Question
Of all the registered automobiles in Colorado, 5% fail the state emissions test. 15 automobiles are selected at random to undergo an emissions test. Find the probability that exactly 3 of them will fail the test is what
Answers
GPT 3.5
To find the probability that exactly 3 of the 15 selected automobiles will fail the test, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n = total number of trials (15 automobiles selected)
- k = number of successes (exactly 3 automobiles fail)
- p = probability of success on each trial (5% or 0.05)
- (n choose k) = n! / (k! * (n-k)!)
Plugging in the values:
P(X = 3) = (15 choose 3) * 0.05^3 * (1-0.05)^(15-3)
P(X = 3) = (455) * 0.000125 * 0.857375
P(X = 3) = 0.0556
Therefore, the probability that exactly 3 out of the 15 selected automobiles will fail the emissions test is approximately 0.0556.
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n = total number of trials (15 automobiles selected)
- k = number of successes (exactly 3 automobiles fail)
- p = probability of success on each trial (5% or 0.05)
- (n choose k) = n! / (k! * (n-k)!)
Plugging in the values:
P(X = 3) = (15 choose 3) * 0.05^3 * (1-0.05)^(15-3)
P(X = 3) = (455) * 0.000125 * 0.857375
P(X = 3) = 0.0556
Therefore, the probability that exactly 3 out of the 15 selected automobiles will fail the emissions test is approximately 0.0556.
Find the probability that fewer than 3 of them fail the test
GPT 3.5
To find the probability that fewer than 3 out of the 15 selected automobiles will fail the test, we need to find the probabilities of exactly 0, 1, or 2 of them failing the test and add them together.
For 0 failures:
P(X = 0) = (15 choose 0) * 0.05^0 * (1-0.05)^15 = 0.2873
For 1 failure:
P(X = 1) = (15 choose 1) * 0.05^1 * (1-0.05)^14 = 0.3966
For 2 failures:
P(X = 2) = (15 choose 2) * 0.05^2 * (1-0.05)^13 = 0.2390
Now, add these probabilities together:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X < 3) = 0.2873 + 0.3966 + 0.2390
P(X < 3) = 0.9229
Therefore, the probability that fewer than 3 out of the 15 selected automobiles will fail the emissions test is approximately 0.9229.
For 0 failures:
P(X = 0) = (15 choose 0) * 0.05^0 * (1-0.05)^15 = 0.2873
For 1 failure:
P(X = 1) = (15 choose 1) * 0.05^1 * (1-0.05)^14 = 0.3966
For 2 failures:
P(X = 2) = (15 choose 2) * 0.05^2 * (1-0.05)^13 = 0.2390
Now, add these probabilities together:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X < 3) = 0.2873 + 0.3966 + 0.2390
P(X < 3) = 0.9229
Therefore, the probability that fewer than 3 out of the 15 selected automobiles will fail the emissions test is approximately 0.9229.
Find the probability that more than 2 of them fail the test
GPT 3.5
To find the probability that more than 2 out of the 15 selected automobiles will fail the test, we can find the complementary probability of having 0, 1, or 2 failures, and then subtract that from 1 to get the probability of more than 2 failures.
For 0, 1, and 2 failures:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X ≤ 2) = 0.2873 + 0.3966 + 0.2390
P(X ≤ 2) = 0.9229
Now subtracting this from 1 gives the probability of more than 2 failures:
P(X > 2) = 1 - P(X ≤ 2)
P(X > 2) = 1 - 0.9229
P(X > 2) = 0.0771
Therefore, the probability that more than 2 out of the 15 selected automobiles will fail the emissions test is approximately 0.0771.
For 0, 1, and 2 failures:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X ≤ 2) = 0.2873 + 0.3966 + 0.2390
P(X ≤ 2) = 0.9229
Now subtracting this from 1 gives the probability of more than 2 failures:
P(X > 2) = 1 - P(X ≤ 2)
P(X > 2) = 1 - 0.9229
P(X > 2) = 0.0771
Therefore, the probability that more than 2 out of the 15 selected automobiles will fail the emissions test is approximately 0.0771.
Would it be unusual for none of them to fail the test? Use a cutoff of 0.05. and what's the probability
GPT 3.5
To determine if it would be unusual for none of the 15 selected automobiles to fail the emissions test, we need to calculate the probability of 0 failures occurring and compare it to the cutoff of 0.05.
The probability of 0 failures:
P(X = 0) = (15 choose 0) * 0.05^0 * (1-0.05)^15
P(X = 0) = 1 * 1 * 0.3193
P(X = 0) = 0.3193
Since 0.3193 is greater than 0.05, it would not be considered unusual for none of the selected automobiles to fail the emissions test based on the cutoff of 0.05.
Therefore, the probability of none of the 15 selected automobiles failing the emissions test is approximately 0.3193.
The probability of 0 failures:
P(X = 0) = (15 choose 0) * 0.05^0 * (1-0.05)^15
P(X = 0) = 1 * 1 * 0.3193
P(X = 0) = 0.3193
Since 0.3193 is greater than 0.05, it would not be considered unusual for none of the selected automobiles to fail the emissions test based on the cutoff of 0.05.
Therefore, the probability of none of the 15 selected automobiles failing the emissions test is approximately 0.3193.