Intuition behind conditional probabilty: $P(A|B)=P(B\cap A)/P(B)$

Question

I've struggled with probability for years. Even the most basic concepts. This is especially something I am not able to understand even after reading for the last 1.5 hours.

Conditional probability is $$ P(A|B)=\frac{P(B\cap A)}{P(B)}. $$ However, I fail to see why $P(A|B)=P(B \cap A)$ cannot be true in itself? Why do we have to divide by $P(B)$?

A video online gave this example. The probability of being a male and an alcoholic is $\sim 2.25\%$. So what is the probability of being an alcoholic given that you are a male? I would say $2.25\%$ but in fact the answer is different. I cannot see how $P(A|B) \neq P(B \cap A)$.

The intuition just isn't there. Is this something I am just supposed to accept and move on?

Answer 1

See, we divide by $P(B)$ since that is the way of restricting the range to only possibilities where $B$ occurs.
If you are an male, the chance might be higher/lower than a non males's combined with the males. Ex: Males have $\frac12$ chance. $\frac13$ of people are males. Non males have $\frac15$ chance. The probability is $\frac13$ given male, but $\frac13\cdot\frac12+\frac23\cdot\frac15=\frac16+\frac2{15}=\frac3{10}$ not given anything.
Take note that the first (given probability) cares about only one type, while the second (probability overall) cares about all types. This is the difference of conditional probability.

Answer 2

"A video online gave this example. The probability of being a male and an alcoholic is ~ 2.25%. So what is the probability of being an alcoholic given that you are a male?" Of course, you are expected to use that the "a-priori" probability of being male is 0.5.

Suppose this involves a population of 10000 people. 5000 of them are males and 5000 females. If the "probability of being male and an alcoholic" is 0.0225 then of the 10000 people, 225 of them are "male and alcholic". You could use that to calculate that "if you are male then the probability you are alcoholic is 225/5000= 0.045".

Answer 3

No, you need to understand the reason behind the formula.

Obviously you do not want to just accept the formula and move on.

Consider a class of $100$ students,where $60$ of them are male and $ 40$ are female.

Let $20$ percent of male students and $30$ percent of females have a grade of $A$ on the first test.

If you randomly pick a student What is the probability that the student is both female and gets an $A$ on the first test?

That is $$ P(A\cap G) = 12/100 = 0.12$$ because there are only $12$ female students who have an $A$ on the first test.

Now what is $P(A|G)$?

This is probability of randomly picking a girl and having an $A$ on the first test. That is $12$ out of $4$0 which is $0$.30.

Note that $$ \frac {P(A\cap G)}{P(G)} = \frac {0.12}{0.4}=0.30$$

Answer 4

Here is an example that shows that in general $P(A|B) =P(A\cap B)$ doesn't hold in general: Imagine you're playing darts and whenever you throw the it's equally likely to land anywhere on the wall. There are two intersecting circles $A$ and $B$ on a huge white wall. Let's say you throw a dart. Suppose someone told you that the dart hit $B$. Now $P(A|B) $ is for the probability that you hit the circle $A$ given the knowledge that it hit $B$.

Now $P(A\cap B)$ is the probability that you hit both circles without any additional knowledge. To see why $P(A\cap B)$ is not always equal to $P(A|B)$ think of the case where both circles $A$ and $B$ are tiny and are exactly of the same size at the same position. So when you hit one of them you always hit the other, so $P(A|B)=1$.

Since the circles are tiny $P(A\cap B)=P(A)=P(B)$ is tiny.

Therefore, $P(A|B) \neq P(A\cap B)$.

This analogy of darts can also explain $P(A|B)P(B)= P(A\cap B)$ since for you hit both circles (now think of them again as being anywhere and not necessarily overlapping) you definitely need to hit $B$ which happens with probability $P(B)$. So if you suppose you hit $B$ what's now the probability that you also hit $A$. It's exactly $P(A|B)$.

See also

Answer 5

However, I fail to see why $P(A|B)=P(B \cap A)$ cannot be true in itself? Why do we have to divide by $P(B)$?

Let's assume that half of all people are men, and that whenever a person flips a coin, the outcome has nothing to do with the person's gender.

I am a man. Given this information, if I flip a coin, then what's the probability that it'll come up heads?

According to you:

$$P(\text{heads}|\text{man}) = P(\text{man} \cap \text{heads}) = 1/4.$$

According to me:

$$P(\text{heads}|\text{man}) = \frac{P(\text{man} \cap \text{heads})}{P(\text{man})} = \frac{1/4}{1/2} = 1/2.$$

Answer 6

The thing to keep in mind is that when solving problems in probability there is mathematical machinery (a model) that underpins the whole deal. When you boil it down, behind it all there must be (sometimes implied) elementary outcomes and a 'thought' experiment.

The following two examples can be worked out using simple counting arguments and do not require plugging into any formulas.

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Example 1: There are 400 people in a big room, 200 are male and 200 are female and the experiment is to randomly pick someone in the room.

A: a person chosen at random is an alcoholic
B: a person chosen at random is a male
F: a person chosen at random is a female

To calculate

$P(A \cap B) $

you count the number of males that are alcoholics and divide by 400. So if

$P (A \cap B)= \text{2.25%}$

then exactly 9 of the people in the room are both male and also alcoholics.

Now if our experiment is restrained by insisting that we only select males, then

$P (A \, | \, B)= \text{4.50%}$.

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Exercise 2: There are 400 people in a big room, 200 are male and 200 are female.

A: a person chosen at random is an alcoholic
B: a person chosen at random is a male
F: a person chosen at random is a female

Assume that exactly 9 of the males are alcoholics and 7 of the females are alcoholics.

Compute both

$P(B \, | \, A)$

and

$P(F \, | \, A)$.

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The OP should regard Example 1 as just going thru the motions. We could have said that 9 of the females were also heavy drinkers and then talk about white noise, but felt better just leaving it alone.

Exercise 2 is much more interesting. If Detective Columbo was looking at a murder case and knew that the suspect was an alcoholic, he would no doubt focus his attention on his male suspects (it is a fact that men are more prone to alcoholism than women).

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The intuition behind this is that when we collapse a sample space so that only certain outcomes are possible, we have to 'uniformly pump up' the probabilities of these remaining outcomes so that the sum of the probabilities of these 'in-focus' outcomes still add up to

$\text{1}$.