1. ## Probability

A TV sports commentator claims that 45 percent of all football injuries are knee injuries, assuming that this claim is true what is the probability that in a game with 5 reported injuries:
a) all are knee injuries?
b) none are knee injuries?
c) at least 3 are knee injuries?
d) no more than 2 are knee injuries?

I am looking for some direction, confused as to where to begin, looking for any kind of help, it'd be much appreciated.

2. Originally Posted by cvaughn13
A TV sports commentator claims that 45 percent of all football injuries are knee injuries, assuming that this claim is true what is the probability that in a game with 5 reported injuries:
a) all are knee injuries?
b) none are knee injuries?
c) at least 3 are knee injuries?
d) no more than 2 are knee injuries?

I am looking for some direction, confused as to where to begin, looking for any kind of help, it'd be much appreciated.
a) P("all are knee injuries") = 0.45*0.45*0.45*0.45*0.45 = 0.45^5

b) P("none are knee injuries") = (1-0.45)^5 = 0.55^5,

because you know P("knee injury") = 0.45
then P("no knee injury") = 1-0.45 = 0.55

c) P("at least 3 knee injuries") = $\begin{pmatrix} 5 \\ 3\end{pmatrix}*0.45^3*0.55^2 + \begin{pmatrix} 5 \\ 4\end{pmatrix}*0.45^4*0.55^1+0.45^5$

I consider "A and B and C has a knee injury, but D and E don't and D and E and A has a knee injury, but player B,C don't" as two different cases. Thats where the 5 over 3 comes from.

3. Sorry, found some errors, see next post.
This is binomial with n=5 and p=P(knee injury)=.45
(d) is P(0 or 1 knee injury)=P(X=0)+P(X=1)

4. There's a typo in (c) that I just noticed. I'm changing the 4 to a 1 here, above the q.

AND (b) is partially right/wrong.

The answer is $.55^5$ which is not equal to $1-0.45^5$

Maybe you meant $(1-0.45)^5$, which is okay, but that's not how I would interprete your math.

Originally Posted by Rapha
a) P("all are knee injuries") = 0.45*0.45*0.45*0.45*0.45 = 0.45^5

b) P("none are knee injuries") = 1-0.45^5 = 0.55^5,

because you know P("knee injury") = 0.45
then P("no knee injury") = 1-0.45 = 0.55

c) P("at least 3 knee injuries") = $\begin{pmatrix} 5 \\ 3\end{pmatrix}*0.45^3*0.55^2 + \begin{pmatrix} 5 \\ 4\end{pmatrix}*0.45^4*0.55^1+0.45^5$

I consider "A and B and C has a knee injury, but D and E don't and D and E and A has a knee injury, but player B,C don't" as two different cases. Thats where the 5 over 3 comes from.