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

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:

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

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value

step-by-step explanation:

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