Question 1:
Consider a stochastic process {X(t)} where X(t) represents the number of customers arriving at a shop in a given time interval t. The arrival of customers follows a Poisson process with an average rate of 5 customers per hour. Calculate the probability of having exactly 3 customers arriving within a 30-minute time interval.
Solution: To solve this problem, we can use the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k! where λ is the average rate of arrival.
In this case, the average rate of arrival is 5 customers per hour, which means λ = 5 * (30 minutes / 60 minutes) = 2.5.
Let's calculate the probability of having exactly 3 customers arriving within a 30-minute time interval:
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P(X = 3) = (e^(-2.5) * 2.5^3) / 3!
P(X = 3) = (e^(-2.5) * 2.5^3) / (3 * 2 * 1)
Calculating this expression, we find:
P(X = 3) ≈ 0.2139
Therefore, the probability of having exactly 3 customers arriving within a 30-minute time interval is approximately 0.2139 or 21.39%.
Question 2:
A manufacturing process produces widgets with a defective rate of 10%. Let's assume that the number of defective widgets produced follows a binomial distribution. If a batch of 100 widgets is produced, what is the probability that exactly 15 of them are defective?
Solution: In this case, we can use the binomial distribution formula to calculate the probability:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k) where n is the number of trials, k is the number of successful outcomes, and p is the probability of success.
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Here, n = 100 (the number of widgets produced), k = 15 (the number of defective widgets), and p = 0.10 (the defective rate).
Let's calculate the probability of having exactly 15 defective widgets:
P(X = 15) = (100 choose 15) * 0.10^15 * (1 - 0.10)^(100 - 15)
Using a calculator or statistical software, we can compute this expression:
P(X = 15) ≈ 0.1303
Therefore, the probability of having exactly 15 defective widgets in a batch of 100 widgets is approximately 0.1303 or 13.03%.
Question 3:
A company receives customer support calls at a rate of 8 calls per hour. The time between successive calls follows an exponential distribution. What is the probability that the company will receive its next call within 10 minutes?
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Solution: To solve this problem, we need to convert the time from minutes to hours since the rate of calls is given in hours. 10 minutes is equal to 10/60 = 1/6 hours.
The time between successive calls follows an exponential distribution with a rate parameter λ. In this case, λ is equal to the average rate of calls, which is 8 calls per hour. Let's calculate the probability that the company will receive its next call within 10 minutes:
P(X < 1/6) = 1 - e^(-λ * (1/6))
P(X < 1/6) = 1 - e^(-8 * (1/6))
Using a calculator or software to compute this expression, we find:
P(X < 1/6) ≈ 0.4276
Therefore, the probability that the company will receive its next call within 10 minutes is approximately 0.4276 or 42.76%.
Question 4: A car rental agency has two types of cars: Economy (E) and Luxury (L).
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The agency has a total of 10 cars, out of which 6 are Economy and 4 are Luxury. A customer arrives at the agency and randomly selects a car. What is the probability that the customer selects a Luxury car?
Solution: To calculate the probability of selecting a Luxury car, we need to consider the proportion of Luxury cars out of the total number of cars.
Let's define: Event A: Selecting a Luxury car. Event B: Selecting any car (Economy or Luxury).
The probability of selecting a Luxury car can be calculated as:
P(A) = P(A | B) * P(B)
P(A | B) is the conditional probability of selecting a Luxury car given that any car is selected. In this case, since the selection is random, the probability of selecting a Luxury car is the proportion of Luxury cars out of the total number of cars.
P(A | B) = Number of Luxury cars / Total number of cars = 4 / 10 = 0.4
P(B) is the probability of selecting any car, which is 1 since the customer will select a car for sure.
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P(A) = P(A | B) * P(B) = 0.4 * 1 = 0.4
Therefore, the probability that the customer selects a Luxury car is 0.4 or 40%.
Question 5: A manufacturing company produces light bulbs. The lifetime of a light bulb follows an exponential distribution with a mean lifetime of 1000 hours. What is the probability that a randomly selected light bulb will last at least 1200 hours?
Solution: The exponential distribution is characterized by a rate parameter, λ, which is the reciprocal of the mean. In this case, the mean lifetime is 1000 hours, so λ = 1/1000.
Let's calculate the probability that a randomly selected light bulb will last at least 1200 hours:
P(X ≥ 1200) = 1 - P(X < 1200)
The cumulative distribution function (CDF) of the exponential distribution is given by:
CDF(x) = 1 - e^(-λx)
Substituting the values, we have:
P(X ≥ 1200) = 1 - (1 - e^(-λ * 1200))
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P(X ≥ 1200) = e^(-λ * 1200)
Substituting the value of λ = 1/1000, we get:
P(X ≥ 1200) = e^(-(1/1000) * 1200)
Using a calculator or software to compute this expression, we find:
P(X ≥ 1200) ≈ 0.3012
Therefore, the probability that a randomly selected light bulb will last at least 1200 hours is approximately 0.3012 or 30.12%.
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