answered by fcas80 7 · 0⤊ 0⤋