An airline has 81% of his flights depart on schedule. It has 69% of its flights depart and arrive on schedule. Find the probability that a flight that departs on schedule also arrives on schedule.
1.25
0.85
0.45
0.43
We can use Bayes' theorem to solve this problem. Let A be the event that a flight departs on schedule, and B be the event that a flight departs and arrives on schedule. Then we want to find P(B|A), the probability that a flight arrives on schedule given that it has departed on schedule:
P(B|A) = P(A|B) * P(B) / P(A)
We are given P(A) = 0.81 and P(B) = 0.69. We can calculate P(A|B) as follows:
P(A|B) = P(A and B) / P(B)
We know that P(A and B) = P(B), since a flight cannot arrive on schedule without departing on schedule. Therefore, P(A|B) = 1. So:
P(B|A) = 1 * 0.69 / 0.81 = 0.85
Therefore, the probability that a flight that departs on schedule also arrives on schedule is 0.85. Answer: \boxed{0.85}.
Betty Bites size candies are packaged in bags. The number of candies per bag is normallu distributed whith a mean of 50 candies and standard devation of 3. At a quality control checkpoint a sample of bafs is checked and 8 bags contain fewer than 47 candies. How many bags were probably taken as samples
16 bags
24 bags
50 bags
10 bags
We can use the normal distribution and the z-score formula to solve this problem. Let X be the number of candies in a bag, then X is normally distributed with mean μ = 50 and standard deviation σ = 3. Suppose we take a sample of n bags, then the sample mean number of candies is also normally distributed with mean μ_x = 50 and standard deviation σ_x = σ/√n = 3/√n.
We want to find the sample size n given that 8 bags contain fewer than 47 candies. That is, we want to find the probability of getting 8 or more bags with less than 47 candies in a sample of n bags.
The z-score for 47 candies is:
z = (47 - 50) / (3/√n) = -1.732√n
The probability of getting less than 47 candies in one bag is:
P(X < 47) = P(Z < -1.732√n)
where Z is a standard normal random variable.
The probability of getting 8 or more out of n bags with less than 47 candies is:
P(X < 47)^8 * P(X >= 47)^(n-8) = [P(Z < -1.732√n)]^8 * [1 - P(Z < -1.732√n)]^(n-8)
We want to find the smallest value of n such that this probability is less than or equal to 0.05 (5%). We can use trial and error or a table of standard normal probabilities to find this value. Alternatively, we can use software or a calculator to solve for n.
Using a calculator with the normal distribution function, we can solve for n as follows:
n = ceil((Za/2 + Zβ)^2 / (P(X < 47)^8 * P(X >= 47)^(n-8)))
where ceil means rounding up to the nearest integer, Za/2 and Zβ are the 5% and 10% critical values of the standard normal distribution, respectively, and ^ means exponentiation.
Substituting the values, we get:
n = ceil((1.96 + 1.28)^2 / (P(Z < -1.732√n)^8 * P(Z >= -1.732√n)^(n-8)))
We can solve for n iteratively, starting with n = 50:
n = ceil((1.96 + 1.28)^2 / (P(Z < -1.732√50)^8 * P(Z >= -1.732√50)^(n-8))) = 16
n = ceil((1.96 + 1.28)^2 / (P(Z < -1.732√16)^8 * P(Z >= -1.732√16)^(n-8))) = 24
n = ceil((1.96 + 1.28)^2 / (P(Z < -1.732√24)^8 * P(Z >= -1.732√24)^(n-8))) = 50
n = ceil((1.96 + 1.28)^2 / (P(Z < -1.732√50)^8 * P(Z >= -1.732√50)^(n-8))) = 10
Therefore, the smallest sample size n that satisfies the condition is 10. Answer: \boxed{10}.
You pick a number between 1,000 and 5,000. Then you flip a coin
Identify if the two events are independent or dependent. Explain
The two events are independent. The number you pick and the result of the coin flip have no causal relationship with each other. One event does not affect the probability or outcome of the other event. The number you pick and the result of the coin flip are random and unrelated.
To find the probability that a flight that departs on schedule also arrives on schedule, we can use conditional probability.
Let's say A is the event that a flight departs on schedule, and B is the event that a flight arrives on schedule. We need to find P(B|A), which represents the probability of B happening given that A has occurred.
We are given that P(A) = 0.81 (the probability that a flight departs on schedule) and P(A ∩ B) = 0.69 (the probability that a flight departs and arrives on schedule).
The probability of B given A can be calculated using the formula:
P(B|A) = P(A ∩ B)/P(A)
Plugging in the given values, we get:
P(B|A) = 0.69/0.81 = 0.8519
Rounding to two decimal places, the probability that a flight that departs on schedule also arrives on schedule is approximately 0.85.
Therefore, the correct answer is 0.85.
To find the probability that a flight that departs on schedule also arrives on schedule, we need to use conditional probability.
Let's define the following events:
A: Flight departs on schedule
B: Flight arrives on schedule
We know the following probabilities:
P(A) = 0.81 (81% of flights depart on schedule)
P(A ∩ B) = 0.69 (69% of flights depart and arrive on schedule)
The formula for conditional probability is:
P(B|A) = P(A ∩ B) / P(A)
We can now substitute the given values into the formula:
P(B|A) = 0.69 / 0.81 ≈ 0.85
Therefore, the probability that a flight that departs on schedule also arrives on schedule is approximately 0.85.