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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.
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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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