I am comfortable with explaining to my high school students that for an event $A$ we have that
$P(A) + P(A^C) = 1$
But what is the best way to help students realize that
$P(A \mid B) + P(A^C \mid B) = 1$
remains true when you have the additional assumption of the event $B$ occuring. To be clear; I understand why this is so, I am just afraid of my explanation being too technical.
If your mother gives you a candy, either you will eat it or you won't. Sure thing.
The key lies in reiterating the definition of probability as $$\frac{\rm favorable {\ } cases}{\rm possible {\ } cases}$$ The conditional probabilities are computed with less possible cases.
For example, say we roll two dice, one after the other (so we can differentiate them), and consider the following statements:
A: "The second die is bigger than or equal to the first"
B: "The sum of rolls is 6"
To compute $P(A)$ and $P(A^c)$, we consider 36 possible cases, with 21 and 15 favorable cases respectively.
But the conditional probabilities $P(A|B)$ and $P(A^c|B)$ only consider as possible those cases in which $B$ occurred, namely 5. Of these (and only these) cases, 3 are favorable to statement $A$, and 2 to its negation.
You can realize probability as the relative area between a figure and an interesting part of that figure.
For instance, if you have a calendar of last year, and you marked all the days where it rained in blue, the probability during the year that it will rain is defined as the number of blue days to the number of days total
But different months rain different amounts. If we look only at the month of April, likely more will be blue that month than on any other month.
Conditional probability here would be linked to looking only at the month of April.
The total probability, then, is the number of days it rained (days marked in blue) and the number of days it didn't (not marked) added together, divided by the number of total days. Of course, there are 30 days in April, and hopefully each day, it either rained or it didn't rain, since any other possibility would be hard to imagine.
Here is another way to look at this.
$P(A)+P(A^c)=1$ aligns with the logical tautology $(A\text{ or }\lnot A)$.
The preposition
$B\rightarrow(A\text{ or }\lnot A)$
is also a logical tautology, and corresponds to
$P(A\vert B)+P(A^c\vert B)=1$.
In my opinion, this can be explained in 2 ways :
We took $P(A'∩B)= P(B) - P(A∩B)$ it the above step. But why? Because $P(A'∩B)$ is the dark gray region in the below Venn diagram [the region is an intersection of A' and B]. We could compute that region by subtracting $P(A∩B)$ from $P(B)$ We subtract $P(A∩B)$ because the compliment of $P(A')$ is every element of sample space without $P(A)$ and the intersection of $P(A')$ with $P(B)$ will give the shaded region.
And by substituting the above value in the equation we'll get the desired answer, as shown in the image below :
Use concrete examples.
Probability and its notation are so abstract to so many students that using context-free logic and symbols is all but guaranteed to not work with a huge share of students. They just don't trigger any related knowledge in students' heads. One of the few truly good rules of thumb in teaching is that you should usually go from concrete to abstract and from prior knowledge to new knowledge. Going backwards rarely works.
Starting here and going backwards rarely works.
$P(A \mid B) + P(A^C \mid B) = 1$
Instead, how about:
(1) The probability of a voter arriving at a voting station then filling in the ballot correctly is 87%. (a) What is the probability of a voter arriving at a voting station and NOT filling the form in correctly? How do you know? (b) Does the given info tell us anything at all about voters who vote by mail?
(2) Among students who start grade 12 before their 19th birthday, 6% fail to graduate within a year. (a) Among those students, what percentage DO not graduate within a year? How do you know? (b) Does this question tell us about students who start grade 12 courses as an adult?
(3) Use let statements and standard probability notation to express your thinking in parts (a) and (b) of the two questions above.
In the spirit of the answer by Tac-Tics, I like the picture with a unit square. One event splits the square horizontally, while the other splits vertically. Your second formula focuses on one strip only. Somehow given B means you can only be in one strip now, not in the whole square.
You may have a look at figure 2, a geometric interpretation of Bayes' Theorem in https://en.wikipedia.org/wiki/Bayes%27_theorem
After a football game, the team always send a member to a nearby pizza shop to pack food. Different member has different preference on different tastes of pizzas, but whoever was on the duty, sum of the possibility of his choosing each taste , shall be 1.
