Notation

The image below shows the common notation for contingent probability . you’ll consider the road as representing “given”. On the left is that the event of interest, and on the proper is that the event we are assuming has occurred.

Conditional probability notation. P(A|B) is read as probability of A given B.

With this notation, you’ll also use words to explain the events. for instance , let’s say you wanted to seek out the probability someone buys a replacement car, once you know they need started a replacement job. this is able to be represented as:

Example employing a table of knowledge

One of the common sorts of problems you’ll see uses a two-way table of knowledge . Here, we’ll check out the way to find different probabilities using such a table.

two-way-table-conditional-probability-example

Example

A survey asked full time and part time students how often that they had visited the college’s tutoring center within the last month. The results are shown below.

Suppose that a surveyed student is randomly selected.

(a) what’s the probability the scholar visited the tutoring center four or more times, as long as the scholar is full time?

Two way table with only the information about full time students highlighted. There were a total of 45 full times students with 8 having visited tutoring four or more times. Therefore P(four or more times |full time) = 8 / 45.

Conditional probability is all about that specialize in the knowledge you recognize . When calculating this probability, we are as long as the scholar is full time. Therefore, we should always only check out full time students to seek out the probability.

(b) Suppose that a student is a component time. what’s the probability that the scholar visited the tutoring center one or fewer times?

This one may be a bit more tricky thanks to the wording. consider it within the following way:

Find: probability student visited the tutoring center one or fewer times

Assume or given: student is a component time (“suppose that a student is a component time”)

tTwo way table with only part time student data highlighted. In a total of 13 part time students, only 2 went to tutoring one or fewer times. Therefore P(one or fewer times | part time) = 2/13.

Since we are assuming (or supposing) the scholar is a component time, we’ll only check out part time students for this calculation.

(c) If the scholar visited the tutoring center four or more times, what’s the probability he or she is a component time?

As above, we’d like to form sure we all know what’s given, and what we are finding.

Find: probability he or she is a component time

Two way table with only the data for students who visited tutoring four or more times highlighted. There were 14 such students, and 6 were part time, so P( part time | four or more visits) = 6/14.

Assume or given: student visited the tutoring center four or more times (“if the scholar visited the tutoring center four or more times…”)

For this question, we are only watching students who visited the tutoring center four or more times.

As you’ll see, when employing a table, you only got to concentrate to which group from the table you ought to specialise in .

The conditional probability formula. P(A give B) = P(A and B) divided by P(B).

Examples of using the formula to seek out contingent probability

In some situations, you’ll got to use the subsequent formula to seek out a contingent probability .

This formula could actually be used with the table data, though it’s often easier to use in problems almost like subsequent example.

Example

In a sample of 40 vehicles, 18 are red, 6 are trucks, and a couple of are both. Suppose that a randomly selected vehicle is red. what’s the probability it’s a truck?

We are asked to seek out the subsequent probability:

P(truck|red)

Applying the formula:

P(truck|red)=P(truck and red)P(red)=2401840=218=19≈0.11

The thinking behind the formula is extremely almost like the thinking used with the table. for instance , notice that what we “know” finishes up on rock bottom of the fraction. we will also apply this to situations where we are given probabilities and not counts.

Example

A parlor game comes with a special deck of cards, a number of which are black, and a few of which are gold. If a card is randomly selected, the probability it’s gold is 0.20, while the probability it gives a second turn is 0.16. Finally, the probability that it’s gold and provides a second turn is 0.08.

Suppose that a card is randomly selected, and it allows a player a second turn. what’s the probability it had been a gold card?

This time, we are given the subsequent probabilities:

“the probability it’s gold is 0.20” -> P(gold) = 0.2

“the probability it gives a second turn is 0.16” -> P(second turn) = 0.16

“the probability that it’s gold and provides a second turn is 0.08” -> P(gold and second turn) = 0.08

We try to calculate:

P(gold|second turn)

We can apply the formula to seek out this probability:

P(gold|second turn)=P(gold and second turn)P(second turn)=0.080.16=0.5

You can see that this works out very nicely if you’re taking a flash to write down down the knowledge given within the problem. In fact, you’ll really say that about any real-life/ word problem in math!

This formula could actually be used with the table data, though it’s often easier to use in problems almost like subsequent example.

Example

In a sample of 40 vehicles, 18 are red, 6 are trucks, and a couple of are both. Suppose that a randomly selected vehicle is red. what’s the probability it’s a truck?

We are asked to seek out the subsequent probability:

P(truck|red)

Applying the formula:

P(truck|red)=P(truck and red)P(red)=2401840=218=19≈0.11

The thinking behind the formula is extremely almost like the thinking used with the table. for instance , notice that what we “know” finishes up on rock bottom of the fraction. we will also apply this to situations where we are given probabilities and not counts.

Example

A parlor game comes with a special deck of cards, a number of which are black, and a few of which are gold. If a card is randomly selected, the probability it’s gold is 0.20, while the probability it gives a second turn is 0.16. Finally, the probability that it’s gold and provides a second turn is 0.08.

Suppose that a card is randomly selected, and it allows a player a second turn. what’s the probability it had been a gold card?

This time, we are given the subsequent probabilities:

“the probability it’s gold is 0.20” -> P(gold) = 0.2

“the probability it gives a second turn is 0.16” -> P(second turn) = 0.16

“the probability that it’s gold and provides a second turn is 0.08” -> P(gold and second turn) = 0.08

We try to calculate:

P(gold|second turn)

We can apply the formula to seek out this probability:

P(gold|second turn)=P(gold and second turn)P(second turn)=0.080.16=0.5

You can see that this works out very nicely if you’re taking a flash to write down down the knowledge given within the problem. In fact, you’ll really say that about any real-life/ word problem in math!