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In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 1/4. If the probability that at most two machines will be out of service on the same day is (3/4)3 k, then k is equal to
- a)4
- b)17/4
- c)17/8
- d)17/2
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
In a workshop, there are five machines and the probability of any one ...
Probability that a machine is out of service = 1/4 = P
Probability that a machine is not out of service
= 1 - 1/4 = 3/4 = q
∴ Probability that at most two machines are out of service = P(X = 0) + P(X = 1) + P(X = 2)
∴
⇒
∴ k =
Probability that a machine is not out of service
= 1 - 1/4 = 3/4 = q
∴ Probability that at most two machines are out of service = P(X = 0) + P(X = 1) + P(X = 2)
∴
⇒
∴ k =
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In a workshop, there are five machines and the probability of any one ...
To solve this problem, we can use the concept of binomial distribution. Let's go step by step to understand the solution.
First, let's calculate the probability that exactly two machines will be out of service on a given day. This can be represented by the binomial distribution formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability that exactly k machines are out of service
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k machines out of n
- p is the probability of a machine being out of service on a given day
- n is the total number of machines
In this case, n = 5 and p = 1/4. So, for k = 2, we have:
P(X = 2) = C(5, 2) * (1/4)^2 * (3/4)^(5-2)
= 10 * (1/16) * (27/64)
= 270/1024
Next, let's calculate the probability that at most two machines will be out of service. This includes the probabilities of 0, 1, and 2 machines being out of service. We can calculate this using the cumulative distribution function (CDF) of the binomial distribution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= C(5, 0) * (1/4)^0 * (3/4)^(5-0) + C(5, 1) * (1/4)^1 * (3/4)^(5-1) + C(5, 2) * (1/4)^2 * (3/4)^(5-2)
= (1/4)^0 * (3/4)^5 + 5 * (1/4)^1 * (3/4)^4 + 10 * (1/4)^2 * (3/4)^3
= 1 * 243/1024 + 5 * 81/1024 + 10 * 27/1024
= 243/1024 + 405/1024 + 270/1024
= 918/1024
Finally, we are given that P(X ≤ 2) = (3/4)^3 * k. Setting this equal to 918/1024, we can solve for k:
(3/4)^3 * k = 918/1024
27/64 * k = 918/1024
k = (918/1024) * (64/27)
k = 17/8
Therefore, the correct answer is option C, k = 17/8.
First, let's calculate the probability that exactly two machines will be out of service on a given day. This can be represented by the binomial distribution formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability that exactly k machines are out of service
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k machines out of n
- p is the probability of a machine being out of service on a given day
- n is the total number of machines
In this case, n = 5 and p = 1/4. So, for k = 2, we have:
P(X = 2) = C(5, 2) * (1/4)^2 * (3/4)^(5-2)
= 10 * (1/16) * (27/64)
= 270/1024
Next, let's calculate the probability that at most two machines will be out of service. This includes the probabilities of 0, 1, and 2 machines being out of service. We can calculate this using the cumulative distribution function (CDF) of the binomial distribution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= C(5, 0) * (1/4)^0 * (3/4)^(5-0) + C(5, 1) * (1/4)^1 * (3/4)^(5-1) + C(5, 2) * (1/4)^2 * (3/4)^(5-2)
= (1/4)^0 * (3/4)^5 + 5 * (1/4)^1 * (3/4)^4 + 10 * (1/4)^2 * (3/4)^3
= 1 * 243/1024 + 5 * 81/1024 + 10 * 27/1024
= 243/1024 + 405/1024 + 270/1024
= 918/1024
Finally, we are given that P(X ≤ 2) = (3/4)^3 * k. Setting this equal to 918/1024, we can solve for k:
(3/4)^3 * k = 918/1024
27/64 * k = 918/1024
k = (918/1024) * (64/27)
k = 17/8
Therefore, the correct answer is option C, k = 17/8.
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In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 1/4. If the probability that at most two machines will be out of service on the same day is (3/4)3 k, then k is equal toa)4b)17/4c)17/8d)17/2Correct answer is option 'C'. Can you explain this answer?
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