mr. and mrs. doran have a genetic history such that the probability that a child being born to them with a…

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 4$ (number of children), $k = 2$ (number of children with the trait), and $p = 0.82$ (probability of a child having the trait), so $1-p=1 - 0.82 = 0.18$.

Step2: Calculate the combination $C(n,k)$

$C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3\times2!}{2!×2!}=\frac{4\times3}{2\times1}=6$.

Step3: Calculate $p^{k}$ and $(1 - p)^{n - k}$

$p^{k}=(0.82)^{2}=0.6724$ and $(1 - p)^{n - k}=(0.18)^{2}=0.0324$.

Step4: Calculate the probability $P(X = 2)$

$P(X = 2)=C(4,2)\times p^{2}\times(1 - p)^{2}=6\times0.6724\times0.0324 = 6\times0.02178576=0.13071456$.

Answer:

$0.131$