# 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?

Result

p =  0.316

#### Solution:

$q = 25 \% = \dfrac{ 25 }{ 100 } = 0.25 \ \\ \ \\ C_{{ 1}}(3) = \dbinom{ 3}{ 1} = \dfrac{ 3! }{ 1!(3-1)!} = \dfrac{ 3 } { 1 } = 3 \ \\ \ \\ p = { { 3 } \choose 1 } \cdot \ q ^1 \cdot \ (1-q)^3 = 3 \cdot \ 0.25 ^1 \cdot \ (1-0.25)^3 = \dfrac{ 81 }{ 256 } \doteq 0.3164 = 0.316$

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