Question

In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.

(Given that e^-0.2 = 0.8187)

Solution:

Here, m = `square` and X − P(m) with parameter m.

The p.m.f. X is:

P(X = x) = `(e^(−m).m^x)/(x!), x = 0, 1, 2,...`

P(X ≥ 3) = 1 − P(X < 3)

= 1 − [`square + square + square`]

= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`

= 1 − [0.8187(1 + 0.2 + 0.02)]

= 1 − `square`

= `square`

Answer 1

Here, m = \[\boxed{0.2}\] and X − P(0.2) with parameter m.

The p.m.f. of X is X ∼ P(m).

P(X = x) = `(e^(−m)  .  m^x)/(x!), x = 0, 1, 2,...`

P(X ≥ 3) = 1 − P(X < 3)

= 1 − [\[\boxed{P(0)}\] + \[\boxed{P(1)}\] + \[\boxed{P(2)}\]]

= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`

= 1 − [0.8187(1 + 0.2 + 0.02)]

= 1 − \[\boxed{0.998814 }\]

= \[\boxed{0.001186}\]