IIT JAM Exam > IIT JAM Questions > Let Xbe a Poisson random variable and P(X= 1)...
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Let X be a Poisson random variable and P(X = 1) + 2 P(X = 0) = 12 P(X = 2). Which one of the following statements is TRUE?
- a)0.40 < P(X = 0) ≤ 0.45
- b)0.45 < P(X = 0) ≤ 0.50
- c)0.50 < P(X = 0) ≤ 0.55
- d)0.55 < P(X = 0) ≤ 0.60
Correct answer is option 'C'. Can you explain this answer?
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Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2)...
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Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2)...
The correct statement is a) 0.40.
Let's solve the equation using the given information:
P(X = 1) = 2P(X = 0)
P(X = 2) = 12P(X = 2)
Using the property of Poisson distribution, we know that P(X = k) = e^(-λ) * λ^k / k!, where λ is the mean of the distribution.
Let λ be the mean of the Poisson distribution.
P(X = 1) = e^(-λ) * λ^1 / 1!
2 * P(X = 0) = 2 * e^(-λ) * λ^0 / 0!
P(X = 2) = e^(-λ) * λ^2 / 2!
From the second equation, we can simplify:
2 * P(X = 0) = 2 * e^(-λ) * λ^0 / 0!
P(X = 0) = e^(-λ) * λ^0 / 0!
Since 0! = 1 and anything raised to the power of 0 is 1, we have:
2 * P(X = 0) = 2 * e^(-λ)
P(X = 0) = e^(-λ)
Dividing the first equation by the second equation, we get:
P(X = 1) / P(X = 0) = 2 * P(X = 0) / P(X = 0)
P(X = 1) = 2
Plugging this value into the third equation, we get:
P(X = 2) = 12 * P(X = 2)
1 = 12
This is not possible, as it leads to a contradiction. Therefore, the given information is not consistent.
Therefore, none of the statements are true.
Let's solve the equation using the given information:
P(X = 1) = 2P(X = 0)
P(X = 2) = 12P(X = 2)
Using the property of Poisson distribution, we know that P(X = k) = e^(-λ) * λ^k / k!, where λ is the mean of the distribution.
Let λ be the mean of the Poisson distribution.
P(X = 1) = e^(-λ) * λ^1 / 1!
2 * P(X = 0) = 2 * e^(-λ) * λ^0 / 0!
P(X = 2) = e^(-λ) * λ^2 / 2!
From the second equation, we can simplify:
2 * P(X = 0) = 2 * e^(-λ) * λ^0 / 0!
P(X = 0) = e^(-λ) * λ^0 / 0!
Since 0! = 1 and anything raised to the power of 0 is 1, we have:
2 * P(X = 0) = 2 * e^(-λ)
P(X = 0) = e^(-λ)
Dividing the first equation by the second equation, we get:
P(X = 1) / P(X = 0) = 2 * P(X = 0) / P(X = 0)
P(X = 1) = 2
Plugging this value into the third equation, we get:
P(X = 2) = 12 * P(X = 2)
1 = 12
This is not possible, as it leads to a contradiction. Therefore, the given information is not consistent.
Therefore, none of the statements are true.
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Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2). Which one of the following statements is TRUE?a)0.40 < P(X= 0) ≤ 0.45b)0.45 < P(X= 0) ≤ 0.50c)0.50 < P(X= 0) ≤ 0.55d)0.55 < P(X= 0) ≤ 0.60Correct answer is option 'C'. Can you explain this answer?
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Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2). Which one of the following statements is TRUE?a)0.40 < P(X= 0) ≤ 0.45b)0.45 < P(X= 0) ≤ 0.50c)0.50 < P(X= 0) ≤ 0.55d)0.55 < P(X= 0) ≤ 0.60Correct answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2). Which one of the following statements is TRUE?a)0.40 < P(X= 0) ≤ 0.45b)0.45 < P(X= 0) ≤ 0.50c)0.50 < P(X= 0) ≤ 0.55d)0.55 < P(X= 0) ≤ 0.60Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Xbe a Poisson random variable and P(X= 1) + 2 P(X= 0) = 12 P(X= 2). Which one of the following statements is TRUE?a)0.40 < P(X= 0) ≤ 0.45b)0.45 < P(X= 0) ≤ 0.50c)0.50 < P(X= 0) ≤ 0.55d)0.55 < P(X= 0) ≤ 0.60Correct answer is option 'C'. Can you explain this answer?.
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