Answer: 0.318528
Step-by-step explanation
Let x be the binomial variable (for success) that represents the children that have a hereditary disease
with parameter p = 0.12 n= 3
Using binomial , we have
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex]
Required probability :-
[tex]P(x\geq1)=1-P(x=0)\\\\=1-^{3}C_{0}(0.12)^{0}(1-0.12)^{3}\\\\=1-(1)(0.88)^{3}}\\\\=0.318528[/tex]
Hence, the probability of the event that at least one child will inherit the disease =0.318528
