Soccer and probability distributions

Question

The USA soccer team is going to play a championship with 7 other tems. The 8 teams, are going to be divided in two groups of 4 each one. From the participants, Brazil is considered the strongest team out of the eigh. The USA soccer manager is analizing the likehood of going further in the championhip. He has estimate that if the team is with Brazil in the group, USA can obtain 3 points with a probability of 0.3, 4 points with probability of 0.3, and 6 points with probability of 0.4. If USA is in a group different from Brazil's, the manager estimate that they can get 4, 6, 7, with a probability of 0.3, 0.5, and 0.2 respectively.

How would you find the probability mass function, and the cummulative distribution function?

I'm honestly unsure how to proceed with this exercise, can anyone please help me out? Thanks!!

Answer 1

The USA soccer team is going to play a championship with 7 other teams. The 8 teams, are going to be divided in two groups of 4 each one.

Assuming no bias in the assignment, the probability of being in the same group as Brazil is: $$\Pr(B) = \frac 3 7$$

(Brazil can either be one the 3 other teams in the same group, or the 4 teams in the other group.)

He has estimate that if the team is with Brazil in the group, USA can obtain 3 points with a probability of 0.3, 4 points with probability of 0.3, and 6 points with probability of 0.4.

Let $N$ be the number of points. We now have a measure of $N$ conditional on $B$.

$$\Pr(N=n\mid B) = \begin{cases}0.3 & : n=3 \\ 0.3 & : n=4 \\ 0.4 & : n=6 \\ 0 & : \text{else}\end{cases}$$

If USA is in a group different from Brazil's, the manager estimate that they can get 4, 6, 7, with a probability of 0.3, 0.5, and 0.2 respectively.

Likewise.

$$\Pr(N=n \mid \neg B) = \begin{cases} 0.3 & : n=4 \\ 0.5 & : n=6 \\ 0.2 & : n=7 \\ 0 & : \text{else} \end{cases}$$

These are your conditionals.

Now use the law of total probability, matching up the values of $n$: $$\begin{align}\Pr(N=n) & = \Pr(N=n\mid B)\Pr(B) + \Pr(N=n\mid \neg B)(1-\Pr(B)) \\ & = \begin{cases}9/70 & : n= 3 \\ \vdots & \vdots & \}\text{fill in the gap for $n=4$ and $n=6$} \\ 4/35 & : n=7 \\ 0 & \text{elsewhere} \\\end{cases} \end{align}$$

That's the probability mass function. $p_N(n) = \Pr(N=n)$

The cumulative mass distribution is: $F_N(n) = \Pr(N\leq n) = \sum_{k=0}^n p_N(k)$

Answer 2

I am assuming that it is equally likely to be in Brazil group or not Brazil group since each group contains 4 teams. This means that probability of getting in Brazil group is $\frac{1}{2}$. (This is wrong as Graham Kemp pointed out it should be $\frac{3}{7}$)

Let $X$ be number of points scored in game. Since this is a discrete random variable (because has finite outcomes) in order to find pmf, $f(x)$, you must find $P(X=x)$ for $x\in\{3,4,6\}$. To do this it may be helpful to note that (letting $B$ represent event that end up in Brazil group) $$P(X=x)=P(X=x|B)P(B)+P(X=x|B^{c})P(B^{c})$$ which is just an application of law of total probability. Thus your pmf should be a peicewise function. Now for CDF you just have to find $P(X\leq x)$ which you would just sum values of pmf for values less than or equal to x