And what's the same for P( C, Neg).

Same for P of B.

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

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

And the same is true for P of E given C.

Optimizing the log is the same as optimizing p because the log is monotonic to p.

Even though the assignment statement didn't include p, the fact that p and q refer to the same object means that it changed the value of p.

And then p times p squared is p to the third.

We're multiplying in p and 1 p. The 1 p stays.

We are one and the same

The important thing here is after the assignment, p and q refer to the same list.

P ... P ...?

Or are they one and the same?

We studied the special case of a fair coin, p 0.5, and now I want to do the same thing for our arbitrary p.

I initialize this with zeros, and I set the same size as my vector p.

Always the same one.

He's the same one

P of not A given the same events, X1, X2, and not X3, which resolves to P of not X3 given not A,

I knew that P of 1 or 2 conditioned on C is the same as P of 2 conditioned on C and 1.

No matter what grid cell we consider, the p(Z) is the same.

And here is one way to implement this, just return p p and for 0.5, it gives me 0.25.

The P, the first one in HlPPO, is for pollution.

So the sentence not P is true when P is false and false when P is true.

And now, same colors, meaning same activities. And one more example.

They're the same p, I just want to show you what's happening algebraically.

Th s p ace a n't gonna be the same w thout Dawn.

I've written down 2 different versions of P amp P is Jane's book one using double quotes and one using an escape sequence.

P. Sher... P. Shirley?

He's got the same one.

P(A) Ʃ p(A B) p(B).

P goes to open P close and then rule two to get rid of the P, and I'm in like Flynn.

And the third one got the same wry smile and said,

Now using the same premise as before, P equals 0.8, cost equals 4 and gamma equals 1,

On p we append 3, and then on q, which is the same list, we pop 3.

With probability P, and P is a number between 0 and 1, the action succeeds, and with 1 P we go into reverse.

P of fair is one and so P of flips given fair is always going to be equal to

So one of his p orbitals right there.

P of A is one value, for which we can derive P of not A.

And that's the same thing as saying, one, two.

P( C) is 0.9, and P(Neg C) is 0.5.

So three two is the same thing is one and one two.

For P and P implies Q, we know that P is true in these bottom 2 cases, and P implies Q, we saw the truth table for P implies Q is true in the first, second, and fourth case.

The same as to for p 1. Once again we know the variance is 0.

Is this one the same as this one?

And we have the variable P(X) with probability of 0.2, and we have 2 conditionals, P(Y_BAR_X) and P(Y_BAR_ X), both 0.6.