Contents of this Precalculus episode:

Conditional probability, A given B.

The real thrill comes now.

Let's see what the probability of A will be, if we know that event B will definitely occur.

Well, we only have 3 cases, because event B will definitely happen,

and the desired event is an odd number, which is only one of these.

Thus, this new probability is 1/3, and the following notation is used:

questionable definite

We read this as A given[e1] B, and it answers the question of what chance does A have, if B definitely occurs.

CONDITIONAL PROBABILITY

The probability of event A, if event B definitely occurs:

Let's see what we can use this for.

In a town, out of 1000 residents, on average, 350 are smoking, 120 has some sort of cardiovascular trouble, and there are 400 who belong to at least one of these groups.

According to a survey, 30% of TV watchers watch the morning news. 90% of TV watchers watch at least one of the morning or evening news.

If a resident has cardiovascular troubles, what is the probability that he is a smoker?

A=smoker

B=cardiovascular trouble

Let's see the problem.

Cardiovascular trouble is definite, smoking is questionable.

There are then these formulas.

Thus, a resident with cardiovascular troubles is a smoker with a 0.583 probability.

Here is another very exciting story.

According to a survey, 90% of TV watchers watch at least one of the morning or evening news. Those who watch the evening news have a 20% chance that they also watched the morning news. The morning news is watched by 30% of all TV watchers.

What is the probability that if someone watches the morning news, then he watches the evening news as well?

A=watches in the morning

B=watches in the evening

Let's try writing the question:

watches in the morning: definite

watches in the evening: questionable

So far so good.

Let's see what we know.

evening for sure, morning 20% chance

There are then these formulas.

90% of TV watchers watch at least one of the morning or evening news.

Precalculus episode