# Problems from Law of total probability

Suppose you are the manager of a team that will play the European Champions League. In the making of the groups for the European Champions League, we can have in our group both the Liverpool or Chelsea with equal probability. Suppose that due to the characteristics of your team, if it plays against Liverpool, there is 60 % of probability of winning, and 15 % of probability of A draw. Instead, If we play against Chelsea, there is 30 % of probability of winning, and 40 % of probability of losing. What probability do we have of, at least, getting a draw?

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### Development:

We represent with $L =$ "to play against the Liverpool", $C =$ "to play against the Chelsea".

We have three possible events: $G =$ "to win", $E =$ "to draw", $P =$ "to lose". We are being asked for the probability $P(G\cup E)$.

We first represent the problem in a tree. Let's observe that, although the statement does not give us all the information explcitly, we can infere it: if against the Liverpool there are $0,6$ of probabilities of winning, and $0,15$ of drawing, then there have to be $0,25$ of probabilities of losing, since only one of three things can happen. For the same motive, the possibility of tying against the Chelsea is of $1 - 0,3-0,4 = 0,3$. We apply the theorem of the total probability. $$P(G\cup E)=P(L)\cdot P(G/L)+P(L)\cdot P(E/L)+P(C)\cdot P(G/C)+ P(C)\cdot P(E/C)$$

In our case,

$$P(G\cup E)=\dfrac{1}{2}\cdot 0,6 + \dfrac{1}{2}\cdot 0,15 + \dfrac{1}{2}\cdot 0,3 + \dfrac{1}{2}\cdot 0,3=0,675$$

Also we could have solved the problem by calculating the probability of losing, which would give us $0,325$, and then calculating the probability of winning or of drawing by doing the complementary

That is, $P(G\cup E)= 1- P(P)=1-0,325=0,675$ (note that we obtain the same result).

### Solution:

The probability is $P(G\cup E)=0,675$.

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