Step by step
The question
Question: X is Poisson with mean 2: find the probability that X is exactly 0.
Poisson, μ = 2: P(X = 0)?
Poisson counts how many times something happens in a fixed time, when we know the average.
Poisson: counts in a time or a space, with a known average
1 Name the numbers
μ, the average, is 2.
μ = 2
x, the count we want, is 0.
x = 0
2 Write the formula
The Poisson formula: μ to the power x, times e to the power minus μ, over x factorial.
P(X = x) = μx × e−μx!
e is a number, about 2.718.
e to the power minus 2 is about 0.1353.
x factorial multiplies down to 1, and 0 factorial is 1.
e ≈ 2.718, e−2 ≈ 0.1353
x! = x × ... × 2 × 1, 0! = 1
Careful!
0 factorial is 1, not 0.
Careful: 0! = 1 (not 0)
3 Put the numbers in
2 to the power 0 is 1.
0 factorial is 1: 1.
20 = 1, 0! = 1 = 1
4 Work it out
1 times 0.1353, over 1, is about 0.1353.
P(X = 0) = 20 × e−20! = 1 × 0.13531 ≈ 0.1353
5 Check
Let's check another way: step up from P of 0.
P of 0 is e to the power minus 2: about 0.1353.
P(0) = e−2 ≈ 0.1353
Let's look back.
First: name the numbers.
Then: write the formula.
Then: put the numbers in.
Then: work it out.
Last: check.
Answer
P(X = 0) = 20 × e−20! = 1 × 0.13531 ≈ 0.1353