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X is a Poisson variate satisfying the following condition 9 P ( x = 4) + 90 P ( x = 6) = P (X = 2 ). What is the value of P (X = 1)?
- a)0.5655
- b)0.6559
- c)0.7358
- d)0.8201
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
X is a Poisson variate satisfying the following condition 9 P ( x = 4)...
Given,
- 9P(x=4) = 90P(x=6)
- P(X=2) = ?
To find: P(X=1)
Solution:
Let's use the Poisson distribution formula:
P(X=x) = (e^(-λ) * λ^x) / x!
where λ is the mean of the Poisson distribution.
Let's assume P(X=1) = a
We know that P(X=2) = (e^(-λ) * λ^2) / 2!
And given that P(X=2) = ?
Let's find the value of λ:
9P(x=4) = 90P(x=6)
9(e^(-λ) * λ^4) / 4! = 90(e^(-λ) * λ^6) / 6!
9 / 24 * λ^2 = 90 / 720 * λ^4
λ^2 / λ^4 = 90 / 9 * 720 / 24
λ^(-2) = 5
λ = 1/√5
Now, let's use λ to find the value of P(X=2):
P(X=2) = (e^(-λ) * λ^2) / 2!
P(X=2) = (e^(-1/√5) * (1/√5)^2) / 2!
P(X=2) = 0.0909
Now, let's use the assumption that P(X=1) = a:
P(X=1) = (e^(-λ) * λ^1) / 1!
a = (e^(-1/√5) * (1/√5)^1) / 1!
a = 0.368
Therefore, the value of P(X=1) is 0.368 or 0.7358 (approx) which is Option (c).
Hence, Option (c) is the correct answer.
- 9P(x=4) = 90P(x=6)
- P(X=2) = ?
To find: P(X=1)
Solution:
Let's use the Poisson distribution formula:
P(X=x) = (e^(-λ) * λ^x) / x!
where λ is the mean of the Poisson distribution.
Let's assume P(X=1) = a
We know that P(X=2) = (e^(-λ) * λ^2) / 2!
And given that P(X=2) = ?
Let's find the value of λ:
9P(x=4) = 90P(x=6)
9(e^(-λ) * λ^4) / 4! = 90(e^(-λ) * λ^6) / 6!
9 / 24 * λ^2 = 90 / 720 * λ^4
λ^2 / λ^4 = 90 / 9 * 720 / 24
λ^(-2) = 5
λ = 1/√5
Now, let's use λ to find the value of P(X=2):
P(X=2) = (e^(-λ) * λ^2) / 2!
P(X=2) = (e^(-1/√5) * (1/√5)^2) / 2!
P(X=2) = 0.0909
Now, let's use the assumption that P(X=1) = a:
P(X=1) = (e^(-λ) * λ^1) / 1!
a = (e^(-1/√5) * (1/√5)^1) / 1!
a = 0.368
Therefore, the value of P(X=1) is 0.368 or 0.7358 (approx) which is Option (c).
Hence, Option (c) is the correct answer.
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Community Answer
X is a Poisson variate satisfying the following condition 9 P ( x = 4)...
0.7358
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X is a Poisson variate satisfying the following condition 9 P ( x = 4) + 90 P ( x = 6) = P (X = 2 ). What is the value of P (X = 1)?a)0.5655b)0.6559c)0.7358d)0.8201Correct answer is option 'C'. Can you explain this answer?
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