Question

Traffic on Rosedale Road in Princeton, NJ, follows a Poisson process with rate6cars per minute. Adeer runs out of the woods and tries to cross the road. It takes the deer a random time (independentof everything else), uniformly distributed between2and5seconds, to cross the road. If there is acar passing while the deer is on the road, it will hit the deer with portability2/3(independent ofeverything else). Find the probability of a collision.

Answer 1

Answer:

Step-by-step explanation:

From the given information:

6 cars pass on the road per minute. This implies that per minute, we will have:

6/60 passes = 0.1 car passing the road per second

Thus, for 5 sec; we have:

5 × 0.1 = 0.5

Let assume X signifies to be a random variable which denote the no of cars in the next  5 minutes and which follows a Poisson distribution.

i.e.

$X \sim Poisson (0.5)$

$\text{P(collision occurs) = 1 - P(no car coming in the following 5 seconds)} \\ \\ =1 - P(X=0)$

$= 1- \dfrac{[e^{-0.5\times 0.5^0}]}{0!}$

$= 1 - e^{-0.5}$

$= 1 - 0.6065$

$= 0.3935$

However, the possibility of collision supposes the deer requires 2 seconds to cross the road can be computed as follows:

Suppose Y denote the arbitrary variable that addresses the no. of cars passing the road in 2 secs, then;

$Y \sim Poisson (0.2)$

$\text{P(collision occurs) = 1 - P(no car coming in the following 2 seconds)} \\ \\ =1 - P(Y=0)$

$= 1- \dfrac{[e^{-0.2\times 0.2^0}]}{0!}$

$= 1 - e^{-0.2}$

$= 1 - 0.8187$

$= 0.1813$