In: Math
One genetic disease was tested positive in both parents of one family. It has been known that any child in this family has a 25% risk of inheriting this disease. A family has three children. The probability of this family having one child who inherited this genetic disease is:
One genetic disease was tested positive in both parents of one family. It has been known that any child in this family has a 25% risk of inheriting this disease. A family has three children.
Now we want to find the probability that this family having one child who inherited this genetic disease.
Let define a random variable X, X : random variable denoting the number of children who inherited this genetic disease.
Here given that, any child in this family has a 25% risk of inheriting this disease.
[ Note:- A discrete probability distribution binomial distribution is frequently used to model the number of successes in a sample. For this problem we assume that one children inherited this genetic disease means one success. So X~ bin(3,0.25) Probability muss function of X is given by, P(x) = , for this problem if we put x=1 then we get the required probability. ]
=> Required probability is,
P(1) = = 3×0.25×0.5625 = 0.421875
Note (using concept):- if X~ bin(n,p) then, P(x) = ; where n= number of independent trials, p= success probability in one independent trial, x= number of success.