Approximately 40% of all U.S. households own at least one dog (according to the humane society). Suppose 15 households are randomly selected for a pet ownership survey.
a. What is the probability that exactly eight of the households have at least one dog?
b. What is the probability that exactly four of the households have at least one dog?
c. What is the probability that more than 10 households have at least one dog?
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To solve these probability questions, we can use the binomial probability formula:
P(X = k) = nCk * p^k * (1-p)^(n-k)
where:
- P(X=k) is the probability of exactly k successes
- n is the total number of trials (sample size)
- k is the number of successful outcomes (households with at least one dog)
- p is the probability of success in a single trial (40%, or 0.4)
- (1-p) is the probability of failure in a single trial (60%, or 0.6)
- nCk is the combination formula, which calculates the number of ways to choose k successes from n trials
Now, let's calculate the probabilities:
a. To find the probability that exactly eight households have at least one dog, we substitute n = 15, k = 8, p = 0.4 into the formula:
P(X = 8) = 15C8 * 0.4^8 * (1-0.4)^(15-8)
By calculating the values, we get:
P(X = 8) = 3003 * 0.4^8 * 0.6^7 ≈ 0.2139
So, the probability that exactly eight households have at least one dog is approximately 0.2139, or 21.39%.
b. To find the probability that exactly four households have at least one dog, we substitute n = 15, k = 4, p = 0.4 into the formula:
P(X = 4) = 15C4 * 0.4^4 * (1-0.4)^(15-4)
By calculating the values, we get:
P(X = 4) = 1365 * 0.4^4 * 0.6^11 ≈ 0.1330
So, the probability that exactly four households have at least one dog is approximately 0.1330, or 13.30%.
c. To find the probability that more than ten households have at least one dog, we can calculate the complement of the probability that ten or fewer households have at least one dog.
P(X > 10) = 1 - P(X ≤ 10)
To find P(X ≤ 10), we sum up the probabilities of having 0 to 10 households with at least one dog:
P(X ≤ 10) = Σ [P(X = k)] for k = 0 to 10
By calculating the values using the binomial formula for each k, we find:
P(X ≤ 10) ≈ 0.7729
Finally, we subtract this value from 1 to get the probability of more than ten households having at least one dog:
P(X > 10) = 1 - 0.7729 ≈ 0.2271
So, the probability that more than ten households have at least one dog is approximately 0.2271, or 22.71%.