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.

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.

We have the probability of rain equals 1 and the probably of sun, S times 0 equals 0.

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.

So, we multiply these 2 things together, and we get 0.72.

What's the probability of rain on day 0, and what are its conditional probabilities?

A sunny day follows a rainy day with 0.6 chance, and a rainy day follows a rainy day please give me your number.

The probability of rain on day 1 is the probability it was rainy on day 0 and it led to a self transition from rain to rain from day 0 to day 1 plus the probability it was sunny on day 0 times the probability that sun led to rain over here.

There's a 0.1 chance I might still be grumpy for some other reason.

I'd like to know from you what's the probability of rain on day 1, the same for day 2, and the same for day 3.

Here is our sequence. There's a couple of sunny days 5 in total a rainy day, 3 sunny days, 2 rainy days.

This guy over here is 0.4 again, but now it's this 0.4 over here.

If you look at the state transition diagram, it's more likely to be sunny afterwards because it's more likely to flip from rain to sun than sun to rain.

These are the transition probabilities as before.

In fact, let's just calculate it.

Thrun

This expression over here is expanded by a Bayes rule as applied over here.

If it's rainy, it stays rainy with a 0.6 chance while with 0.4 it becomes sunny.

But if you understood Laplacian smoothing, you might want to give it a try.

And then let's assume that a sunny day follows a sunny day with 0.8 chance, and a rainy day follows a sunny day with well

Suppose I observe that I'm happy on day 1.

Let's interpret the 0.229 in the context of the question we asked.

Again, the parent, Cloudy, is positive, so we're looking at this part of the table.

Multiplying these 2 things gives us 0.624.

And if you plug in all these values, we get 0.6 times 0.6 plus rain following sun which is this arc over here, 0.2, times 0.4 as the prior, and this results in 0.44.

And I get 0.0217, which is a really small thing.

Rain. is you?

Suppose there are 2 types of weather rainy, which we call R, and sunny, which we call S and suppose we have the following state transition diagram.

Suppose you look out of the window and see sequences of rainy days followed by sunny days followed by rainy days and you wonder what numbers to put here, here, here, and here.

So for example, we observed that we always have a single first day, and this becomes our prior probability.

We transition 8 times out of a sunny state 1, 2, 3, 4, 5, 6, 7, 8 twice into a rainy state, and therefore 6 times we remain in a sunny state, so the probability of sun to sun is ¾, whereas sun to rain is ¼.

Now at summer, I live in California and it truly doesn't rain in the summer.

Given that we have a probability of rain of 0.2 on a given day, what is the probability of having rain on at least two days during the week?

In fact, we worked out that the probability of rain at a time step later was only 0.4, so it was 0.6 sunny.

Decorative Rain

Rain

Rain!

We know that at time 0 it was raining with half a chance.

Given that we have a probability of rain on any given day of 0.2, what is the probability that over the course of a week it rains on exactly 2 days?

There was a light rain yesterday.

RainDrop

Raindrops...

Drizzle

Raindrops

Ah! I don't have much time. I'm leaving.