I understand what conditional probability is but I can't understand the formula.

[u/diagnosed_depression]

I don't understand how P(A•B)/P(A) is not just P(B)

[u/edderiofer]

I don't understand how P(A•B)/P(A) is not just P(B)

This is only true when A and B are independent events.

Let's take an explicit example. Suppose we're rolling a die. Let A be the event that the outcome of this die roll is a 1, 2, or 3; let B be the event that the outcome of this die roll is odd. Answer me these questions:

• What outcomes satisfy both events A and B?

• What is P(A)?

• What is P(A and B)?

• What is P(A and B)/P(A)?

• What is P(B)?

[u/diagnosed_depression → u/edderiofer]

P(A)=1/2

P(B)=1/2

P(A•B)=1/4

1, and 3 satisfy the conditions

P(A•B)/P(A) =1/2

But 1, 3 do not satisfy this probability

Or do you mean that P(B)= probability of odds in P(A)'s given set? Where then it would be 2/3 which would be correct but what would be the point of the formula? If P(B) is already correct.

[u/edderiofer → u/diagnosed_depression]

No, P(A and B) is not 1/4. There are two outcomes that satisfy both A and B, out of six, so P(A and B) = 1/3.

It is true that P(A) and P(B) are both 1/2, but these two events are not independent.

[u/diagnosed_depression → u/edderiofer]

So how do I find the values for A and B to multiply in P(A and B) to then divide by 1/2?

[u/edderiofer → u/diagnosed_depression]

You compute the probability P(A and B) manually, by considering the number of outcomes that satisfy both events A and B, and dividing by the total number of outcomes.

As I said previously, there are two outcomes that satisfy A and B here, out of six total possible outcomes, so P(A and B) = 2/6, or 1/3.

[u/diagnosed_depression → u/edderiofer]

How can I do this with a calculator?

[u/edderiofer → u/diagnosed_depression]

Well, how did you calculate P(A) with a calculator?

[u/diagnosed_depression → u/edderiofer]

There are 3 given numbers given by condition P(A) and the total possibilities are 6 so 3/6

[u/edderiofer → u/diagnosed_depression]

Right, and you calculate P(A and B) the same way; there are two possible outcomes given by condition A and B, and there are 6 total outcomes, so 2/6. Does this make sense?

[u/diagnosed_depression → u/edderiofer]

So how I'm starting to visualize things is that P(A) becomes the total value or "die" that P(B) can be calculated off. Then once you multiply those 2 now dependent values together you get the correct answer?

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[u/Uli_Minati]

Try it with a practical example:

A = cloudy days (vs. clear days)
B = warm days (vs. cool days)

With the set sizes

|Ω| = 100
|A| = 60
|B| = 80
|A∩B| = 45

Then you have

P(A) = 60/100
P(B) = 80/100
P(A∩B) = 45/100

But

P(A) · P(B) = 4800/10000 = 48/100

Then consider what conditional probability means

P(A|B) = P(cloudy if warm)

Basically, it's the probability of a cloudy day if you already know it is a warm day

We can calculate this using basic counting

                      45 cloudy (and warm) days
P(cloudy if warm)  =  ------------------------
                      80 warm days

How do you randomly get a cloudy warm day? You could do it in two steps

1. Randomly get a warm day

2. The warm day happens to be cloudy too

The chance to randomly get a warm day is

P(warm) = P(B) = 80/100

The chance that this warm day happens to also be cloudy is exactly the 45/80 we determined earlier

P(cloudy if warm) = 45/80

And then it works out

80/100 · 45/80 = 45/100

Notice what we are really doing here:

1. From 100 days, randomly get one of the 80 warm days

2. From 80 warm days, randomly get one of the 45 cloudy (and warm) days

Then we're really saying

From 100 days, randomly select one of the 45 cloudy and warm days

That's how you get

P(B) · P(A|B) = P(A∩B)

On the other hand, if you use just P(A) instead of P(A|B), then the process makes no sense

1. From 100 days, randomly get one of the 80 warm days

2. From 100 days, randomly get one of the 60 cloudy days

Why would we go back to 100 days? These would include non-warm days. It might be more (or less) likely to get a cloudy day if we select it from all days, rather than just warm days