5.1 Question 13
The Poisson distribution with parameter λ = .3 has been assigned for the outcome of an experiment. Let X be the outcome function. Find P (X = 0), P (X = 1), and P (X > 1).
The Poisson distribution is as the following:
\[P(X = x) = \lambda^xe^{-\lambda}/x!\]
# P (X = 0)
lambda_dist <- 0.3
x <- 0
x_0 <- lambda_dist**x * exp(-lambda_dist) / factorial(x)
print(paste0("P(X = 0) = ", x_0))## [1] "P(X = 0) = 0.740818220681718"# P (X = 1)
lambda_dist <- 0.3
x <- 1
x_1 <- lambda_dist**x * exp(-lambda_dist) / factorial(x)
print(paste0("P(X = 1) = ", x_1))## [1] "P(X = 1) = 0.222245466204515"Now, we must remember that the Poisson Distribution still follows the laws of probabilities. In other words, P(0) + P(1) + P(2) + … P(\(\infty\)) = 1. Therefore, P (X > 1) = 1 - P(0) - P(1).
# Reference: https://onlinecourses.science.psu.edu/stat414/node/82
# P (X > 1)
P_greater_1 <- 1 - x_0 - x_1
print(paste0("P (X > 1) = ", P_greater_1))## [1] "P (X > 1) = 0.0369363131137668"