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?
To solve this problem, we can use the concept of probability and the binomial distribution.
Let's start with part (a), where we need to find the probability that none of the restaurants are in violation of the health code.
The probability that a single restaurant is NOT in violation of the health code is given by (1 - 0.30) = 0.70. Since there are six restaurants selected, and each selection is independent, we can multiply the probabilities together to find the probability that none of them are in violation of the health code.
So, the probability that none of the restaurants are in violation of the health code is:
P(none) = (0.70) * (0.70) * (0.70) * (0.70) * (0.70) * (0.70)
Now, let's calculate this probability:
P(none) = 0.70^6 ≈ 0.1176
Therefore, the probability that none of the restaurants are in violation of the health code is approximately 0.1176.
Moving on to part (b), we need to find the probability that one of the restaurants is in violation of the health code.
To calculate this probability, we need to consider all the possible combinations of exactly one restaurant being in violation of the health code out of the six selected. Since the probability of a single restaurant being in violation of the health code is 0.30, and it can occur at any of the six restaurants, we can calculate the probability using the binomial coefficient formula.
The probability that exactly one restaurant is in violation of the health code is:
P(one) = (6 Choose 1) * (0.30)^1 * (1 - 0.30)^(6-1)
Now, let's calculate this probability:
P(one) = (6! / (1! * (6-1)!)) * (0.30)^1 * (1 - 0.30)^(6-1)
P(one) = 6 * 0.30^1 * 0.70^5 ≈ 0.3241
Therefore, the probability that one of the restaurants is in violation of the health code is approximately 0.3241.
Lastly, for part (c), we need to find the probability that at least two of the restaurants are in violation of the health code.
To calculate this probability, we need to consider all the possible combinations of having two or more restaurants in violation of the health code out of the six selected. This means we need to calculate the probabilities of having exactly two, three, four, five, or six restaurants in violation and then sum up these individual probabilities.
P(at least two) = P(two) + P(three) + P(four) + P(five) + P(six)
Using similar reasoning as in part (b), we can calculate each individual probability as:
P(two) = (6 Choose 2) * (0.30)^2 * (1 - 0.30)^(6-2)
P(three) = (6 Choose 3) * (0.30)^3 * (1 - 0.30)^(6-3)
P(four) = (6 Choose 4) * (0.30)^4 * (1 - 0.30)^(6-4)
P(five) = (6 Choose 5) * (0.30)^5 * (1 - 0.30)^(6-5)
P(six) = (6 Choose 6) * (0.30)^6 * (1 - 0.30)^(6-6)
Now, let's calculate these probabilities and sum them up:
P(at least two) = [P(two) + P(three) + P(four) + P(five) + P(six)]
Therefore, we need to calculate each of these probabilities individually and then sum them up to find the final result.