Answer:
The probability that a child is born with this genetic trait is 0.03.
Explanation:
The probability that a woman is a carrier of a particular genetic trait is,
P (W) = 0.15.
The probability that a man is a carrier of a particular genetic trait is,
P (M) = 0.20.
The two events are independent of each other, since the event of a woman being a carrier is not affected by the man being a carrier and vice-versa.
It is provided that this trait can only be inherited by a child if both parents are carriers.
Compute the probability that both parents are carriers of the trait as follows:
![P(W\cap M)=P(W)* P(M)](https://img.qammunity.org/2021/formulas/mathematics/college/b1evtov2qa84f8uxs2kt594wwvm9svdaed.png)
![=0.15* 0.20\\=0.03](https://img.qammunity.org/2021/formulas/mathematics/college/bx6qgmwd908vsckiv0r5nebp51oek4lnyr.png)
Thus, the probability that a child is born with this genetic trait is 0.03.