Genetic disease

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 the disease. A family has 3 children. What is the probability of this family having one child who inherited this genetic disease?

Correct result:

p =  0.4219

Solution:

$$\begin{gathered} C_1(3)=\left({\displaystyle \genfrac{}{}{0px}{}{3}{1}}\right)={\displaystyle \frac{3!}{1!(3-1)!}}={\displaystyle \frac{3}{1}}=3 \\ q=25\mathrm{\%}={\displaystyle \frac{25}{100}}=0.25 \\ n=3 \\ k=1 \\ p=\left(\genfrac{}{}{0px}{}{n}{k}\right)\cdot q^k\cdot (1-q{)}^{n-k}=3\cdot 0.25^1\cdot (1-0.25{)}^{3-1}={\displaystyle \frac{27}{64}}=0.4219 \end{gathered}$$

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