P(S2)=3/5 (since there are 3 lists containing S=2) of the total 5 lists

P(F1)=2/5 (since there are 2 lists containing F=1) of the total 5 lists

P(F1|S2) = 1/3 ( since given S=S2, means there are only 3 lists, and out of which F=F1 is only 1 list)

There you go!

So you’ve just proven to yourself that these two things are not independent.

Now. Your problem gives you a list of lists. You dont know what they are, but you know what they are made up of: For each element, F will be 1,2,3,4,or 5, and S will be 1,2 or 3.

You know how to find P(S2). The only thing you’re missing from your formula at this point is P(F1 n S2), and then you can find P(F1|S2), right?

Well, what is P(F1 n S2) ?

Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true.

P(F1∩S2) = ?

Here I am not sure if it is dependant or independant.

*Ask yourself, does one event in any way affect the outcome (or the odds) of the other event?* If yes, it is dependent.

I think its dependent.

P(F1)=**2/5** (since there are 2 lists containing F=1) of the total 5 lists

P(F1|S2) = **1/3** ( since given S=S2, means there are only 3 lists, and out of which F=F1 is only 1 list)

It’s impossible for A and B to be independent if an outcome of one of the events changes the probability of the other event.

P(F1 n S2) can be read as: “The Probability that a randomly chosen list has both F=1 and S=2”.

```
[F,S]:
[[1,3],[1,2],[2,4],[2,2],[3,2]]
```

What is P(F1 n S2) ?

P(F1 n S2) = 1/5 (since there is only one list where the list has both F=1 and S=2, out of 5 lists)

Right.

So, let’s go back to your first formula for P(F1|S2).

Lets see if the formula agrees with what you told me.

In our example above, you told me:

```
P(F1 n S2) = 1/5.
P(S2) = 3/5.
```

So.

```
P(F1 n S2) / P(S2) =
(1/5) / (3/5) =
(1/5) * (5/3) (multiplicative inverse) =
(1*5) / (5*3) = (cancel the 5's with each other...)
1/3 = P(F1|S2)
```

And 1/3 is what you told me. So the formula agrees with what you intuited.

So lets take a look at that again.

P(S2)=3/5 (since there are 3 lists containing S=2) of the total 5 lists

P(F1 n S2) = 1/5 (since there is only one list where the list has both F=1 and S=2, out of 5 lists)

P(F1|S2) = P(F1 n S2) / P(S2)

Let’s abstract that a little bit:

`P(SX)`

= `I/K`

there are `I`

lists containing `S=X`

) of the total `K`

lists

`P(FY n SX)`

= `J/K`

there are `J`

lists where the list has both `F=Y`

and `S=X`

, out of `K`

lists)

`P(FY|SX)`

= `P(FY n SX) / P(SX)`

You’ve demonstrated enough python skills to me so far that you should now be able to code this.

For a given X and Y, and a list of lists:

Find K.

Find I.

Find J.

Find P(SX) using I and K.

Find P(FY n SX) using J and K.

Find P(FY|SX) using P(SX) and P(FY n SX).

EDIT: Cleaning up formatting. silly symbols.

You are right, I am getting it. You are awesome, excellent.

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