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.

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

Step-by-step 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 %

no one knows lol

step-by-step explanation: