How to compute probability distribution functions? Q: shooting hoops. Player 1 has a 0.5 probability of scoring a point, player 2 has a prob of 0.7 and player 3's prob is 0.9 of successfully scoring.
Compute the prob. distribution function of number of successful shots.
I'm not really sure how to approach this. My understanding is that you'd first calculate the total number of possible outcomes (2^3, since there's two outcomes (score or miss) and there's 3 players). Then, you have a denominator of 8, and each player would have a different numerator to reflect the different probabilities? Summing to 1. But, I'm not too sure - any clarification appreciated!
 A: Just write down all the possible combinations for every number of successful shots. Let $A_i$ denote a successful shot of player $i$, $i=1,2,3$. Now,  let $X$ be the number of successful shots, thus $X=0,1,2,3$, and $P(X=x)=P_X(x)$ can be calculated for each value
$$
P_X(0) = \prod_{i=1}^3 P(A^c_i) =  0.5\times0.3\times0.1,
$$ 
for $X=1$ you have to be slightly more careful
$$
P_X(1)= P(A_1)P(A^c_2)P(A^c_3) + P(A^c_1)P(A_2)P(A^c_3) + P(A^c_1)P(A^c_2)P(A_3).
$$
for $X=3$ you'll have
$$
P_X(3) = 0.5\times0.7\times 0.9,
$$
and the $P_X(2)$ you can calculate by subtracting all the rest, i.e., 
$$
P_X(3)=1 - \sum_{j=0,1,3} P(X=j).
$$
A: Here is the approach you should take.
Let $H$ denote a hit and $M$ denote a miss. Then your probability space looks like this
\begin{align}
1&&2&&3\\
M&&M&&M\\
M&&M&&H\\
M&&H&&M\\
M&&H&&H\\
H&&M&&M\\
H&&M&&H\\
H&&H&&M\\
H&&H&&H
\end{align}
You can regroup these according to the number of hits.


*

*$0$ hits
\begin{align}
M&&M&&M
\end{align}

*$1$ hit
\begin{align}
M&&M&&H\\
M&&H&&M\\
H&&M&&M
\end{align}

*$2$ hits
\begin{align}
M&&H&&H\\
H&&M&&H\\
H&&H&&M
\end{align}

*$3$ hits
\begin{align}
H&&H&&H
\end{align}


For each player you know the probability of a hit and the probability of a miss. So you should be able to find the probability of each of the separate eight events in the space. Then you should be able to find the probability of each of the four possible totals.
