Question

Suppose that at a busy traffic junction, the probability p of an individual car having an accident is 0.0001. During a certain part of the day, 100 cars pass through the junction. What is the probability of two or more cars being involved in an accident within this period?

Answer 1

Let X = the random variable denoting the number of cars being involved in an accident during the certain part of the day, X = 0(1)100

We have, the probability p of an individual car having an accident is 0.0001.

Then X ~ bin(n = 100, p = 0.0001)

Here we see that,

(i) n = 100 is large i.e. n $\to\infty$

(ii) p = 0.0001 is small i.e. p $\to$ 0

(iii) np = 0.01 is finite

Therefore, the Binomial distribution can be approximated by Poisson distribution with parameter $\lambda$ = np = 0.01.

Hence, X ~ P(λ = 0.01)

The p.m.f. of X is given by,

P(X = x) = $\frac{e^{-0.01}(0.01)^x}{x!}$ , x = 0(1)100

= 0 , otherwise

Therefore, the probability of 2 or more cars being involved in an accident within the certain period

= P(X $\geq$ 2)

= 1 - P(X < 2)

= 1 - [P(X = 0) + P(X = 1)]

= 1 - [$\frac{e^{-0.01}(0.01)^0}{0!} + \frac{e^{-0.01}(0.01)^1}{1!}$ ]

= 1 - 0.9901 x (1 + 0.01) = 0.0000

Answer: The probability of 2 or more cars being involved in an accident within the certain period is 0.