Suppose 20% 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.
To determine the probability that exactly two of the six randomly selected restaurants are in violation of the health code, we can use the concept of probability and combinations.
Step 1: Calculate the probability of a restaurant being in violation of the health code.
Since 20% of the restaurants are in violation, the probability of a randomly selected restaurant being in violation is 0.20 or 20%.
Step 2: Calculate the probability of a restaurant not being in violation.
The probability of a restaurant not being in violation is complementary to the probability of it being in violation. So, the probability of a randomly selected restaurant not being in violation is 1 - 0.20 = 0.80 or 80%.
Step 3: Determine the number of ways to choose exactly two restaurants out of six.
To calculate this, we can use the concept of combinations. The number of combinations of choosing two items out of a total of six items can be calculated using the formula:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of items (in this case, 6) and r is the number of items chosen (in this case, 2).
C(6, 2) = 6! / (2!(6-2)!)
= (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * 4 * 3 * 2 * 1)
= 15
So, there are 15 different combinations of choosing exactly two restaurants out of the six.
Step 4: Calculate the probability of exactly two restaurants being in violation out of the six selected.
To calculate this probability, we need to use the binomial probability formula:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
where n is the total number of trials (in this case, 6), x is the number of successful trials (in this case, 2), p is the probability of success (in this case, the probability of a restaurant being in violation, which is 0.20), and (1-p) is the probability of failure.
P(2) = (C(6, 2)) * (0.20^2) * ((1-0.20)^(6-2))
= 15 * 0.20^2 * 0.80^4
= 0.09216
Therefore, the probability that exactly two of the six randomly selected restaurants are in violation of the health code is 0.09216, or approximately 9.22%.