Answer:
Option B.
Step-by-step explanation:
It is given that there is a 42% chance that the child becomes infected with the disease.
Let A be the event that the child becomes infected with the disease.
[tex]P(A)=42%=0.42[/tex]
A' be the event that the child is not infected.
[tex]P(A')=1-P(A)=1-0.42=0.58[/tex]
Assume that the infections of the three children are independent of one another.
Let X be the number of children get the disease from their mother.
The probability that all three children are free from disease is
[tex]P(X=0)=P(A')\cdot P(A')\cdot P(A')=0.58\cdot 0.58\cdot 0.58=0.195112[/tex]
We need to find the probability that at least one of the children get the disease from their mother.
[tex]P(X\ge 1)=1-P(X=0)[/tex]
[tex]P(X\ge 1)=1-0.195112[/tex]
[tex]P(X\ge 1)=0.804888[/tex]
[tex]P(X\ge 1)\approx 0.805[/tex]
The probability that at least one of the children get the disease from their mother is 0.805.
Therefore, the correct option is B.
