NEW 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 28%. If they have seven children, what is the probability that at most one of their seven children will have that trait? Round your answer to the nearest thousandth.
To find the probability that at most one of their seven children will have the certain trait, we need to calculate the probability of zero and one child having the trait and then add them together.
Let's first calculate the probability that all seven children will not have the trait.
P(0 children have the trait) = (1 - 0.28)^7
Next, let's calculate the probability that exactly one child will have the trait. We can choose which child that will be in 7 different ways, and each has a probability of 28% of having the trait, and the rest have a probability of 72% of not having the trait.
P(1 child has the trait) = 7 * (0.28) * (0.72)^6
Now, let's add the two probabilities together.
P(at most one child has the trait) = P(0 children have the trait) + P(1 child has the trait)
P(at most one child has the trait) = (1 - 0.28)^7 + 7 * (0.28) * (0.72)^6
Calculating this expression, the answer is approximately 0.838 (rounded to the nearest thousandth).
To find the probability that at most one of their seven children will have the specific trait, we can use the concept of binomial probability. Let's break down the problem into two parts:
Part 1: The probability that exactly zero children have the trait.
Part 2: The probability that exactly one child has the trait.
First, let's calculate the probability of having exactly zero children with the trait.
Given that the probability of a child having the trait is 28%, the probability of a child not having the trait is 1 - 0.28 = 0.72.
Using the binomial probability formula, we can calculate the probability of having exactly zero children with the trait:
P(X = 0) = C(n, x) * p^x * q^(n-x)
where n is the number of trials (number of children in this case), x is the number of successful outcomes (children having the trait), p is the probability of success (0.28 in this case), q is the probability of failure (0.72 in this case), and C(n, x) is the binomial coefficient.
Applying the formula:
P(X = 0) = C(7, 0) * 0.28^0 * 0.72^(7-0)
Now, let's calculate the probability of having exactly one child with the trait.
P(X = 1) = C(n, x) * p^x * q^(n-x)
P(X = 1) = C(7, 1) * 0.28^1 * 0.72^(7-1)
To find the probability of at most one child having the trait, we add the probabilities of exactly zero children and exactly one child having the trait:
P(at most one child has the trait) = P(X = 0) + P(X = 1)
Finally, we can calculate the probability.
P(at most one child has the trait) = [C(7, 0) * 0.28^0 * 0.72^7] + [C(7, 1) * 0.28^1 * 0.72^6]
Now, let's calculate this value:
P(at most one child has the trait) = [1 * 1 * 0.028973] + [7 * 0.28 * 0.028973]
Calculating further:
P(at most one child has the trait) = 0.028973 + 0.5705332
Rounding the answer to the nearest thousandth:
P(at most one child has the trait) ≈ 0.599
Therefore, the probability that at most one of their seven children will have the specific trait is approximately 0.599 or 59.9% (rounded to the nearest thousandth).
interpretation:
"most one of their seven children"
to me means, 4, 5, 6, or 7 children
do it the same way I just used in your previous post
or
exclude case of 0, 1, 2, or 3 children with the trait,
that is, subtract that sum from 1
(actually more steps than my first method)