Probability tree diagrams

Question

Exercice : True or false

Let $E$ and $F$ be two events of an experiment.

$$\mathbb{P}(E| \bar{F})=1-\mathbb{P}(E| F)$$

Solution :

Flase due to the Law of total probability we've:

$$\mathbb{P}(E| \bar{F}) +\mathbb{P}(E| F)=\mathbb{P}(E)$$ which is not necessarily equal $1$

Could anyone explain me this by using Probability tree diagrams they can helps me to understand that very well.

Answer 1

Nope.

The Law of Total Probability is: $$\Bbb P(\bar F\mid E)+\Bbb P(F\mid E)= 1$$

And by Bayes' Rule: $$\frac{\Bbb P(E\mid\bar F)\Bbb P(\bar F)}{\Bbb P(E)}+\frac{\Bbb P(E\mid F)\Bbb P(F)}{\Bbb P(E)} = 1$$

So:$$\Bbb P(E\mid\bar F)\Bbb P(\bar F)+\Bbb P(E\mid F)\Bbb P(F) = \Bbb P(E)$$

But this still does not really disprove the original claim, does it?

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Hint: Try a disproof by counter example.   Can you find two events $E,F$ such that $\,\Bbb P(E\mid\bar F)+\Bbb P(E\mid F)\neq 1\,$ ?

Answer 2

Let's take the experiment being the roll of f single die, and let the events be: $F$ - the result is even. $E$ - the result is 6. Now, clearly $P(E)=\frac{1}{6}$, $P(E|F)=\frac{1}{3}$ and $P(E|\bar{F})=0$.

Two things can be observed now, first of all that indeed $P(E|\bar{F})+P(E|F)\neq 1$ as required, but also that $P(E|\bar{F})+P(E|F)\neq P(E)$ contrary to your claim.

You can easily draw the probability tree for that example

Answer 3

Take $E = \Omega$. Then $P(E|F) + P(E|\overline F) = 2$.