# Law of Total Probability and Conditional Probability

There is something that's a bit confusing for me in the topic of Conditional Probabilities. Assume we have the Universal Set, which is partitioned into $$F$$ and $$F^c$$. Now if we want to calculate the Probability of $$E$$ happening; The total rule of probability says that $$Pr(E) = Pr(EF) + Pr(EF^c) = Pr(E|F)·Pr(F) + Pr(E|F^c)·Pr(F^c),$$ which sounds pretty reasonable.

But from another point of view, we can say that Probability of E happening equals = E happens given F has already happened + E happens given F has not happened. Why this argument doesn't lead us to the right answer?

• How does that differ from what you wrote?
– lulu
Commented Jul 21 at 13:58
• @lulu My argument is just $Pr(E) = Pr(E|F) + Pr(E|F^c)$ Commented Jul 21 at 13:59
• But that's wrong. Suppose $E$ is certain when $F$ occurs and that $E$ has a $\frac 12$ chance of happening when $F$ does not occur. Then you are saying that $P(E)=1+\frac 12>1$. Conditional probability doesn't mean what you think it means.
– lulu
Commented Jul 21 at 14:00
• @VacationDue20000 Your argument would prove that $1=1+1$: just let $E$ be the whole whole space and $F$ any event with probability not $0$ or $1$. Commented Jul 21 at 17:59
• This is essentially about interpreting English phrases of the form “X given Y”; there's a strong tendency in natural language to interpret a statement of assumptions as implying that they're absolutes rather than contingencies. Commented Jul 22 at 0:15

Consider an extreme example. You are about to enter a boxing match against Mike Tyson in his prime.

Let $$E$$ be the event that you win the boxing match.

Let $$F$$ be the event that Mike Tyson has some sort of medical emergency during the match which prevents him from winning.

You are saying, why is not true that $$P(E)=P(E\mid F)+ P(E\mid \neg F)$$? Well, if Mike Tyson has a medical emergency, you will probably win by default, so $$P(E\mid F)=0.9$$, say. Given there is no medical emergency, you have basically no chance of winning, so $$P(E\mid \neg F)=0$$. If probabilities worked the way you suggested, that would mean the probability you best Mike Tyson was $$P(E)=0.9+0=0.9$$. This is clearly wrong.

Basically, it is worthless to consider conditional probabilities given nearly impossible events. If you had done the calculation correctly, we would need to factor in $$P(F)$$, the probability of Tyson having a medical emergency, which itself is close to zero. This would have ensured that our calculated $$P(E)$$ value is small, as it should be.

You are forgetting to "weight" the conditional probabilities $$P (E | F), P (E | F^c)$$ by their respective priors. In particular, we have that $$P (E \cap F) = P (E | F) \color{red}{P (F)}$$ and $$P (E \cap F^c) = P (E | F^c) \color{red}{P (F^c)}$$.

It is obviously not true in general that $$P (E \cap F) = P (E | F)$$. The left side is the probability that both $$E$$ and $$F$$ occur. The right side is the probability that, given $$F$$ occurs, $$E$$ also occurs.

As an example, suppose we flip two fair coins. Let $$E$$ be the event that the first coin lands heads, and let $$F$$ the event that the second coin lands heads. Clearly, $$P (E) = P (F) = \frac{1}{2}$$. Since the two flips are independent, we have that $$P (E | F) = P (E) = \frac{1}{2}$$. And we know that the probability of getting both heads is $$P (E \cap F) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \ne \frac{1}{2}$$.