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One genetic disease was tested positive in both parents of one family. It has been known...

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:

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Expert Solution

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.


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