Question

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter λ = 3.

(a) Find the probability that 3 or more accidents occur today.

(b) Repeat part (a) under the assumption that at least 1 accident occurs today.

Short answer

The probability that at least $3$ accidents occur is $0.57681$ while the probability that at least $3$ accidents occur under the assumption that at least $1$ accident occurs today is $0.60703$.

Step-by-step solution

Step 1:Given Information (Part a)

The number of accidents occurring on a highway each day is a Poisson random variable with parameter$\lambda =3$.

Step 2:Calculation (Part a)

$P(X\ge 3)=1-\sum _{i=0}^2\frac{3^i}{i!e^{\lambda }}$

$=1-\left(\frac{3^0}{0!e^3}+\frac{3^1}{1!e^3}+\frac{3^2}{2!e^3}\right)$

$=0.57681$.

Final answer (Part a)

The probability that at least $3$accidents occur is$0.57681$.

Given Information (Part b)

The number of accidents occurring on a highway each day is a Poisson random variable with parameter $\lambda =3$.

Calculation (Part b)

$P\left\{\right(X\ge 3)\mid (X\ge 1\left)\right\}=\frac{P\left\{\right(X\ge 3)\cap (X\ge 1\left)\right\}}{P\{X\ge 1\}}$

$(X\ge 1)\subset (X\ge 3)\Rightarrow \frac{P\left\{\right(X\ge 3)\cap (X\ge 1\left)\right\}}{P\{X\ge 1\}}=\frac{P\{X\ge 3\}}{P\{X\ge 1\}}$

Substitute the given expression,

$\frac{P\{X\ge 3\}}{P\{X\ge 1\}}=\frac{0.57681}{1-P\{X=0\}}$

$=\frac{0.57681}{1-e^{-3}}$

$=0.60703$.

Final answer (Part b)

The probability that at least 1 accidents occur is $0.60703$.