Question

A husband and wife discover that there is a 12​% probability of their passing on a hereditary disease to any of their children. If they plan to have three ​children, what is the probability of the event that at least one child will inherit the​ disease?

Answer 1

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

$P(x)=^nC_xp^x(1-p)^{n-x}$

Required probability :-

$P(x\geq1)=1-P(x=0)\\\\=1-^{3}C_{0}(0.12)^{0}(1-0.12)^{3}\\\\=1-(1)(0.88)^{3}}\\\\=0.318528$

Hence, the probability of the event that at least one child will inherit the​ disease =0.318528