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.
![P(A)=42%=0.42](https://tex.z-dn.net/?f=P%28A%29%3D42%25%3D0.42)
A' be the event that the child is not infected.
![P(A')=1-P(A)=1-0.42=0.58](https://tex.z-dn.net/?f=P%28A%27%29%3D1-P%28A%29%3D1-0.42%3D0.58)
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
![P(X=0)=P(A')\cdot P(A')\cdot P(A')=0.58\cdot 0.58\cdot 0.58=0.195112](https://tex.z-dn.net/?f=P%28X%3D0%29%3DP%28A%27%29%5Ccdot%20P%28A%27%29%5Ccdot%20P%28A%27%29%3D0.58%5Ccdot%200.58%5Ccdot%200.58%3D0.195112)
We need to find the probability that at least one of the children get the disease from their mother.
![P(X\ge 1)=1-P(X=0)](https://tex.z-dn.net/?f=P%28X%5Cge%201%29%3D1-P%28X%3D0%29)
![P(X\ge 1)=1-0.195112](https://tex.z-dn.net/?f=P%28X%5Cge%201%29%3D1-0.195112)
![P(X\ge 1)=0.804888](https://tex.z-dn.net/?f=P%28X%5Cge%201%29%3D0.804888)
![P(X\ge 1)\approx 0.805](https://tex.z-dn.net/?f=P%28X%5Cge%201%29%5Capprox%200.805)
The probability that at least one of the children get the disease from their mother is 0.805.
Therefore, the correct option is B.