We subtract 4 x 0.1, which is 0.4, which is 0.6.

If I keep expanding this, I get the following solution.

least maybe it doesn't make intuitive sense just yet, but I showed you that the probability of a and b is equal to the probability of a given b times the probability of b.

You might say OK, that's the probably of getting exactly 1 times the probability of getting 2 out of 3 plus the probability of getting 3 out of 3.

I suppose so, but it might mean that you It means that, statistically speaking, one of the most impressive records of failure is destined to be broken.

And we get this, the probability of b given a is equal to this, probability of a given b.

And we can denote that by the probability that we get a Result state, S prime given that we start in state, S, and apply action, A.

We now apply Bayes rule.

So let's look at this, let's look at a population where the probability of success we'll define success as 1 as having a probability of p, and the probability of failure, the probability of failure is 1 minus p.

So this is plus the probability of getting, of getting a cube.

Time the probability of two sided coin.

Frack the odds.

Let's use the Laplacian smoother with K 1 to calculate the few interesting probabilities

Then the value of the state for the action go up would be obtained as follows.

And these are mutually exclusive events, you can't have both of them

Probability

Phil, the odds against

And this is the probability of four out of six heads, given an unfair coin.

That it's the probability of the data given the null.

That's what probability of getting 5 out of 5 heads given that I picked a normal coin out.

On the second state we know that the probability of R is 0.6 and therefore, the probability of sun is 0.4, and we compute the probability of rain on day 2 using total probability.

Similarly for tails, if you took a die

What's the probability of the joint X, Y?

And it makes much more sense to talk about the probability or random variable equaling a value, or the probability that it is less than or greater than something or the probability that is has some property

Then we get, by virtue of our expansion of the state at time T, the probability of transitioning from rain to rain is 0.6, the probability of having it rain is X again, the probability of transitioning from sun to rain is 0.2, and the probability of having sun before is 1 minus X, so we get X equals 0.4X plus 0.2.

Question 1 In the first question, I'm going to ask you some very basic probability questions.

And here is my answer. You can really read off the formula that I just gave you.

Thrun As usual, we can resolve this using Bayes' rule.

It applies to coin flips and many of the other examples we discussed before.

But if we're in A, we stay in A with a 0.5 chance. So you put this together.

What are the odds?

Well if you played this game many, many, many, many times

And coin 2 is also loaded.

Just how would you rate our chances, Doctor?

Now, just one more thing.

The motions don't move at all, move right, move down, move down, and move right again.

And we know that our variance is essentially the probability of success times the probability of failure.

Now, I'm going to make it really difficult. I'm going to give you a coin let's call it loaded.

From that you can easily calculate the probability of measuring 'white', if a particle falls on a black square.

The probability of any random variable Y can be written as probability of Y given that some other random variable X assumes value i times probability of X equals i, sums over all possible outcomes i for the (inaudible) variable X.

This is being answered using Bayes rule, so this is the probability of being happy given that it rains times the probability that it rains over the probability of being happy.

Yes or no.

Now let's ask, and this is probably the most interesting thing to ask,

Divided by the probability of in general, the probability of getting 5 out of 5 heads.

Most of the times that you're dealing with, as in the case right here, a discrete random variable let me make it clear this one over here is also a discreet random variable.