Question

There are 4 accidents, on average, at an intersection. Assume the variable follows a Poisson distribution. Find the probability that there will be less than 2 accidents at this intersection.

Answer 1

P [ X < 2 ] = 0,091 or P [ X < 2 ] = 9,1 %

Explanation:

The random variable follows a Poisson Distribution then:

P [ X = x ] = λˣ *e₋∧-λ / x!

λ = 4

The probability of X less than 2 is:

P [ X < 2 ] = P[ X = 0 ] + P[ X = 1 ]

P[ X = 0 ] = 4⁰ * e ∧ -4 / 0!

P[ X = 0 ] = e⁻⁴ / 0!

P[ X = 0 ] = e⁻⁴ /1

P[ X = 0 ] = 0,01831

P[ X = 0 ] = 0,01831 or P[ X = 0 ] = 1,8 %

Now

P[ X = 1 ] = 4¹ * e⁻⁴ / 1!

P[ X = 1 ] = 4 * 0,01831

P[ X = 1 ] = 0,073 or P[ X = 1 ] = 7,3 %

Then

P [ X < 2 ] = 0,01831 + 0,073

P [ X < 2 ] = 0,091 or P [ X < 2 ] = 9,1 %