Thirty percent of all automobiles undergoing a headlight inspection at a certain inspection station fail the inspection.
Among 15 randomly selected cars, what is the probability that at most four fail tha inspection?
To calculate the probability that at most four cars fail the inspection, we need to calculate the probability of each possible outcome and then sum them up.
Let's use the binomial probability formula to solve this problem. The formula is:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes (in this case, failing the inspection)
n is the number of trials (in this case, 15 cars)
k is the number of successes (that is, the number of cars failing the inspection)
p is the probability of success on a single trial (in this case, 30% or 0.3)
(1-p) is the probability of failure on a single trial (that is, passing the inspection in this case)
(n choose k) is the binomial coefficient, which can be calculated as n! / (k!(n-k)!), where ! denotes factorial.
Now, let's calculate the probabilities for each possible outcome:
P(X=0) = (15 choose 0) * 0.3^0 * (1-0.3)^(15-0)
P(X=1) = (15 choose 1) * 0.3^1 * (1-0.3)^(15-1)
P(X=2) = (15 choose 2) * 0.3^2 * (1-0.3)^(15-2)
P(X=3) = (15 choose 3) * 0.3^3 * (1-0.3)^(15-3)
P(X=4) = (15 choose 4) * 0.3^4 * (1-0.3)^(15-4)
Finally, we sum up these probabilities to get the probability that at most four cars fail the inspection:
P(at most 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
Now, let's calculate these probabilities step-by-step:
P(X=0) = (15 choose 0) * 0.3^0 * (1-0.3)^(15-0)
= 1 * 1 * 0.7^15
≈ 0.036756
P(X=1) = (15 choose 1) * 0.3^1 * (1-0.3)^(15-1)
= 15 * 0.3 * 0.7^14
≈ 0.122520
P(X=2) = (15 choose 2) * 0.3^2 * (1-0.3)^(15-2)
= (15! / (2!(15-2)!)) * 0.3^2 * 0.7^13
≈ 0.204200
P(X=3) = (15 choose 3) * 0.3^3 * (1-0.3)^(15-3)
= (15! / (3!(15-3)!)) * 0.3^3 * 0.7^12
≈ 0.233474
P(X=4) = (15 choose 4) * 0.3^4 * (1-0.3)^(15-4)
= (15! / (4!(15-4)!)) * 0.3^4 * 0.7^11
≈ 0.200120
Now, we can sum up these probabilities to get the final result:
P(at most 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
≈ 0.036756 + 0.122520 + 0.204200 + 0.233474 + 0.200120
≈ 0.797070
Therefore, the probability that at most four cars fail the inspection among 15 randomly selected cars is approximately 0.797070 or 79.71%.
To find the probability that at most four cars fail the inspection, we need to calculate the probability of each possible outcome (0, 1, 2, 3, or 4 cars failing) and then add them together.
Given that the probability of a car failing the headlight inspection is 30%, the probability of a car passing the inspection is 100% - 30% = 70%.
To calculate the probability for each possible outcome, we will use 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 (in this case, the number of cars)
- k is the number of successful trials (in this case, the number of cars passing the inspection)
- p is the probability of success in a single trial (in this case, the probability of a car passing the inspection)
- (n C k) is the number of combinations of n things taken k at a time, also known as the binomial coefficient.
Let's calculate the probability for each possible outcome and sum them up.
P(0 cars fail) = (15 C 0) * (0.3)^0 * (0.7)^(15 - 0)
= 1 * 1 * 0.7^15
≈ 0.00034
P(1 car fails) = (15 C 1) * (0.3)^1 * (0.7)^(15 - 1)
= 15 * 0.3 * 0.7^14
≈ 0.00363
P(2 cars fail) = (15 C 2) * (0.3)^2 * (0.7)^(15 - 2)
= 105 * 0.3^2 * 0.7^13
≈ 0.01725
P(3 cars fail) = (15 C 3) * (0.3)^3 * (0.7)^(15 - 3)
= 455 * 0.3^3 * 0.7^12
≈ 0.05157
P(4 cars fail) = (15 C 4) * (0.3)^4 * (0.7)^(15 - 4)
= 1365 * 0.3^4 * 0.7^11
≈ 0.10313
Now, we sum up these probabilities:
P(at most 4 cars fail) = P(0 cars fail) + P(1 car fails) + P(2 cars fail) + P(3 cars fail) + P(4 cars fail)
= 0.00034 + 0.00363 + 0.01725 + 0.05157 + 0.10313
≈ 0.17592
Therefore, the probability that at most four cars fail the inspection among the 15 randomly selected cars is approximately 0.17592 or 17.592%.