a husband and wife discover that there is a 4% probability of their passing on a hereditary disease to 3 of their children if they plan to have their children what is the probability of the event that at least one child will inherit the disease

"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

multiply then divide

step-by-step explanation: