We now apply Bayes rule.

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

The probability of observing X and Y depends on what state the hidden markov model is in.

For A, it's 0.1 for X and 0.9 for Y.

So now I want to ask you a really tricky question

For coin X, we know that the probability of heads is 0.3.

And here's my code, this implements Bayes rule.

Now that one comes up with heads at 0.9.

And coin 2 is also loaded.

So there's 2 total events.

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.

So let's say I want to figure out the probability I'm going to flip a coin eight times and it's a fair coin.

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

Probability

Phil, the odds against

So 0.5 times 0.95 gives you 0.45 whereas the unfair coin, the probability of tails is 0.1 multiply by the probability of picking it at 0.5 gives us 0.05

I'm going to take this coin and I'm going to flip it twice. ... the probability of getting a heads and then getting another heads.

There's a 0.5 chance of taking coin 2.

Suppose the probability of heads is a quarter, 0.25.

And we could do the same thing on the other side. What if O had to go first?

Let's now go to the extreme, and this is a challenging probability question.

Secret carries 1 3, is 1 9, and secret 1 3 again.

Now, this statement is not true.

The same argument going on this side.

What are the odds?

And see what happens. It multiplies.

It's a loaded coin, and the reason is, well, each of them come up with a certain probability.

Now, what you do, you add those up and then normally don't add up to one.

Then we compute the probability for those 2 cases.

Let's do it with the second case.

Consider a coin that has a probability of landing on heads of 0.7.

let's look at some actual random variable definitions.

The coin can come up heads or tails, and my question is the following

Suppose the probability for heads is 0.5.

Then we can flip and get heads or tails for the coin we've chosen. Now what are the probabilities?

However, if I now divide, that is, I normalize those non normalized probabilities over here by this factor over here,

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

So this case over here, which indeed has tails, tails. We have 0 probability.

It has no impact on what happens on the next roll.

This can be resolved as follows.

Now, I suppose the robot sees red.

Probability of success

Probability of failure

So let's just figure out the probability of rolling it each of the times.