A random variable X is modeled by the Poisson distribution with parameter > > 0: P(X = k) = = e 14 k! -, k = 0, 1, However, the parameter X turns out to be from a random variable A with probability density given by -0.3X f₁(A) = 0.3e if x ≥ 0, 0 Otherwise. 1. Find the unconditioned PMF px (k) of X and evaluate P(X = 3). 2. Given X = k, find the conditional PDF £x (λ | k) and evaluate ƒÃ¡x (0.22 | 3)). Px (3), fax (0.22 | 3)) = 0.1050,0.0038
A random variable X is modeled by the Poisson distribution with parameter > > 0: P(X = k) = = e 14 k! -, k = 0, 1, However, the parameter X turns out to be from a random variable A with probability density given by -0.3X f₁(A) = 0.3e if x ≥ 0, 0 Otherwise. 1. Find the unconditioned PMF px (k) of X and evaluate P(X = 3). 2. Given X = k, find the conditional PDF £x (λ | k) and evaluate ƒÃ¡x (0.22 | 3)). Px (3), fax (0.22 | 3)) = 0.1050,0.0038
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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How were the correct answers of 0.1050 and 0.0038 attained?

Transcribed Image Text:A random variable X is modeled by the Poisson distribution with parameter > > 0:
P(X = k) =
th
k!
However, the parameter X turns out to be from a random variable A with probability density given by
f(x) = 0.3e-0.3A
if X≥ 0,
Otherwise.
,k= 0, 1,
0
(Px (3), fax (0.22 | 3)) = 0.1050,0.0038
1. Find the unconditioned PMF px (k) of X and evaluate P(X = 3).
2. Given X = k, find the conditional PDF fx (X k) and evaluate fax (0.22 | 3)).
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