Question

The MIT soccer team has 2 games scheduled for one weekend. It has a 0.4 probability of not losing the first game. and a 0.7 probability of not losing the second game, independent of the first. If it does not lose a particular game, the team is equally likely to win or tie. independent of what happens in the other game. The MIT team will receive 2 points for a win, 1 for a tie. and 0 for a loss. Find the PMF of the number of points that the team earns over the weekend.

Answer 1

Answer:

Step-by-step explanation:

To determine the PMF of the points that the team receive, we must determine the probabilities of winning each of the games.

Consider the firt game. We know that the probability of not losing is 0.4. Since the team can either lose the game or not, we have that the probability of losing it is 0.6. We are told that given that the team does not lose, the probability of winning or having a tie is the same. Recall that, given events A, B, the conditional probability P(A|B) (which means probability of A given B) is given as

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

in our case, we have that B:= not losing the game. Consider A:= To win the game, C:= To tie the game. We know that P(A|B)=P(C|B). When the team doesn't lose, then the team either ties or wins, this means that

$P(A|B)+P(C|B) = 1$

So we get that P(A|B) = P(C|B) = 1/2. Let us find P(A)

$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A)}{P(B)}=\frac{1}{2}$

Then $P(A) = \frac{P(B)}{2} = 0.2$. By an analogous calculation we have that P(A) = P(C) = 0.2.

We will define the following events: $W_i$ is the event that the team wins the game i, $T_i$ is the event that the team ties the game i and $L_i$ is the event that the team loses the game i. In this notation, we have that

$P(W_1) = P(T_1) = 0.2, P(L_1) = 0.6$

By reasoning the same way for the second game, we can calculate that

$P(W_2)=P(T_2)=0.35, P(L_2) = 0.3$

To calculate the PMF we must check all the possible outcomes of the results of both games. Since we can either win (W), tie(T) or lose(L) a game, we will express the outcome as follows TW(3) means that the team ties  the first game and wins the second one, for a total of points. In this way, the possible outcomes are

WW(4), WT(3), WL(2), TW(3), TT(2), TL(1), LW(2), LT(1), LL(0)

We see that the number of possible points are 0,1,2,3,4. Let X be the number of points, to get the pmf of X is to give the probability P(X=k) for k=0,1,2,3,4 explicitly. Note that since each game is independent of each other, to get the probability of one of the outcomes it suffices to multiply the probability of each event. For example we have that $P(WW) = P(W_1)\cdotP(W_2) = 0.07$

Then, the pmf is given by

$P(X=0) = P(LL) = 0.18$

$P(X=1)=P(LT)+P(TL) = 0.27$

$P(X=2)=P(LW)+P(TT)+P(WL) = 0.34$

$P(X=3)=P(WT)+P(TW) = 0.14$

$P(X=4)=P(WW)= 0.07$