While p is not empty, we append onto q what we pop from p.

P might be an empty relation, so this is satisfiable.

P of H given R times P of R over P of H.

P of not X3 given A, P X2 given A, P X1 given A times P of A.

P ... P ...?

We start with the empty list. We go through all the elements in p.

P of X2 given not A, P of X1 given not A, and P of not A.

The problem specifically asks for you to deal with the empty list case, and if you noticed, if p is empty we don't execute any of this.

That's P of H.

P. Sher... P. Shirley?

P(A) Ʃ p(A B) p(B).

And the sentences are P or not P, P and not P,

P of H given R comma S times P of R plus P of H given not R and S times P of not R, which is total probability.

P of D given C is 2 parameters for P of D given C and P of D given not C.

Then you pay a monthly payment of p, so minus p.

Empty we'll empty it out!

And if you want to factor a p out of this, this is going to be equal to p times, if you take p divided p you get a 1, p square divided by p is p.

P

p

p.

P

P .

P ...?

P.

P.

P...

Same for P of B.

P of A equals 0.5.

Now, each has exactly the same probability of P for heads x (1 P) x (1 P).

This is just our encoding of start symbol goes to P, P goes to open parentheses P,

Force formatting of empty DVDs

Background color of empty place

Most of these are empty.

Empty

Empty

empty

Empty

Empty

empty

Empty

Empty.

Empty!

Empty.

Empty?

Empty.