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)
Christian’s Response:
To solve this problem, we can apply the Poisson probability for mass function and substitute the values for X and λ when solving (X=0 and x=1).
\(f(x) = \frac{λ^{x}e^{-λ}}{x!}\)
When x = 0
\(f(0) = \frac{.3^{0}e^{-.3}}{0!}= \frac{1*.7408}{1}=7408\)
When x = 1
\(f(1) = \frac{.3^{1}e^{-.3}}{1!}= \frac{.3*.7408}{1}=.2222\)
Finally to solve for when P(x>1) we simply subtract 1 from P(x=0) and P(x=1) to get the probability that there is more than one event in the interval
\(1 - 0.7408-0.2222 = 0.037\)
To solve in R, we can use the dpois function to apply to Poisson probability for mass function and carry out the same steps performed above:
prob_0 <- dpois(0,0.3)
prob_1 <- dpois(1,0.3)
more_than1 <- 1 - prob_0 - prob_1
cat("The value of P(x>0) is", more_than1)## The value of P(x>0) is 0.03693631