Two events, A and B, are independent if and only if P(A|B) = P(A), or equivalently, P(B|A) = P(B). That is, the occurrence of one event has no influence on the probability of the occurrence of the other event.

In more detail, whether or not B occurs will have no effect on the probability of A and vice versa. Thus, there will be no difference between P(A|B) and P(A), and similarly there will be no difference between P(B|A) and P(B).

For example, suppose you flip a coin twice. The event of getting heads on the first flip does not affect the probability of getting heads on the second flip. Therefore, the event of getting heads on the second flip is independent of the event of getting heads on the first flip.

When two events are not independent, they must be dependent: the occurrence of one is related to the probability of the occurrence of the other.

If we are trying to forecast one event, information about a dependent event will be useful but information about an independent event will not be useful.

Example 1

If C = {the price of insurance share C goes up} and D = {the price of insurance share D goes up}, then clearly C and D are dependent events, because the market as a whole might be bullish or the insurance sector alone might be having a good day. Although the increase in share price C might not affect share price D at all, there is clearly a good chance that the two shares will move in the same direction.

Example 2

If A = {a person in Europe drives a red car} and B = {a person in Australia drives a white car}, then events A and B are clearly independent, as the one almost certainly has no bearing upon the other.

Remember that if A and B are independent, Ac and Bc will also be independent.

A and B are two events. If A and B are independent, the probability that events A and B both occur is:

P(A and B) = P(A) x P(B)

In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. This relationship is known as the multiplication rule for independent events.

What is the probability that a fair coin will come up with heads twice in a row? Two events must occur: heads on the first toss and heads on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 x 1/2 = 1/4.

Now consider a similar problem: Someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)? Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is: 1/52 x 1/4 = 1/208.

Similarly, for any number of independent events E₁, E₂.....E_n, the probability that all of them occur is:

P(E₁ and E₂..... and E_n) = P(E₁) x P(E₂) x ..... x P(E_n)

Example

In a bullish market, three shares, chosen from different sectors of the market, have probabilities of 0.6, 0.5 and 0.8 that their share prices will rise on any particular day. Let's call the shares K, L and M respectively.

If we make the assumption that prices of shares on successive days are independent, and also that the price movement of one of the shares above is independent of the others, we can then carry out the following calculations.

Note that although this may seem slightly unrealistic, the fact that the shares come from different sectors of the market lends some credence to the assumption, and also simplifies our calculations considerably.

P (share K has a price rise on two consecutive days) = 0.6 x 0.6 = 0.36
P (shares L and M both rise in price on the same day) = 0.5 x 0.8 = 0.4

Thus, the independent assumptions make our work much easier.

To calculate the probability that share M rises in price on four consecutive days, we can use the results above to calculate this probability as: 0.8 x 0.8 x 0.8 x 0.8 = 0.4096 (i.e., 0.8 multiplied four times, once for each day).

Although the chance that the share will rise in price on any one day is 80%, the chance that this will happen for four days in a row is just over half of this total, or 40.96%.

If we wish to calculate the probability that all three shares will rise in price on the same day, we can use the results above to get: 0.6 x 0.5 x 0.8 = 0.24 (i.e., the individual probabilities multiplied together)

Warning: It is important to note that multiplying individual probabilities together can only be done if the events that make up those probabilities are independent. If the events are dependent, this process is not valid.

To calculate the probability that, of the three shares above, none will have a price rise on a particular day, we can multiply the probabilities of the complementary events together to get: 0.4 x 0.5 x 0.2 = 0.04.

Thus, there is only a 4% chance that no shares will have a price rise on a particular day.

You might be wondering, if the chance that all three shares rise in price is 24% and the chance that no shares rise in price is 4%, about what has happened to the remaining 100% - 24% - 4% = 72%. The answer is that this percentage covers situations where some shares rise in price while others don't. You'll see this type of example in more detail in later lessons.

Practice Question 1

Decide if the following events are mutually exclusive and whether or not they are independent or dependent.

Event A: drawing a queen from a standard deck of 52 cards
Event B: drawing a heart or diamond from a standard deck of 52 cards

A. not mutually exclusive, independent
B. mutually exclusive, independent
C. mutually exclusive, dependent

Correct Answer: A

Practice Question 2

For the tree diagram shown with P(A) = 0.3 and P(B) = 0.6, which of the following is false?

A. P(not A) = 0.7
B. Events A and B are dependent.
C. P(not B) = 0.4
D. Events A and B are independent.

Correct Answer: D

We know events A and B are NOT independent events because P(B | A) = 2/3 and it was given that P(B) = 0.6. Because P(B | A) is not equal to P(B), the events are dependent.

Practice Question 3

For events A and B, which of the following is false?

A. If events A and B are independent, P(A and B) = P(A) x P(B)
B. P(A and B) = P(A) x P(B | A)
C. P(A | B) = P(A and B)
D. If events A and B are independent, P(A | B) = P(A)

Correct Answer: C

The statement P(A | B) = P(A and B) is false.

Practice Question 4

A jar contains 3 red and 5 yellow balls; two balls are randomly selected from the jar. Which of the following is true?

A. This is a binomial situation.
B. The events first pick and second pick are independent events.
C. P(red ball and yellow ball) = 0
D. The events first pick and second pick are dependent events.

Correct Answer: D

Practice Question 5

Suppose that P(A and B) = 0.08 and P(B) = 0.2. If A and B are independent, what is P(A)?

A. 0.12
B. 0.30
C. 0.40

Correct Answer: C

If A and B are independent then P(A) = P(A|B) = P(AB)/P(B) = 0.08/0.20 = 0.40.

Practice Question 6

If 15% of the population is left-handed, what is the probability that in a randomly selected group of five people, all five people are left-handed?

A. 0.0300
B. 0.3333
C. 0.0000759

Correct Answer: C

p = 0.15⁵ = 0.0000759

Practice Question 7

A multiple-choice test has three questions, each with four choices for the answer, of which only one of the choices is correct. What is the probability of guessing correctly on at least one of the questions?

A. 0.422
B. 0.578
C. 0.250

Correct Answer: B

The probability of guessing all questions wrong: 0.75 x 0.75 x 0.75 = 0.4218. The probability of guessing at least one correctly is 1 - 0.4218 = 0.5781.

Practice Question 8

If two six-sided dice are rolled once, what is the probability of rolling doubles (same number on each die) and a sum of six?

A. 0.278
B. 0.250
C. 0.028

Correct Answer: C

That is, to get 3 and 3! (1/6) x (1/6) = 0.0277

Practice Question 9

Joan has a 60% probability of making a free-throw. If she shoots two free-throws, what is the probability she misses both of them?

A. 48%
B. 16%
C. 60%

Correct Answer: B

Drawing a tree diagram, the event Miss AND Miss is represented by the bottom branch. The probability of traveling this branch is (0.4)(0.4) = 0.16 = 16%. Thus, P(Joan misses both free-throws) = 16%.

Practice Question 10

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

A. 1/2
B. 1/6
C. 1/12

Correct Answer: C

In a tree diagram, the event 5 AND heads is represented by the top branch. The probability of traveling this branch is (1/6)(1/2) = 1/12. Thus, P(5 and heads) = 1/12.

Practice Question 11

Three pharmaceutical companies have contracted with the government for the delivery of three different types of vaccines by January 1, 2012. Because of other work commitments and procedural problems with the FDA, contracting firms cannot always meet the agreed delivery date. The probability that the first vaccine will be delivered on time is 0.7. The probability that the second vaccine will be delivered on time is 0.6, and the probability that third vaccine will be delivered on time is 0.8. If the deliveries are independent of one another, find the probability that all three pharmaceutical companies will meet their deadlines.

A. 0.144
B. 0.336
C. 0.980

Correct Answer: B

Let A denote the event that the first vaccine is on time, B denote the event that the second vaccine is on time and C denote the event that the third vaccine is on time. Since the deliveries are independent, the joint probability is P(ABC) = P(A)P(B)P(C) = (0.7)(0.6)(0.8) = 0.336.

Practice Question 12

Affirmative action commitments by industrial organizations have led to an increase in the number of women in executive positions. Satellite Office Systems has vacancies for two executives, which it will fill from among four women and six men. What is the probability that two woman are selected?

A. 3/5
B. 3/4
C. 2/15

Correct Answer: C

Probability of both positions being filled by women = 4/10*3/9 = 12/90

Practice Question 13

Suppose that the probability of A is 0.80 and the probability of B is 0.60. If A and B are independent events, what is the joint probability of A and B?

A. 0.48
B. 0.75
C. 0.80

Correct Answer: A

Since A and B are independent events, P(A) = P(A|B) = 0.8 and P(AB) = P(A|B)P(B) = 0.80 x 0.60 = 0.48.

Practice Question 14

Suppose that the probability of A is 0.80 and the probability of B is 0.60. If A and B are independent events, what is the probability of A or B?

A. 0.48
B. 0.80
C. 0.92

Correct Answer: C

Since A and B are independent events, P(A) = P(A|B) = 0.8 and P(AB) = P(A|B)P(B) = 0.80 x 0.60 = 0.48, so P(A or B) = P(A) + P(B) - P(AB) = 0.80+0.60-0.48=0.92.

Practice Question 15

Suppose we roll a die and let A be the event that the number of spots showing is 6. We then toss a coin and let B be the event that the coin comes up heads. The P(A and B) is ______.

A. 1/12
B. 4/6
C. 7/12

Correct Answer: A

The two events are independent because the outcome of the die cannot influence the outcome of the coin. In this case, P(A) = 1/6 and P(B) = 1/2. Because the events are independent, we can use the multiplication rule for independent events to find P(A and B) = P(A) x P(B).

Practice Question 16

Main Street contains 8 traffic lights. The lights work independently of each other and the probability that one of the lights is green is 60%. The probability of travelling Main Street without stopping for a light is ______% (to the nearest 0.1%).

A. 16.7%
B. 0.1%
C. 1.7%

Correct Answer: C

Let x count the number of green lights. The random variable is binomial with N = 8 and p = 0.6. To go through town without stopping for a light, we need 8 green lights, P(x = 8) = P(8). Using the calculator and finding the distribution for x, we get P(8) = 0.0168 = 1.7% (to the nearest 0.1%).

Practice Question 17

Suppose that stocks A, B, C, and D are independent with respect to their price movement and have probabilities of increasing of 0.25, 0.50, 0.40, and 0.30. What is the probability that stocks A and C will increase in price while stocks B and D will fail to increase?

A. 10%
B. 2.5%
C. 3.5%

Correct Answer: C

If these events are independent, the joint probability of them occurring together is just the product of the individual probabilities. So, P(AC) = 0.25 * 0.40 = 10%. To this, we multiply the probabilities of B and D failing to increase: 10% * (1 - 0.50)*(1 - 0.30) = 3.5%.

Practice Question 18

Suppose that events A, B, C, and D are independent, and have probabilities of 0.25, 0.50, 0.40, and 0.30, respectively. What is P(ABCD)?

A. 0.15
B. 1.0%
C. 1.5%

Correct Answer: C

If these events are independent, the joint probability of them occurring together is just the product of the individual probabilities. P(ABCD) = 0.25 * 0.50 * 0.40 * 0.30 = 1.5%

Practice Question 19

An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most had satisfactory weights.

Weight: % of Total
Underweight: 2.5
Satisfactory: 90.0
Overweight: 7.5

Three packages are selected from the food processing line.

What is the probability of selecting and finding that all three of them are satisfactory?

A. 0.729
B. 0.810
C. 0.075

Correct Answer: A

P(all three satisfactory) = 0.9*0.9*0.9 = 0.729.

Practice Question 20

Two events, A and B, are independent if ______.

A. P(A and B) = P(A) P(B)
B. P(A and B) = P(A) + P(B)
C. P(A and B) = 1

Correct Answer: A

This is the definition of independence.

Practice Question 21

Each salesperson in a large department store chain is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated either fair, good, or excellent with respect to potential for advancement. These traits for 500 salespeople were cross classified into the following table.

Sales Ability Potential for Advancement

What is the probability that a sales person selected at random will have below-average sales ability and fair potential for advancement?

A. 0.16
B. 0.032
C. 0.10

Correct Answer: B

Prob. of selecting a below-average person: (16 + 12 + 22)/500 = 1/10

Prob. of selecting fair potential amongst those in the below-average group: 16/50 = 8/25

1/10*8/25 = 4/125

Practice Question 22

An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most had satisfactory weight.

Weight % of Total
Underweight 2.5
Satisfactory 90.0
Overweight 7.5

Three packages are selected from the food processing line. What is the probability of selecting and finding that all three of them are underweight?

A. 0.0000156
B. 0.075
C. 0.0000001

Correct Answer: A

P(all three underweight) = 0.025*0.025*0.025 = 0.0000156

Practice Question 23

A fundamental analyst studying 100 potential companies for inclusion in her stock portfolio uses the following three screening criteria:

Screening Criterion | Number of Companies meeting the screen
Market-to-Book Ratio > 4 | 20
Current Ratio >2 | 40
Return on Equity >10% | 25

Assuming that the screening criteria are independent, the probability (in %) that a given company will meet all three screening criteria is closest to ______.

A. 2.0
B. 6.5
C. 8.5

Correct Answer: A

The joint probability of the three independent criteria is calculated as: 0.2 x 0.4 x 0.25 = 0.02 or 2% of the 100 companies.