Suppose that 20% of men and 15% of women are carriers of a particular genetic trait. This trait can only be inherited by a child if both parents are carriers. What is the probability that a child is born with this genetic trait

Answer :

Answer:

The probability that a child is born with this genetic trait is 0.03.

Step-by-step 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:

[tex]P(W\cap M)=P(W)\times P(M)[/tex]

                 [tex]=0.15\times 0.20\\=0.03[/tex]

Thus, the probability that a child is born with this genetic trait is 0.03.

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