18 3.1 The Terminology of Probability
Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment.
Example of an experiment: Flipping one fair coin twice.
A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space.
Example: if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes.
An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head.
Example: The probability of an event A is probability of getting at most one head. It is also written as P(A).
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, 0 [latex]\leq[/latex] probability of an event [latex]\leq[/latex] 1).
- P(A) = 0 means the event A can never happen.
- P(A) = 1 means the event A always happens.
- P(A) = 0.5 means the event A is equally likely to occur or not to occur.
Example: If you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability.
Example:
- If you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face.
- If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur.
- If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space.
For examples:
- If you toss a fair dime and a fair nickel, you will see four possible outcomes. These 4 outcomes will form a sample space. Therefore, the sample space is {HH, TH, HT, TT} where T = tails and H = heads.
If event A = getting one head, then there are two outcomes that meet this condition {HT, TH}.
The probability of event A, P(A) = [latex]\frac{\text{number of outcome with only one head}}{total of possible outcomes}[/latex] = [latex]\frac{2}{4}[/latex] = 0.5.
- Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five.There are two outcomes {5, 6}.
P(E) = [latex]\frac{\text{number of outcome that rolling a number that is at least five}}{total of possible outcomes}[/latex] = [latex]\frac{2}{6}[/latex] as the number of repetitions grows larger and larger.
This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)
This video gives more examples of basic probabilities.
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased.
Examples:
- Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias.
- Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.
“OR” Event
An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B.
For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}.
A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. (Notice that 4 and 5 are NOT listed twice.)
“AND” Event
An outcome is in the event A AND B if the outcome is in both A and B at the same time.
For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, respectively. Then
A AND B = {4, 5}.
Complimentary Event
The complement of event A is denoted A′ (read “A prime”). A′ consists of all outcomes that are NOT in A.
P(A) + P(A′) = 1.
For example:
We have sample space S = {1, 2, 3, 4, 5, 6}.
If event A = {1, 2, 3, 4}. Then, event A’={5, 6}.
P(A) = [latex]\frac{{4}}{{6}}[/latex] and P(A’) = [latex]\frac{{2}}{{6}}[/latex]
P(A) + P(A’) = [latex]\frac{{4}}{{6}}+\frac{{2}}{{6}}={1}[/latex]
Conditional Probability of an Event
The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is
[latex]\displaystyle{P}{({A}{|}{B})}=\frac{{{P}{({A}\text{ AND } {B})}}}{{{P}{({B})}}}[/latex] where P(B) is greater than zero.
For example:
Suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}.
Let event A = face is 2 or 3 and B = event that face is even.
Event A = {2, 3}, Event B ={2, 4, 6}.
To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S).
We get the same result by using the formula. Remember that S has six outcomes.
A and B = {2} (as 2 appears in both event A and event B.)
[latex]\displaystyle{P}{({A}{|}{B})}=\frac{{{P}{({A}\text{ AND } {B})}}}{{{P}{({B})}}}=\frac{{\frac{{\text{the number of outcomes that are in both event A and event B}}}{{total outcomes}}}}{{\frac{{\text{the number of outcomes in event B}}}{{total outcomes}}}}=\frac{{\frac{{1}}{{6}}}}{{\frac{{3}}{{6}}}}=\frac{{1}}{{3}}[/latex]
Understanding Terminology and Symbols
It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.
Example 1
The sample space S is the whole numbers starting at one and less than 20.
- S = _____________________________Let event A = the even numbers and event B = numbers greater than 13.
Show Answer
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
- A = _____________________, B = _____________________
Show Answer
A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19}
- P(A) = _____________, P(B) = ________________
Show Answer
[latex]\displaystyle{P}{({A})}=\frac{{9}}{{19}},{P}{({B})}=\frac{{6}}{{19}}[/latex]
- A AND B = ____________________, A OR B = ________________
Show Answer
A AND B = {14,16,18}, A OR B = 2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}
- P(A AND B) = _________, P(A OR B) = _____________
Show Answer
[latex]\displaystyle{P}{({A}\text{ AND } {B})}=\frac{{3}}{{19}},{P}{({A}\text{ OR } {B})}=\frac{{12}}{{19}}[/latex]
- A′ = _____________, P(A′) = _____________
Show Answer
[latex]\displaystyle{A'}={1},{3},{5},{7},{9},{11},{13},{15},{17},{19};{P}{({A}′right)}=\frac{{10}}{{19}}[/latex]
- P(A) + P(A′) = ____________
Show Answer
P(A)+P(A’)=[latex]\displaystyle{1}{\left(\frac{{9}}{{19}}+\frac{{10}}{{19}}\right)={1}}[/latex]
- P(A|B) = ___________, P(B|A) = _____________; Are the probabilities equal?
Show Answer
P(A|B) = [latex]\displaystyle\frac{{{P}{({A}\text{ AND } {B})}}}{{{P}{({B})}}}=\frac{{3}}{{6}}[/latex],P(B|A)=[latex]\frac{{{P}{({A}\text{ AND } {B})}}}{{{P}{({A})}}}=\frac{{3}}{{9}}[/latex], No
Try It
The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).
- S = _____________________________Let event
A = the sum is even and event B = the first number is prime. - A = _____________________, B = _____________________
- P(A) = _____________, P(B) = ________________
- A AND B = ____________________, A OR B = ________________
- P(A AND B) = _________, P(A OR B) = _____________
- B′ = _____________, P(B′) = _____________
- P(A) + P(A′) = ____________
- P(A|B) = ___________, P(B|A) = _____________; are the probabilities equal?
Show Answer
- S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
- A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)}
B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} - [latex]\displaystyle{P}{({A})}=\frac{{1}}{{2}},{P}{({B})}=\frac{{2}}{{3}}[/latex]
- A AND B = {(2,2), (2,4), (3,1), (3,3)}
A OR B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} - P(A and B)=[latex]\displaystyle\frac{{1}}{{3}}[/latex], {P(A or B)=[latex]\frac{{5}}{{6}}[/latex]
- B’={(1,1),(1,2),(1,3),(1,4)}, P(B’)=[latex]\frac{{1}}{{3}}[/latex]
- P(B) + P(B′) = 1
- P(A|B)=[latex]\displaystyle=\frac{{2}}{{3}}[/latex], No.
Example 2
A fair, six-sided die is rolled. Describe the sample space
S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).
- Event T = the outcome is two.
Show Answer
T = {2}, P(T) =[latex]\frac{{1}}{{6}}[/latex]
- Event A = the outcome is an even number.
Show Answer
A = {2, 4, 6} , P(A)=[latex]\frac{{1}}{{2}}[/latex]
- Event B = the outcome is less than four.
Show Answer
B = {1, 2, 3},P(B)=[latex]\frac{{1}}{{2}}[/latex]
- The complement of A.
Show Answer
A’ = {1, 3, 5}, P(A’)=[latex]\frac{{1}}{{2}}[/latex]
- A GIVEN B
Show Answer
(A | B)={2},P(A | B) =[latex]\frac{{1}}{{3}}[/latex]
- B GIVEN A
Show Answer
(B | A) ={2},P(B|A)=[latex]\frac{{1}}{{3}}[/latex]
- A AND B
Show Answer
(A and B)={2}, P(A and B)=[latex]\frac{{1}}{{6}}[/latex]
- A OR B
Show Answer
(A orB)={1, 2, 3, 4, 6},P(A or B)=[latex]\frac{{5}}{{6}}[/latex]
- A OR B′
Show Answer
(A or B’)={2, 4, 5, 6},P(A or B’)=[latex]\frac{{2}}{{3}}[/latex]
- Event N = the outcome is a prime number.
Show Answer
N = {2, 3, 5}, P(N)=[latex]\frac{{1}}{{2}}[/latex]
- Event I = the outcome is seven.
Show Answer
A six-sided die does not have seven dots. P(7) = 0.
Example 3
Table describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed.
| Right-handed | Left-handed | Total | |
|---|---|---|---|
| Males | 43 | 9 | 43 + 9 = 52 |
| Females | 44 | 4 | 44 + 4 = 48 |
| Total | 43 + 44 = 87 | 9 + 4 = 13 | 100 |
Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities:
- P(M)
Show Answer
P(M) = [latex]\frac{\text{amount of male}}{\text{total}}[/latex] = [latex]\frac{52}{100}[/latex] = 0.52
- P(F)
Show Answer
P(F) = [latex]\frac{\text{amount of female}}{\text{total}}[/latex] = [latex]\frac{48}{100}[/latex] = 0.48
- P(R)
Show Answer
P(R) = [latex]\frac{\text{amount of right-handed}}{\text{total}}[/latex] = [latex]\frac{87}{100}[/latex] = 0.87
- P(L)
Show Answer
P(L) = [latex]\frac{\text{amount of left-handed}}{\text{total}}[/latex] = [latex]\frac{13}{100}[/latex] = 0.13
- P(M AND R)
Show Answer
P(M AND R) = [latex]\frac{\text{amount of right-handed male}}{\text{total}}[/latex] = [latex]\frac{43}{100}[/latex] = 0.43
- P(F AND L)
Show Answer
P(F AND L) = [latex]\frac{\text{amount of left-handed female}}{\text{total}}[/latex] = [latex]\frac{4}{100}[/latex] =0.04
- P(M OR F)
Show Answer
P(M OR F) = [latex]\frac{\text{amount of male + amount female}}{\text{total}}[/latex] = [latex]\frac{52 + 48}{100}[/latex] = 1
- P(M OR R)
Show Answer
P(M OR R) = [latex]\frac{\text{amount of male + amount of right-handed}}{\text{total}}[/latex] = [latex]\frac{43 + 9 + 44}{100}[/latex] = 0.96
- P(F OR L)
Show Answer
P(F OR L) = [latex]\frac{\text{amount of female + amount of left-handed}}{\text{total}}[/latex] = [latex]\frac{9 + 4 + 44}{100}[/latex] = 0.57
- P(M’)
Show Answer
P(M’) = 1 – P(M) = 1 – 0.52 = 0.48
- P(R|M)
Show Answer
P(R|M) = [latex]\frac{number of right-handed male}{number of male}[/latex] = [latex]\frac{43}{52}[/latex] = 0.8269 (rounded to four decimal places)
- P(F|L)
Show Answer
P(F|L) = [latex]\frac{number of left-handed female}{number of left-handed}[/latex] = [latex]\frac{4}{13}[/latex] 0.3077 (rounded to four decimal places)
- P(L|F)
Show Answer
P(L|F) = [latex]\frac{number of left-handed female}{number of female}[/latex] = [latex]\frac{4}{48}[/latex] = 0.0833
References
“Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).
Concept Review
In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.
Formula Review
A and B are events
P(S) = 1 where S is the sample space
0 ≤
P(A) ≤ 1
P(A|B)=[latex]\displaystyle\frac{{{P}{({A}\text{ AND } {B})}}}{{{P}{({B})}}}[/latex]