Question

Now suppose that Tommy is very adept at avoiding cars but will be injured if a truck passes during his attempt to cross the highway. The proportion of trucks to cars on Highway 6 is approximately 1 to 4. Assume that vehicles arrive in accordance with a Poisson process with rate λ = 4 per minute, and that a given vehicle that arrives has a 0.2 probability of being a truck and a 0.8 probability of being a car. What is the probability that Tommy will remain unhurt if it takes him c seconds to cross? Calculate specific probabilities for c = 3,5, 15.

Answer 1

Tommy will remain unhurt if no vehicles arrive in time or if any vehicles arrive all are cars.

The number of customers arriving in seconds can be modelld by Poisson distribution.

The Poisson PMF is

$P\left ( X=x \right )=e^{-\lambda c}\frac{\left (\lambda c \right )^x}{x!};x=0,1,2,3,....$ . Here $\lambda =4/60=1/15,c=3,5,15$

The probability of no vehicles is

$P\left ( X=0 \right )=e^{-\lambda c}\frac{\left (\lambda c \right )^0}{0!}=e^{-\lambda c}$

The probability of vehicles in seconds and all are cars is

$0.8^xP\left ( X=x \right )=0.8^xe^{-\lambda c}\frac{\left (\lambda c \right )^x}{x!};x=1,2,3,....$

Thus the probability of not being hit is

$P\left ( \textup{unhurt} \right )=P\left ( X=0 \right )+\sum_{x=1}^{\infty }0.8^xe^{-\lambda c}\frac{\left (\lambda c \right )^x}{x!}\\ P\left ( \textup{unhurt} \right )= P\left ( X=0 \right )+\sum_{x=1}^{\infty }e^{-\lambda c}\frac{\left (0.8\lambda c \right )^x}{x!}\\ P\left ( \textup{unhurt} \right )=P\left ( X=0 \right )+e^{-\lambda c}\sum_{x=1}^{\infty }\frac{\left (0.8\lambda c \right )^x}{x!}\\ P\left ( \textup{unhurt} \right )=e^{-\lambda c}+e^{-\lambda c}\left ( e^{0.8\lambda c}-1 \right )\\ P\left ( \textup{unhurt} \right ) =e^{-0.2\lambda c}$

For $c=3,5,15$ , the probabilities are

${\color{Blue} P\left (\textup{unhurt} \right )=0.9607894, 0.9355070 ,0.8187308}$