If we have two events, A and B, that we are interested in, we often want to know the probability that either A or B occurs. Note the use of the word "or," the key to this rule. The "or" is what we call an "inclusive or." In other words, either one event can occur or both events can occur.

Such probabilities are calculated using the addition rule for probabilities.

P(A or B) = P(A) + P(B) - P(AB)

The logic behind this formula is that when P(A) and P(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, P(AB) is subtracted.

If events A and B are mutually exclusive, the joint probability of A and B is 0. Consequently, the probability that either A or B occurs is simply the sum of the unconditional probabilities of A and B: P (A or B) = P(A) + P(B).

What is the probability that a card selected from a deck will be either an ace or a spade? The relevant probabilities are:
P(ace) = 4/52; P(spade) = 13/52
The only way in which an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so:
P(ace and spade) = 1/52.
The probability of an ace or a spade can be computed as:
P(ace) + P(spade) - P(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

Consider the probability of rolling dice twice and getting a 6 on at least one of the rolls. The events are defined in the following way:
Event A: 6 on the first roll: p(A) = 1/6
Event B: 6 on the second roll: p(B) = 1/6
P(A and B) = 1/6 x 1/6
P(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36

The same answer can be computed using the following (admittedly convoluted) approach: Getting a 6 on either roll is the same thing as not getting a number from 1 to 5 on both rolls. This is equal to: 1 - P(1 to 5 on both rolls).
The probability of getting a number from 1 to 5 on the first roll is 5/6. Likewise, the probability of getting a number from 1 to 5 on the second roll is 5/6. Therefore, the probability of getting a number from 1 to 5 on both rolls is: 5/6 x 5/6 = 25/36. This means that the probability of not getting a 1 to 5 on both rolls (getting a 6 on at least one roll) is: 1-25/36 = 11/36.

Despite the convoluted nature of this method, it has the advantage of being easy to generalize to three or more events. For example, the probability of rolling dice three times and getting a six on at least one of the three rolls is: 1 - 5/6 x 5/6 x 5/6 = 0.421

Practice Question 1

Suppose that A and B are independent events. The probability of A is 0.6 and the probability of B is 0.8. What is the probability either A or B will occur?

A. 0.14
B. 0.48
C. 0.92

Correct Answer: C

If A and B are independent events, then the probability of both events occurring is the product of the probability of A times the probability of B. So, Pboth = .60 x .80 = .48.

The probability of either event occurring is given by the formula: Peither = PA + PB - Pboth
where PA = probability of event A
and PB = probability of event B

Therefore, the probability of either event occurring is: Peither = .60 + .80 - .48 = .92

Practice Question 2

A six-sided die is rolled once and the outcome is observed. Define the events A and B as follows:

Event A: Number observed is even
Event B: Number observed is greater than 3

Find P(A or B).

Correct Answer: 2/3

P(A) = 0.5, P(B) = 0.5, and P(A and B) = P(4 or 6) = 1/3. Then P(A or B) = P(A) + P(B) - P(A and B) = 2/3.

Practice Question 3

In a town, 36% of households own a dog, 20% own a cat, and 60% own neither a dog nor a cat. If we select a household at random, the chance that it owns both a dog and a cat is ______.

Correct Answer: 0.16

Let A be the event that the selected household owns a dog and B be the event that the selected household owns a cat. The event A or B indicates the selected household owns either a dog or a cat. (P(A or B) = P(own either a dog or a cat) = 1 - P(own neither a dog nor a cat) = 1 - 0.6 = 0.4.
The general addition theorem says P(A or B) = P(A) + P(B) - P(A and B), or equivalently,
P(A and B) = P(A) + P(B) - P(A or B) = 0.36 + 0.2 - 0.4 = 0.16.

Practice Question 4

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 is a woman or an engineering major?

A. 0.0396
B. 0.5604
C. 0.6000

Correct Answer: B

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. According to the general addition rule, P(A or B) = P(A) + P(B) - P(A and B).
Since P(A and B) = P(B) x P(A|B) = 0.18 x 0.22 = 0.0396
P(A or B) = 0.42 + 0.18 - 0.0396 = 0.5604.

Practice Question 5

Students at University X must be in one of the class ranks: freshman, sophomore, junior, or senior. At University X, 35% of the students are freshman and 30% are sophomores. If a student is selected at random, the probability that he or she is either a junior or a senior is ______.

A. 35%
B. 65%
C. 30%

Correct Answer: A

Using the addition rule, because freshman and sophomore are disjoint, P(freshmen or sophomore) = P(freshman) + P(sophomore) = 0.35 + 0.3 = 0.65.
The probability of the event junior or senior is 1 minus the probability that this event does not occur, namely 1 minus the probability of the event freshman or sophomore. Its probability is thus 1 - 0.65 = 0.35.

Practice Question 6

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

A. 5/12
B. 1/12
C. 7/12

Correct Answer: C

P(5) = 1/6, P(heads) = 1/2, P(A and heads) = 1/6 x 1/2 = 1/12
P(5 or heads) = P(5) + P(heads) - P(5 and heads) = 7/12

Practice Question 7

A new and relatively obscure pharmaceutical company Nucomp is competing for two large government contracts to produce vaccines for two different diseases. The probability that the company is awarded the first contract is 0.85, the probability that it is awarded the second contract is 0.80, and the probability that it is awarded both contracts is 0.78. What is the probability that Nucomp is awarded at least one of the contracts?

A. 0.82
B. 0.87
C. 1.65

Correct Answer: B

Let A denote the event that Nucomp is awarded the first contract and B denote the event that Nucomp is awarded the second contract. P(A or B) = P(A) + P(B) - P(AB) = 0.85 + 0.80 - 0.78 = 0.87

Practice Question 8

Which of the following is (are) true?

I. If P(A or B) = P(A) + P(B), then A and B are independent.
II. If P(A and B) = P(A) x P(B), then A and B are mutually exclusive.
III. If P(A and B) = P(A) + P(B), then P(A) = P(B) = 0.

A. III only
B. II & III
C. I only

Correct Answer: A

If A and B are independent, P(A and B) = P(A) x P(B) and vice versa. If A and B are mutually exclusive, P(A and B) = 0.

Note that P(A or B) = P(A) + P(B) - P(A and B). Hence, if P(A and B) = P(A) + P(B), then P(A or B) = 0. This implies that neither A nor B can occur, giving P(A) = P(B) = 0.

Practice Question 9

If two events, A and B, are mutually exclusive, what does the special rule of addition state?

A. P(A and/or B) = P(A) +P(B)
B. P(A and B) = P(A) + P(B)
C. P(A or B) = P(A) + P(B)

Correct Answer: C

The key information is that these events are mutually exclusive. So P(A or B) = P(A) + P(B). The probabilities stand alone.

Practice Question 10

The probability that events A and B do not occur simultaneously equals 0.77. The probability of neither A nor B occurring equals 0.38. If P(A) equals 0.26, the probability of B occurring equals ______.

A. 0.43
B. 0.59
C. 0.62

Correct Answer: B

The probability that events A and B occur simultaneously equals one minus the probability that events A and B do not occur simultaneously. Thus, P(A and B) = 1 - 0.77 = 0.23.

The probability of either A or B occurring equals one minus the probability of neither A nor B occurring. Thus, P(A or B) = 1 - 0.38 = 0.62.

Now, P(A or B) = P(A) + P(B) - P(A and B). Therefore, P(B) = P(A or B) - P(A) + P(A and B) = 0.62 - 0.26 + 0.23 = 0.59.

Practice Question 11

A survey of top executives revealed that 35% regularly read Time magazine, 20% read Newsweek, and 40% read U.S. News & World Report. Ten percent read both Time and U.S. News & World Report. What is the probability that a particular top executive reads either Time or U.S. News & World Report regularly?

A. 0.85
B. 0.06
C. 0.65

Correct Answer: C

35% + 40% - 10% = 65%

Practice Question 12

A student's grade in a course is determined by her score on two exams. The probability that she passes the first exam is 0.85, the probability that she passes the second exam is 0.80, and the probability that she passes both exams is 0.78. What is the probability that she passes at least one of the exams?

A. 0.05
B. 0.83
C. 0.87

Correct Answer: C

Let A denote the event that she passes the first exam and B denote the event that she passes the second exam. P(A or B) = P(A) + P(B) - P(A and B) = 0.85 + 0.80 - 0.78 = 0.87

Practice Question 13

Of letters sent through a certain post office, 20% have insufficient postage, 48% have the wrong zip code, and 58% have one or both of these problems. What is the probability that a letter has both insufficient postage and the wrong zip code?

A. 0.10
B. 0.30
C. 0.68

Correct Answer: A

Let A denote the event that the letter has insufficient postage and B denote the event that it has the wrong zip code. We know that P(A) = 0.20, P(B) = 0.48 and P(A or B) = 0.58. Solving using the general rule of addition for P(A and B), one obtains P(A and B) = P(A) + P(B) - P(A or B). Thus, the probability of a letter having both problems is P(A and B) = 0.20 + 0.48 - 0.58 = 0.10.

Practice Question 14

The manager of a pension fund determines that during the past five years 85 percent of the stocks in the portfolio have paid a dividend and 40 percent of the stocks have announced a stock split. If 95 percent of the stocks have paid a dividend and/or announced a stock split, the joint probability of a stock paying a dividend and announcing a stock split is closest to ______.

A. 25%
B. 30%
C. 40%

Correct Answer: B

The probability that at least one of two events will occur is the sum of the probabilities of the separate events less the joint probability of the two events.

P(A or B) = P(A) + P(B) - P(AB)
95% = 85% + 40% - P(AB); therefore P(AB) = 30%