The probability of a positive test and having cancer.

Let's say the probability of having this cancer is 0.1.

So we can see here the probability of having a loaded coin times the probability of the flips given the loaded coin is 0.06561 and the probability of having the same flips with a fair coin times its probability is 0.05625.

What do you think is the probability of having that specific type of cancer?

Let's assume there are two coins equally distributed, same probability of having each.

And look what it does to my probability of having received a raise.

And this expression equals 0.6080 and so the probability that we have the same coin is simply the probability that we both have type 1 or fair coins, plus the probability of my having coin 2 times the probability of your having coin 2.

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 if you get a positive test result you're going to raise the probability of having cancer relative to the prior probability.

The probability of a negative test and having cancer, and this is not conditional anymore.

The probability of having tails given that it's a certain coin is just going to be 1 minus the probability with being heads.

Probability of success

Probability of failure

Let's assume that we know that the probability of actually having a loaded coin is zero.

We know this simply because the probability of having a loaded coin at all is zero.

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.

What is the probability of having heads on the first flip or on the second flip if we have one coin with a probability of heads of 0.7 and a second coin with a probability of heads of 0.5?

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?

Now let me ask you if the probability of having a loaded coin is 0.1, what happens to the probability of fair given flips in each of these cases.

Times the probability of b divided by the probability of a.

Then given the dataset of how your computers in your data center usually behave, you can model the probability of x, so you can model the probability of these machines having different amounts of memory use or probability of these machines having different numbers of disc accesses or different CPU loads and so on.

We'd like some way of searching through them efficiently without having to consider the probability of every possible segmentation.

This thing over here is the same as probability of test 2 to be positive, which I'm going to abbreviate with a 2 over here, conditioned on test 1 being positive and me having cancer times the probability of me having cancer given test 1 was positive plus the probability of test 2 being positive conditioned on test 1 being positive and me not having cancer times the probability of me not having cancer given that test 1 is positive.

The non normalized probability according to Bayes Rule is obtained as follows the probability of observing H for the fair coin is 0.5, and the probability of having grabbed the fair coin is 0.5 as well.

Probability of finishing before

So that tells us the probability of a given b times the probability of b, is equal to the probability of b given a, times the probability of a.

So the probability of a given b could be the probability of a.

So the probability of having four heads in a row from a loaded coin is going to be 0.6561.

I now multiply these 2 things together to get my non normalized probability of having cancer given the plus.

We just applied Bayes Rule to compute a really involved probability of having cancer after seeing a test result.

Probability of X2 equals heads.

What's the probability of Y?

The probability of D2 sunny.

Probability

The probability of success in this example was 0.6, probability of failure was 0.4.

Now heads has a probability of 0.2 and tails has a probability of 0.8.

What's the probability of cancer now written in short form probability of C given ?

That's equal to the probability of a occurring given b, times the probability of b, which is also equal to the probability of b occurring given a, times the probability of a.

Here you get 0.72, which is the product of not having cancer in the first place 0.9 and the probability of getting a negative test result under the condition of not having cancer.

So, there's the probability of no cancer, probability of negative, which is negation of positive, given C, and probability of negative positive given not C.

My resulting probability will be 1 α of the non normalized probability.

So what's the probability of b, or the probability of getting four out of six heads?

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

This is the formal definition of total probability, which is the probability of A1, given previously in A0, times the probability of being A0.

Perhaps the probability of it being sunny is 0.7, probability of a raise is 0.01.