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The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b = 2.
Que : The probability that more than 3 cars will arrive during any 10 min period is
- a)0.249
- b)0.143
- c)0.346
- d)0.543
Correct answer is option 'B'. Can you explain this answer?
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Solution:
Given, the number of cars arriving at ICICI bank drive-in window during a 10-minute period is a Poisson random variable X with parameter λ = 2.
To find the probability that more than 3 cars will arrive during any 10-minute period, we need to calculate P(X > 3).
Step 1: Find the probability of having 0, 1, 2, and 3 cars
- The probability mass function of a Poisson random variable is given by P(X = k) = (e^(-λ) * λ^k) / k!, where k is the number of cars.
- Using the parameter λ = 2, we can calculate the probabilities for 0, 1, 2, and 3 cars:
- P(X = 0) = (e^(-2) * 2^0) / 0! = e^(-2) ≈ 0.1353
- P(X = 1) = (e^(-2) * 2^1) / 1! = 2e^(-2) ≈ 0.2707
- P(X = 2) = (e^(-2) * 2^2) / 2! = 2e^(-2) ≈ 0.2707
- P(X = 3) = (e^(-2) * 2^3) / 3! = (8e^(-2))/6 ≈ 0.1805
Step 2: Calculate the probability of having more than 3 cars
- To find P(X > 3), we need to sum the probabilities of having 4, 5, 6, and so on, up to infinity.
- P(X > 3) = 1 - P(X ≤ 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
≈ 1 - (0.1353 + 0.2707 + 0.2707 + 0.1805)
≈ 1 - 0.8572
≈ 0.1428
Therefore, the probability that more than 3 cars will arrive during any 10-minute period is approximately 0.1428, which is closest to option B (0.143).
Hence, the correct answer is option B.
Given, the number of cars arriving at ICICI bank drive-in window during a 10-minute period is a Poisson random variable X with parameter λ = 2.
To find the probability that more than 3 cars will arrive during any 10-minute period, we need to calculate P(X > 3).
Step 1: Find the probability of having 0, 1, 2, and 3 cars
- The probability mass function of a Poisson random variable is given by P(X = k) = (e^(-λ) * λ^k) / k!, where k is the number of cars.
- Using the parameter λ = 2, we can calculate the probabilities for 0, 1, 2, and 3 cars:
- P(X = 0) = (e^(-2) * 2^0) / 0! = e^(-2) ≈ 0.1353
- P(X = 1) = (e^(-2) * 2^1) / 1! = 2e^(-2) ≈ 0.2707
- P(X = 2) = (e^(-2) * 2^2) / 2! = 2e^(-2) ≈ 0.2707
- P(X = 3) = (e^(-2) * 2^3) / 3! = (8e^(-2))/6 ≈ 0.1805
Step 2: Calculate the probability of having more than 3 cars
- To find P(X > 3), we need to sum the probabilities of having 4, 5, 6, and so on, up to infinity.
- P(X > 3) = 1 - P(X ≤ 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
≈ 1 - (0.1353 + 0.2707 + 0.2707 + 0.1805)
≈ 1 - 0.8572
≈ 0.1428
Therefore, the probability that more than 3 cars will arrive during any 10-minute period is approximately 0.1428, which is closest to option B (0.143).
Hence, the correct answer is option B.
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The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b = 2.Que : The probability that more than 3 cars will arrive during any 10 min period isa)0.249b)0.143c)0.346d)0.543Correct answer is option 'B'. Can you explain this answer?
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