Conditional probability for an event is computed under the assumption that some other event is going to occur or already has occurred..

1. Know when to compute the conditional probability.

The conditional probability for an event is computed under the assumption that some other event is going to occur or already has occurred and is written Pr[A|B] or the Pr[A] given Pr[B]..

The following is a conditional probability statement:

The probability that it will rain given it is cloudy is 0.70?

An is written in the form: Pr[rain | cloudy] = 0.70.

The event rain is listed first , and its Probability is 0.70. The second event is the given event or conditional event and it appears after the vertical bar, |.

The Pr[cloudy] is called the unconditional probability for the conditional probability statement.

Another Pr[rain | low humidity] = 0.35.

Both cloudy and low humidity are conditional statements and if no conditions apply the probability of rain is stated simply as : Pr[rain].

2. Know how to compute the conditional probability

Count and Divide Method:

Simple count the number of possibilities for both the first and second events and eliminate extraneous possibilities.

Example: Find the probability for drawing a Jack from among the face cards:

Since there are 4 jacks and 12 face cards,

Conditional Probability Identity:

The conditional probability is found by dividing the joint probability by the unconditional probability, Pr[B] for the given event.

This formula can only be used if the appropriate probabilities are known: Pr[A and B] and P[B].

Example: Pr[Jack|Face] = Pr[Jack and Face] = 4/52 and Pr[Face]= 12/52,

So 

3. Know how to tell when events are conditional or independent.

Since the probability of independent (mutually exclusive) events are unaffected by the occurrence of the other.

Then Events A and B are independent if Pr[A] = P[A|B]

Events A and B are dependent events whenever

So to determine if two events are independent or dependent, compare their conditional and unconditional probabilities.

Is the Pr[Queen] and Pr[Hearts] independent or dependent?

Pr[Queen] = 4/52. Pr[Queen|Heart] = Pr[Queen and Heart] / Pr[Heart]=

Since Pr[Queen] = 1/13 = Pr[Queen|Heart] = 1/13, Queens and Hearts are independent events.

However, Pr{Queen] and Pr[Face] are dependent events since 

4. Know how to construct and calculate conditional probabilities using a joint probability table.

The joint probability of A and B is the probability that both A and B occurred at the same time and is written as Pr[A and B].

Most researcher will use the joint probability table to compute conditional, marginal and joint probabilities.

Given the frequency table:

By Example: The following table shows the number of State University Students on scholarship classified in terms of three sex and educational levels.
 

Sex	Educational Level		Total
	Undergraduate (U)	Graduate (G)
Man (M)	24	16	40
Women (W)	18	14	32
Total	42	30	72

The joint probability table is found by dividing the frequency of each cell by the total number of all cells.

Joint Probability Table:

Sex	Educational Level		Marginal Probability
	Undergraduate (U)	Graduate (G)
Man (M)	Pr(M and U) =24/72	Pr(M and G) =16/72	Pr(M) = 40/72
Women (W)	Pr(W and U)=18/72	Pr(W and G)=14/72	Pr(W)=32/72
Marginal Probability	Pr(U) =42/72	Pr(G)=30/72	1

The probability rows and column totals, margins are refer to as the marginal probabilities.

For Example, Pr[W] for 2nd row above of joint probability table.

To compute the conditional probabilities from the Joint Probability table use the formula:

For example the Joint Probability of 

Example: Find Pr[Jack|Face] using the Joint Probability table.
 

Face	Jack		Total
	Number of Jack (J)	Number of Non Jack (N)
Number of Face (F)	4 (Pr[J and F] = 4/52)	8	12 (Pr[F] =12/52)
Number of Non - Face (O)	0	40	40
Total	4	48	52

Pr[Jack|Face]=Pr[J|F] = Pr[ Jack and Face] / Pr[Face] =

5. Know how to compute the joint or unconditional probabilities from the conditional probability.

Knowing the conditional probability of an event, Pr[A|B], then:

The joint probability, Pr[ A and B] = Pr[A|B] x P[B]

And the unconditional probability, Pr[B] = Pr[A and B] divided by Pr[A|B].

Example if the Conditional Probability, Pr[Jack|Face] is 1/3 and Pr[Face] is 12/52

Then the Pr[ A and B] = Pr[Jack|Face] x Pr[Face] = 

Workshop Problem (from textbook) Conditional Probability Table

A change was proposed in the mathematics curriculum at a college. The mathematics majors were asked whether they approved of the proposed change. The results of the survey follow:
 

Student Sex	Survey Response			Total
	Approved	Not Approve	No Opinion
Male	21	12	6
Female	14	7	10
Total

Suppose that a mathematics major is selected by chance. Find the probability that.

(a) The student is female, given no opinion

(b) The student approves of the proposed change, given the student is male.

(c) The student is male, given the student does not approved of the proposed change.

(d) The student is male and approves of the proposed change. (Use the Multiplication Rule).

Worksheet - Joint Probability Table
 

Event B	Event A (joint probabilities) - Pr[A and B]				Total (marginal probabilities) Pr[B]
	Level 1	Level 2	Level 3	Level 3
Level 1
Level 2
Level 3
Level 4
Total (marginal probabilities) Pr[A]

6. Know how conditional probabilities are related to sampling without replacement.

Suppose you drew 12 cards and they were all the face cards (12 in all) then the remaining cards in the deck or non face cards and the Pr[face] is 0 if you were to draw another card.

However, if you had replaced the 12 cards and drew a 13th card, then the probability would be the same as if you had not drawn any card.

What if there were 5 defective parts in a shipment of 100 item and your first sample was defective, then you have 4 more defective parts left in the remaining 99 shipment items.

So given that your first sample was defective what is the proportion defective in the remaining shipment or 99 item? The answer is 4 out of 99. Another way of asking this question is using conditional probability:

Pr{defective| first was defective) = 4/99.

If you sample a population or set of data and replace the sample, then the condition of the data set would be as if you did not sample - sample with replacement. (assuming that sampling did not change the sample of data set in anyway that would introduce bias.

Sampling without replacement is sampling without replacing the sample into the data set or population. This can results in a conditional probability statement.