Proof that 1-P(B|C)=P(~B|C). Is everything correct?

Question

Proof:

Suppose we have a set of students. Different students study different subjects, with some of them studing several subjects at once. Suppose among many other subjects we have Biology and Chemistry. Then $P(B|C)$ would be probability of selecting a Biology student out of set of Chemistry students.

Now we need to ask ourselves: what would be the complement of event "selecting a Biology student out of set of Chemistry students"?. Remember that we select a Biology student out of set of Chemistry students, NOT from all students. In other words, our sample space shrunk to the size of the set of Chemistry students. The complement event would be selecting a Chemistry student who does NOT study Biology (although the person can study other subjects along with Chemistry).

The probability of selecting such person out of set of Chemistry students can be written as $P(\tilde B|C)$.

Thus $1-P(B|C)=P(\tilde B|C)$ because $P(B|C)$ and $P(\tilde B|C)$ are complements of each other.

Answer 1

Your are correct, and the proof is rather simple (not requiring the wall of text you wrote :)

$$\begin{align}P(\neg B|C)&=\frac{P(\neg B \land C)}{P(C)} &\text{by definition} \\&= \frac{P(C) - P(B\land C)}{P(C)} & \text{Because $B\land C$ and $\neg B\land C$ form a partition of $C$} \\&=\frac{P(C)}{P(C)}-\frac{P(B\land C)}{P(C)}&\text{Algebraic manipulation} \\&=1-P(B|C)&\text{by definition}\end{align}$$

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Note: I assume here that $P(C)>0$, i.e. that $C$ is not an impossible event. Things can get complicated quickly if we look at a more general solution.

Answer 2

Your argument is correct. Here is another way you could think about it:

$$P(B^c|C) =\frac{P(B^c\cap C)}{P(C)}$$ $$=\frac{P(C)-P(B\cap C)}{P(C)}$$ $$=1-P(B|C)$$

Draw a Venn diagram to see that $P(B^c \cap C) = P(C)-P(B\cap C)$.

Answer 3

Let $(\Omega,\mathcal A,P)$ denote a probability space and let $C\in\mathcal A$ with $P(C)>0$

Then the function $P(-\mid C):\mathcal A\to\mathbb R$ is a probability measure on measurable space $(\Omega,\mathcal A)$.

It is prescribed by $A\mapsto P(A\cap C)/P(C)$ and it is not difficult to show that this is indeed a probability measure on measurable space $(\Omega,\mathcal A)$.

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For every probability measure $\mu$ it is true that: $$1-\mu(B)=\mu(B^{\complement})\text{ for every set }B\text{ in the domain of }\mu$$

So what is said above ensures that indeed:$$1-P(B\mid C)=P(B^{\complement}\mid C)$$