Is $P(E | \bar F) + P(\bar E | \bar F) = 1$?

Question

Is $P(E \, | \, \bar{F}) + P(\bar{E} \, | \, \bar F) = 1$?

I tried using Venn diagram and here's what I get:

The red region shows $P(E \, | \, \bar{F})$ and the blue region shows $P(\bar{E} \, | \, \bar{F})$.

Clearly it is not $1$ but my book says it is. Where have I misunderstood?

Answer 1

Note that

• $(E \cap \bar F) \cup (\bar E \cap \bar F) = \bar F$ and
• $(E \cap \bar F) \cap (\bar E \cap \bar F) = \emptyset$

Answer 2

Since $E$ and $\bar{E}$ are disjoint events we have:

$$\mathbb{P}(E|\bar{F}) + \mathbb{P}(\bar{E}|\bar{F}) = \mathbb{P}(E \cup \bar{E}|\bar{F}) = \mathbb{P}(\Omega|\bar{F}) = 1.$$

The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).

Answer 3

An intuitive view, to make the parallel with your Venn diagram is to realize that in your case $1$ (=your "universe") is not the "full picture" by only $\bar{F}$. This is apparent from the formula: $$ P(E|\bar{F})=\frac{P(E,\bar{F})}{P(\bar{F})} $$ To reword what I said, your normalization factor is now $\bar{F}$.

Obviously you have: \begin{align*} P(E|\bar{F})+P(\bar{E}|\bar{F}) &=\frac{P(E,\bar{F})}{P(\bar{F})}+\frac{P(\bar{E},\bar{F})}{P(\bar{F})} \\ &=\frac{P(E,\bar{F})+P(\bar{E},\bar{F})}{P(\bar{F})} \\ &=\frac{P(\bar{F})}{P(\bar{F})} \\ &= 1 \end{align*}