Suppose 30% of the restaurants in a certain part of a town are in violation of the health code. A health inspector randomly selects six of the restaurants for inspection. (Round your answers to four decimal places.)
(a) What is the probability that none of the restaurants are in violation of the health code?
(b) What is the probability that one of the restaurants is in violation of the health code?
(c) What is the probability that at least two of the restaurants are in violation of the health code?
.117649
To answer these questions, we can use the binomial probability formula:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
Where:
P(x) is the probability of getting exactly x successes
n is the number of trials
p is the probability of success in one trial
(1-p) is the probability of failure in one trial
nCx is the combination formula: n! / (x!(n-x)!)
In this case:
p = 0.30 (probability of a restaurant being in violation)
n = 6 (number of restaurants inspected)
(a) Probability that none of the restaurants are in violation:
x = 0 (number of successes)
P(0) = (6C0) * (0.30^0) * (0.70^(6-0))
Using the combination formula:
(6C0) = 6! / (0! * (6-0)!) = 6! / (0! * 6!) = 1
P(0) = 1 * 1 * (0.70^6) = 1 * 1 * 0.1176 = 0.1176
Therefore, the probability that none of the restaurants are in violation is 0.1176.
(b) Probability that one of the restaurants is in violation:
x = 1
P(1) = (6C1) * (0.30^1) * (0.70^(6-1))
Using the combination formula:
(6C1) = 6! / (1! * (6-1)!) = 6! / (1! * 5!) = 6
P(1) = 6 * 0.30 * (0.70^5) = 6 * 0.30 * 0.16807 = 0.30204
Therefore, the probability that one of the restaurants is in violation is 0.30204.
(c) Probability that at least two of the restaurants are in violation:
P(at least 2) = 1 - P(0) - P(1)
P(at least 2) = 1 - 0.1176 - 0.30204 = 0.58036
Therefore, the probability that at least two of the restaurants are in violation is 0.58036.
To find the probability in this scenario, we will be using the binomial probability formula:
P(X=k) = (n C k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes
n is the number of trials
k is the number of successes
p is the probability of success in a single trial
In this case, the probability of a restaurant being in violation of the health code is 30% or 0.3, and the number of trials is 6 (since the health inspector randomly selects six restaurants for inspection).
(a) What is the probability that none of the restaurants are in violation of the health code?
To find the probability of none of the restaurants being in violation, we want to find P(X=0).
P(X=0) = (6 C 0) * 0.3^0 * (1-0.3)^(6-0)
Calculating this:
P(X=0) = (6 C 0) * 0.3^0 * 0.7^6
Using the combination formula (6 C 0) = 1:
P(X=0) = 1 * 1 * 0.7^6
P(X=0) = 0.7^6
Calculating this gives:
P(X=0) = 0.1176
Therefore, the probability that none of the restaurants are in violation of the health code is approximately 0.1176.
(b) What is the probability that one of the restaurants is in violation of the health code?
To find the probability of exactly one restaurant being in violation, we want to find P(X=1).
P(X=1) = (6 C 1) * 0.3^1 * 0.7^(6-1)
Calculating this:
P(X=1) = (6 C 1) * 0.3^1 * 0.7^5
Using the combination formula (6 C 1) = 6:
P(X=1) = 6 * 0.3 * 0.7^5
Calculating this gives:
P(X=1) ≈ 0.3025
Therefore, the probability that one of the restaurants is in violation of the health code is approximately 0.3025.
(c) What is the probability that at least two of the restaurants are in violation of the health code?
To find the probability of at least two restaurants being in violation, we want to find 1 - P(X=0) - P(X=1).
P(at least two) = 1 - P(X=0) - P(X=1)
Substituting the values we found earlier:
P(at least two) = 1 - 0.1176 - 0.3025
Calculating this gives:
P(at least two) ≈ 0.5799
Therefore, the probability that at least two of the restaurants are in violation of the health code is approximately 0.5799.