Question

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 0.8. If they have three children, what is the probability that exactly two of their three children will have that trait? Round your answer to the nearest thousandth

Answer 1

The probability that exactly two of the three children of Mr. and Mrs. Doran will have a certain trait is approximately 0.384.

Step-by-step explanation:

To find the probability that exactly two out of three children of Mr. and Mrs. Doran will have a certain trait, we can use the binomial probability formula:

P(x=k) = C(n, k) * p^k * (1-p)^(n-k)

In this case, n = 3 (number of children), k = 2 (number of children with the trait), and p = 0.8 (probability of having a child with the trait).

Substituting these values into the formula, we get:

P(x=2) = C(3, 2) * 0.8^2 * (1-0.8)^(3-2)

Calculating the values, we have:

P(x=2) = 3 * 0.8^2 * 0.2^1 = 0.384

Therefore, the probability that exactly two out of three children will have the trait is approximately 0.384, rounded to the nearest thousandth.

Answer 2

0.384 = 38.4% probability that exactly two of their three children will have that trait.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they have the trait, or they do not. The probability of a child having the trait is independent of any other child. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 0.8.

This means that
`p = 0.8`

Three children:

This means that
`n = 3`

What is the probability that exactly two of their three children will have that trait?

This is P(X = 2).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(3,2).(0.8)^(2).(0.2)^(1) = 0.384`

0.384 = 38.4% probability that exactly two of their three children will have that trait.