Poisson mean 2, P(X >= 1) Step by Step | The Poisson distribution

Step by step

The question

Question: X is Poisson with mean 2: find the probability that X is at least 1.

Poisson, μ = 2: P(X ≥ 1)?

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

At least 1: x can be 1, 2, 3, and so on, with no end.

x = 1, 2, 3, ...

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) = μ^x × 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 Use the rest

x can go on with no end, so we cannot add them all.

All the probabilities add up to 1: so we take away the ones below 1.

P(X ≥ 1) = 1 − P(X < 1)

For 0: 2 to the power 0 is 1, and 0 factorial is 1.

1 times 0.1353, over 1, is about 0.1353.

P(X = 0) = 2⁰ × e⁻²0! = 1 × 0.13531 ≈ 0.1353

1 minus 0.1353 is about 0.8647.

P(X ≥ 1) = 1 − 0.1353 ≈ 0.8647

4 Check

Let's check the key number another way: e to the power 2 is about 7.3891.

1 divided by it is about 0.1353: the same as e to the power minus 2.

So it is right!

e² ≈ 7.3891, 1 ÷ 7.3891 ≈ 0.1353

Let's look back.

First: name the numbers.

Then: write the formula.

Then: use the rest.

Last: check.

Answer