Question

26%10:00 p.m × Question A worker at a fuel station knows from his experience that average nom of cars that arrive fir refiling at the station is 2 per minute. what is the probability that: a) there will be more than 2 cars in the next one minute. b) there will be atleat one car in the next 5 minutes

Answer 1

Whenever we have to estimate the probabilty of the number of cars arriving (arrival process), we model the situation using a Poisson distribution.

(a)

Let X denote the random variable representing the no. of cars arriving in a minute or arriving in the next one minute.

Now, we assume: $X \sim Poi(\lambda)$

$\Rightarrow E(X) = \lambda$

But we are given that the average number of cars arriving at the station in a minute is 2 cars.

Thus,

$\begin{align*} E(X) &= 2 \\ \Rightarrow \ \ \ \ \ \lambda &= 2 \end{align*}$

Thus, the probability mass function of X is given by:

$P(X=x) = \frac{e^{-\lambda} \lambda ^x}{x!}=\frac{e^{-2}*2^x}{x!} \ \ \ ; x=0,1,2,...$

Now, the required probability:

$\begin{align*} \text{P(more than 2 cars in next minute)} &= P(X>2) \\ &= 1 - P(X \leq 2) \\ &= 1 - [P(X=0)+P(X=1)+P(X=2)] \\ &= 1 - [\frac{e^{-2}*2^0}{0!} + \frac{e^{-2}*2^1}{1!} + \frac{e^{-2}*2^2}{2!}] \\ &= 1 - [e^{-2} + 2e^{-2}+ 2e^{-2}] \\ &= 1-5e^{-2} \\ &= 1 - 5*0.135335 \\ &= 0.323324 \end{align*}$ which is the required answer.

(b)

Let Y denote the random variable representing the no. of cars arriving in 5 minutes or arriving in the next 5 minutes.

Now, we assume: $Y \sim Poi(\mu)$

$\Rightarrow E(Y) = \mu$

But we are given that the average number of cars arriving at the station in a minute is 2 cars.

Thus, the average number of cars arriving at the station in the next 5 minutes = 5*2 = 10 cars.

Thus,

$\begin{align*} E(Y) &= 10 \\ \Rightarrow \ \ \ \ \ \mu &= 10 \end{align*}$

Thus, the probability mass function of Y is given by:

$P(Y=y) = \frac{e^{-\mu} \mu^y}{y!}=\frac{e^{-10}*10^y}{y!} \ \ \ ; y=0,1,2,...$

Now, the required probability:

$\begin{align*} \text{P(atleast one car in next 5 minutes)} &= P(Y \geq 1) \\ &= 1 - P(Y < 1) \\ &= 1 - P(Y=0) \\ &= 1 - \frac{e^{-10}*10^0}{0!} \\ &= 1- e^{-10} \\ &= 1 - 0.0000454 \\ &= 0.9999546 \end{align*}$

which is the required answer.

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