Question

Traffic fatalities and sporting events. The relationship betweenclose sporting events and game-day traffic fatalities was investigated in the Journal of Consumer Research (December 2011). The researchers found that closer football and basketball games are associated with more traffic fatalities. The methodology used by the researchers involvedmodeling the traffic fatality count for a particular game as a Poisson random variable. For games played at the winner’s location (home court or home field), the mean number of traffic fatalities was .5. Use this information to find the probability that at least three game-day traffic fatalities will occur at the winning team’s location.

Short answer

The probability that at least three game-day traffic fatalities will occur is 0.0144.

Step-by-step solution

Given information

To the Journal of Consumer Research (December 2011), The traffic fatality count for a particular game as a Poisson random variable

Calculation for the probability that at least three game-day traffic fatalities will occur

The random variable x is the number of game-day traffic fatalities in the winning location.

Here, x follows a Poisson distribution with a mean is 0.5

The probability that at least three game-day traffic fatalities will occur at the winning team’s location,

The probability mass function of x is,

$f\left(x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$where$\lambda =0.5$

$\begin{array}{rcl}P\left(x⩾3\right)& =& 1-P\left(x<3\right)\\ & =& 1-P\left(x=0\right)-P\left(x=1\right)-P\left(x=2\right)\\ & =& 1-\frac{e^{-0.5}{0.5}^0}{0!}-\frac{e^{-0.5}{0.5}^1}{1!}-\frac{e^{-0.5}{0.5}^2}{2!}\\ & =& 1-e^{-0.5}\left[\frac{1}{1}+\frac{0.5}{1}+\frac{0.25}{2}\right]\\ & & \end{array}$

$\begin{array}{rcl}& =& 1-0.6065\left[1+0.5+0.125\right]\\ & =& 1-\left(0.6065\times 1.625\right)\\ & =& 1-0.9855625\\ & =& 0.0144375\\ & \approx & 0.0144\\ & & \end{array}$

$P\left(x⩾3\right)=0.0144$

Therefore, the probability that at least three game-day traffic fatalities will occur is 0.0144.