1. The number of injuries at a college football game has the following probability distribution:

    $$\begin{array}{|l:l|} \hline \textrm{Number of injuries} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hdashline \textrm{Probability} & 0.14 & 0.33 & 0.23 & 0.17 & 0.09 & 0.04 \\ \hline \end{array}$$

    The coach must file a report whenever there is more than one injury in any game.

    a. Calculate the expected number of reports filed in the next 20 games.

    b. Calculate the standard deviation of the number of reports filed in the next 20 games.

    c. Calculate the expected number of football games to be played before the first game in which there are at least four injuries.


closed as off-topic by Graham Kemp, heropup, Claude Leibovici, Watson, colormegone Apr 4 '16 at 7:14

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  • $\begingroup$ Note: This does not involve Poisson distribution at all. $\endgroup$ – Graham Kemp Apr 4 '16 at 3:44
  • $\begingroup$ Rather than just cut-and-posting the problem, it is best practice to provide an indication of what you can do and where you are having difficulties. What have you attempted so far? People are more willing to help out if they can see that you have tried to do your own work. $\endgroup$ – Graham Kemp Apr 4 '16 at 3:48
  • $\begingroup$ I am trying to figure out what strategy I should use. Is this a binomial or poissson distribution? $\endgroup$ – user3777772 Apr 4 '16 at 3:58
  • $\begingroup$ Neither:-) To start can you count expected value of injuries in one game? $\endgroup$ – iiivooo Apr 4 '16 at 4:19
  • $\begingroup$ .14(0)+.33(1)+.23(2)+.17(3)+.09(4)+.04(5)=1.86 $\endgroup$ – user3777772 Apr 4 '16 at 4:48


  1. The probability that coach must fill the report after a game is the probability that there is more than one injury, count it and denote it $p$. Since obviously the numbers of injuries in different games are independent, number of filled report in 20 games has Binomial distribution with parameters $20$ and $p$ (sorry I made mistake in comments, did read your question too fast).

  2. You already should know the distribution of number of reports.

  3. If we say that $4$ and more injuries is failure, then number of games before the first failure has negative binomial distribution.

Denote $X$ the number of injured players in one game, then $$P[X>1]=P[X=2] + P[X=3] + P[X=4] + P[X=5] = 0.53$$

The probability of at least four injured players in one game is $$P[X\geq 4]= P[X=4] + P[X=5] = 0.13,$$ number of games played before the first one with at least four injured has $NB(1,0.87)$ distribution. Its expected value is $\frac{0.87}{0.13}$.

  • $\begingroup$ Can you break this down a bit more, I still do not get you $\endgroup$ – user3777772 Apr 4 '16 at 5:16
  • $\begingroup$ What do you don't get. Can you count $p$? $\endgroup$ – iiivooo Apr 4 '16 at 5:20
  • $\begingroup$ .23(2)+.17(3)+.09(4)+.04(5) = 1.53? $\endgroup$ – user3777772 Apr 4 '16 at 5:22
  • $\begingroup$ $p$ is probability, so it cannot exceed 1! I included the correct computation in my answer. Also please use MathJax for writing equations. $\endgroup$ – iiivooo Apr 4 '16 at 5:30
  • $\begingroup$ ok so then the standard deviation would be: (2-1.53)^2 * .23 + (3-1.53)^2 * .17 + (4-1.53)^2 * .09 + (5-1.53)^2 * .04 = 1.204? the answer to b? and the answer to a would be 10.6? $\endgroup$ – user3777772 Apr 4 '16 at 5:45

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