Respuesta :
"4% probability of their passing on a hereditary disease to 3 of their children"
I'm not sure if this means that the possibility of them passing the disease to one child is 4%, or if the probability of them passing the disease to the collective three children is 4%.
In case its the first one:
To make this a little easier, if you think about it the only possible way for the condition "at least one child will inherit the disease" to be untrue is if all 3 children do not have the disease. Therefore we just need to calculate the probability of all 3 children not having the disease.
if the chance of having the disease is 0.04, then the chance of not having it is 0.96
for three children: 0.96*0.96*0.96
= 0.884736
to find the chance of this not happening:
1 - 0.884736
= 0.115264 ≈ 12%
there is a 12 percent chance at least one of the children will inherit the disease
In case its the second one:
the probability of all 3 children having the disease is 0.04,
the probability of 1 child getting the disease is unknown, x.
this means
x*x*x = 0.04
x ≈ 0.342
now we just basically repeat the steps above
1 - 0.342
= 0.658 (probability that one child does not get the disease)
0.658 * 0.658 * 0.658
= 0.284890312 (probability that all 3 children do not get the disease)
1 - 0.284890312
= 0.715109688 ≈ 72%
there is a 72 percent chance at least one of the children will inherit the disease
