Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 67%. If they have five children, what is the probability that at least three of their five children will have that trait? Round your answer to the nearest thousandth.
To find the probability that at least three out of five children will have the given trait, we can use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * (p^x) * (q^(n-x))
where:
P(x) is the probability of getting exactly x successes,
n is the total number of trials,
p is the probability of success in one trial,
q is the probability of failure in one trial, which is equal to 1 - p, and
nCx is the number of ways to choose x successes from n trials.
In this case, the probability of a child having the trait is 0.67, so p = 0.67. The probability of a child not having the trait is 1 - 0.67 = 0.33, so q = 0.33. The number of trials is 5, as they have five children.
Now, we need to calculate the probability of getting three, four, or five children with the trait. This is a cumulative probability, so we need to calculate the probabilities for each case separately and then add them up.
P(3 children have the trait) = (5C3) * (0.67^3) * (0.33^(5-3))
P(4 children have the trait) = (5C4) * (0.67^4) * (0.33^(5-4))
P(5 children have the trait) = (5C5) * (0.67^5) * (0.33^(5-5))
Using combinatorics, we can calculate the values for the binomial coefficients:
(5C3) = 5! / (3! * (5-3)!) = 10
(5C4) = 5! / (4! * (5-4)!) = 5
(5C5) = 5! / (5! * (5-5)!) = 1
Using these values, we can now calculate the probabilities:
P(3 children have the trait) = 10 * (0.67^3) * (0.33^2)
P(4 children have the trait) = 5 * (0.67^4) * (0.33^1)
P(5 children have the trait) = 1 * (0.67^5) * (0.33^0)
Finally, we add up these probabilities to find the overall probability of at least three out of five children having the given trait:
P(at least 3 children have the trait) = P(3 children have the trait) + P(4 children have the trait) + P(5 children have the trait)
Calculate these values and round the answer to the nearest thousandth to find the probability that at least three out of five children will have the trait.
at least 3 of 5
----> could be 3, or could be 4, or could be 5
Prob(trait) = .67
prob(not trait) = .33
prob(your event)
= C(5,3)(.67^3)(.33^2) + C(5,4)(.67^4)(.33) + C(5,5)(.67^5)
= 10(.67^3)(.33^2) + 5(.67^4)(.33) + (.67^5)
= .....