The complement of event A is the event that A does not occur. It is expressed as A^c. The probabilities of an event and its complement must have a sum of 1: P(A) + P(A^c) = 1. Note that event A and its complement, A^c, are mutually exclusive (there is no overlapping of their possible outcomes) and exhaustive (these two events include all possible outcomes). For example, event A refers to an increase in interest rates. The probability of event A is 0.7. A^c is the event that interest rates will not increase. P(A^c) = 1 - 0.7 = 0.3.

Probabilities are either unconditional or conditional.

Unconditional probability, also called marginal probability, is simply the probability of an event occurring. It refers to the probability of an event that is not conditioned on the occurrence of another event. For example, what is the probability that a stock earns a return above the risk-free rate? An unconditional probability can be considered as a stand-alone probability. It is expressed as P(A).

A conditional probability is the probability of an event given that another event has occurred. It is denoted as P(A | B) ("the probability of A given B").

For example, what is the probability that the total of two dice will be greater than 8 if the first dice is a 6? This can be computed by considering only outcomes which could occur if the first dice is a 6 and determining the proportion of these outcomes that total more than 8. There are six outcomes for which the first dice is a 6, and of these, there are four that total more than 8 (6,3; 6,4; 6,5; 6,6). The probability of a total greater than 8, given that the first dice is 6, is therefore 4/6 = 2/3. More formally, this probability can be written as: P(total>8 | Dice 1 = 6) = 2/3. In this equation, the expression to the left of the vertical bar represents the event and the expression to the right of the vertical bar represents the condition. Thus, it would be read as "The probability that the total is greater than 8, given that Dice 1 is 6, is 2/3." In more abstract form, P(A|B) is the probability of event A given that event B occurred.

A joint probability is the probability of both events A and B happening together. It is denoted as P(AB) ("the probability of A and B"). For example, Kevin is assessing the probability that both airfare and oil prices increase. Such a probability is a joint probability.

If two events are mutually exclusive, then they cannot occur together, so the joint probability of two mutually exclusive events is 0.

A joint probability function of two random variables, X and Y, gives the probability of joint occurrences of the values of X and Y.

If we know the conditional probability P(A|B) and we want to know the joint probability P(AB), we can use the following multiplication rule for probabilities:

P(AB) = P(A|B) x P(B)

Example 1

If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17 and the probability of A and B is: 1/13 x 1/17 = 1/221.

Example 2

The probability of an increase in oil price, P(B), is 0.4. The probability of an increase in airfare given an increase in oil price, P(A|B), is 0.3. The joint probability of an increase in both oil price and airfare, P(AB), is 0.3 x 0.4 = 0.12.

Hint:

• Look out for the words "given that" or "you are told that," which will help you know that the probability is conditional. In the absence of such information, the probability will be unconditional.
• The letter after the | is the event that we know has definitely occurred, whereas the letter before the | is the event whose probability we are trying to calculate.

Practice Question 1

For the tree diagram shown below:

P(not B | A) = ______

A. 0.1
B. 2/3
C. 1/3

Correct Answer: C

From the tree diagram, we know that 0.3 times P(B | A) = 0.2. Now, 0.3x = 0.2 yields x = 0.2/0.3 = 2/3. Thus, P(B | A) = 2/3. We have: P(B | A) + P(not B | A) = 1. Finally, 2/3 + P(not B | A) = 1 and P(not B | A) = 1 - 2/3 = 1/3.

Practice Question 2

For the tree diagram shown at the right,

P(B | A) = ______

A. 1/3
B. 3/7
C. 2/3

Correct Answer: C

From the tree diagram, we know that 0.3 times P(B | A) = 0.2. Now, 0.3x = 0.2 yields x = 0.2/0.3 = 2/3. Thus, P(B | A) = 2/3.

Practice Question 3

Given that P(Jane goes to Miami) = 50% and P(Jane attends the Super Bowl given she goes to Miami) = 20%, then P(Jane goes to Miami AND attends the Super Bowl) = ______.

A. 10%
B. 45%
C. 40%

Correct Answer: A

From the tree diagram, P(Jane goes to Miami AND she attends the Super Bowl) = P(Jane goes to Miami) times P(Jane attends Super Bowl given she goes to Miami) = (0.5)(0.2) = 0.1. So, P(Jane goes to Miami AND attends Super Bowl) = 10%.

Practice Question 4

Roll a die and flip a coin, P(5 given heads) = ______.

A. 1/2
B. 1/6
C. 1/8

Correct Answer: B

Because these events are independent events (what happens with the coin has no effect on what happens with the die), P(5 | heads) = P(5). Now, roll a die P(5) = 1/6. So, P(5 | heads) = 1/6.

Practice Question 5

The meaning of P(B | A) is ______.

A. probability of IF event B THEN event A
B. probability of IF event A THEN event B
C. probability of event A given event B

Correct Answer: B

The notation P(B | A) means the probability of If event A then event B. Another wording is probability of event B given event A.

Practice Question 6

The probability of the occurrence of an airplane crash and a successful resolution to a labor strike is called ______.

A. an unconditional probability
B. a joint probability
C. a conditional probability

Correct Answer: B

A joint probability takes the form of P(AB), the probability that event A (an airplane crash) and event B (the resolution of a strike) both happen.

Practice Question 7

Two cards are randomly selected, without replacement, from a standard deck of playing cards. What is the probability of first picking a five and then picking a card other than a five?

A. 0.071
B. 0.004
C. 0.072

Correct Answer: C

p = (4/52) x (48/51) = 0.072398

Practice Question 8

At a certain university, 42% of the students are women, and 18% are engineering majors. Of the engineers, 22% are women. If a student at this university is selected at random, what is the probability that the selected person will be a woman engineering major?

A. 0.0396
B. 0.0924
C. 0.6400

Correct Answer: A

Let A be the event that a woman is selected and B be the event that an engineering major is selected. We are given the information that P(A) = 0.42, P(B) = 0.18, and P(A|B) = 0.22. An application of the general multiplication rule gives P(A and B) = P(B)P(A|B) = 0.18 x 0.22 = 0.0396.

Practice Question 9

Sal has a 60% probability of being Salesman of the Month. He has a 40% probability of getting a raise. Sal has a 10% probability of being Salesman of the Month and getting a raise. What is P(Sal gets Salesman of the Month given he gets the raise)?

A. 25%
B. 30%
C. 16.7% to nearest 0.1%

Correct Answer: A

In a tree diagram, the event Salesman of Month given raise is represented by the top branch of the second set of branches. To find the probability of this branch we use: 0.4x = 0.1. Now, x = 0.1/0.4 = 1/4 = 0.25. Thus, P(Sal gets Salesman of the Month given raise) = 25%.

Practice Question 10

Sal has a 60% probability of being Salesman of the Month. He has a 40% probability of getting a raise. Sal has a 10% probability of being Salesman of the Month and getting a raise. What is P(Sal gets raise given he gets Salesman of the Month)?

A. 16.7% to nearest 0.1%
B. 50%
C. 40%

Correct Answer: A

In a similar tree diagram, the event raise given Salesman of Month is represented by the top branch of the second set of branches. To find the probability of this branch we use: 0.6x = 0.1. Now, x = 0.1/0.6 = 1/6 = 0.1666. Thus, P(Sal gets raise given he gets Salesman of the Month) = 16.7% to nearest 0.1%.

Practice Question 11

A dormitory on campus houses 200 students. 120 are male, 50 are upper-division students, and 40 are upper-division male students. A student is selected at random. The probability of selecting a lower-division student, given the student is female, is ______.

A. 7/8
B. 7/20
C. 2/5

Correct Answer: A

Practice Question 12

The probability that a mutual fund will generate a positive return in the next 12 months is called ______.

A. a conditional probability
B. an unconditional probability
C. a joint probability

Correct Answer: B

An unconditional probability takes the form of P(A), the probability that an event (A) will happen. It is unconditional because it is not conditioned on any other event.

Practice Question 13

If an analyst is trying to estimate the value of a stock, using the formula E(Y | E) = [y_1 * P(y_1 |E) + y_2 * P(y_2|E) + ... y_n * P(y_n|E), the analyst is making use of ______.

A. subjective probability
B. the multiplication rule for independent events
C. conditional expectation

Correct Answer: C

Since the stock value is conditioned on an uncertain event and the likelihood of certain stock price outcomes given the occurrence of that uncertain event, the analyst is using conditional expectation. The expected value is conditioned upon the occurrence of the event.

Practice Question 14

Supposing that P(A) = 0.7 and P(AB) = 0.42, find the probability of B given A, P(B|A).

A. 0.294
B. 0.600
C. 0.829

Correct Answer: B

The correct answer is P(B|A) = P(AB)/P(A) = 0.42/0.7 = 0.60.

Practice Question 15

The formula for joint probability is given by ______.

A. P(AB) = P(A | B) / P(B)
B. P(AB) = P(A | B) * P(B)
C. P(AB) = P(A | B) / [P(B) * P(A)]

Correct Answer: B

A joint probability takes the form of P(AB) = P(A | B) * P(B). Note that this is just a rearranged form of the formula for conditional probability.

Practice Question 16

Supposing that P(A) = 0.2 and P(B|A) = 0.6, find P(AB).

A. 0.12
B. 0.30
C. 0.80

Correct Answer: A

The correct answer is P(AB) = P(B|A) P(A) = 0.6 x 0.2 = 0.12.

Practice Question 17

Suppose that the probability that Company AirCo gets a government contract to produce military jets is 0.65 and the probability that AirCo lays off 5,000 employees if they don't get the contract is 0.80. What is the probability that AirCo does not get the government contract and lays off 5,000 employees?

A. 0.280
B. 0.438
C. 0.520

Correct Answer: A

Let A denote the event that AirCo does not get the government contract and B denote the event that 5,000 AirCo employees are laid off. The probability of A is P(A) = 1 - 0.65 = 0.35 and the probability of B given A is P(B|A) = 0.8. The probability of A and B is P(AB) = P(B|A) P(A) = 0.80 x 0.35 = 0.28.

Practice Question 18

Suppose that P(A) = 0.9, P(B) = 0.8, and P(B|A) = 0.6. Find P(A|B).

A. 0.480
B. 0.675
C. 0.720

Correct Answer: B

First, one finds that P(AB) = P(B|A)P(A) = 0.54, then P(A|B) = P(AB)/P(B) = 0.54/0.80 = 0.675.

Practice Question 19

The joint probability of three events is given as 35%. Their individual probabilities are 50%, 60%, and 80%. What should their joint probability be if they are independent?

A. 35%
B. 24%
C. 63.3%

Correct Answer: B

For the events to be independent, their joint probability should equal the product of their individual probabilities = 0.50 x 0.60 x 0.80 = 0.24, or 24%. Since this is different from 35%, the events are not independent.

Practice Question 20

Country A is a favorite for the final soccer match. However, country B has a stronger chance of making it to the final if country A wins its semi-final match. The probability that both A and B make it to the final is estimated at 30% and the probability of that country A reaches the final is 60%. What is the conditional probability of country B making it to the final if country A wins its semi-final match?

A. 50%
B. 90%
C. 18%

Correct Answer: A

P(A&B) = P(A|B) x P(B)

Prob(Country B in final|Country A is in final) = Prob(Countries A and B in final)/Prob(Country A in final) = 0.30/0.60 = 0.50, or 50%.