We begin by defining some terms.
Random Experiment: any activity with a random (unpredictable) result that can be measured.
Trial: one repetition of a random experiment.
Outcome: a result of a random experiment.
Event: a group of outcomes that are somehow related.
Universe: the set of all outcomes for an experiment. Also called the Sample Space.
Examples:
[1.] If the experiment is rolling a die, then the Universe is the numbers 1 through 6. An event might be "Even," which contains the outcomes 2, 4 and 6. Another event could be "Prime," which would contain the outcomes 2, 3, and 5.
[2.] If the experiment is asking a person how many years of school that they have completed, then the Universe contains numbers that are greater than zero. An event might be "Completed College," which would probably include the numbers from 14 through 18. Another event might be "Dropped out of High School," which would contain the numbers 8 through 11.
It is important to write out the Universe so that each outcome is equally likely—the only general advice concerning how to do this is to be as specific as possible in listing the outcomes.
Example:
[3.] If we roll two dice, and list the possible outcomes, we get 36 of them—illustrated in the diagram below. The outcome "two fives" is circled.
With these, we may now define the Probability of an Event: . P(E) means "the probability of event E,", |E| means "the number of outcomes in event E," and |U| means "the number of outcomes in the universe."
Examples:
[4.] A bag contains 3 red, 4 blue and 6 white marbles. If you reach in and pull out one marble, what is the probability that the marble is red?
There are 13 outcomes (13 marbles), and the event "red marble" has 3 outcomes (3 marbles). The probability is .
[5.] In a certain study hall, there are 8 Freshmen, 12 Sophomores, 18 Juniors and 22 Seniors. A student in the room is randomly selected and expelled (isn't school fun!). What is the probability that the student is a Senior?
There are 8 + 12 + 18 + 22 = 60 outcomes, and the event "a Senior is selected" contains 22 outcomes. The probability is .
[6.] A Roulette wheel has 18 red, 18 black, and two green spaces. The wheel is spun, and small ball is spun around the rim of the wheel—the ball is equally likely to land in any of the 38 spaces. What is the probability that the ball lands on red?
There are 38 outcomes in the universe, and 18 outcomes in the event ("Lands on red"). The probability is .
Finally, the Complement of an Event: the set of outcomes that do not belong to the event. Think of this as the opposite of the event. The key word to look for is not.
There is the connection between the probability of an event, and the probability of the event's complement: P(not A) = 1 - P(A).
Examples:
[7.] When two dice are rolled, what is the probability that you do not roll doubles?
P(not doubles) = 1 - P(doubles); P(doubles) = (seen easily in the chart earlier). Thus, P(not doubles) = 1 - = .
[8.] When a roulette wheel is spun, what is the probability that the ball does not land on black?
P(not black) = 1 - P(black). P(black) is the same as P(red), since there are an equal number of red and black spaces; P(black) = . So P(not black) = 1 - = .
The idea of independence is important in probability & statistics. Here's a definition.
Two events are independent if knowing whether or not one event has occurred does not change the probability that another event will occur.
Knowing whether or not one event has occurred is a powerful thing. Consider the following examples:
What is the probability that a randomly selected person is female? Would your answer change if I told you that the person just walked out of the Women's Restroom?
What is the probability that a randomly selected book is a mathematics textbook? Would your answer change if I told you that the book came out of Mr. Holloman's room?
What is the probability that I will need an umbrella today? Would your answer change if I told you that it's raining outside?
If you've got two events—call them A and B—knowing that A has definitely occurred (or not occurred) might change the probability that B will occur. When this happens, the events are not independent. If the probability of the future event doesn't change, then the events are independent. Consider a few more examples:
What is the probability that a randomly selected person is female? Would your answer change if I told you that the person has red hair?
What is the probability that a randomly selected book is a mathematics textbook? Would your answer change if I told you that the book was 5cm thick?
What is the probability that I will need an umbrella today? Would your answer change if I told you that I'm using a DELL computer?
The idea of independence is rooted in the idea of conditional probability. Here's the notation: The probability that B will occur, if we know that A has occurred, is P(B|A). With this notation, we can now rewrite the definition of independence: events A and B are independent if P(A|B) = P(A).
Sometimes you'll be told that two events are independent—sometimes, you won't. If you're not told, you'll have to think about whether or not the two things could possibly be related.
Sometimes we are interested in the probability that two events occur simultaneously—the probability that A and B both occur. The fancy word for this is intersection. The key word to look for is and.
P(A and B) = P(A ∩ B) = P(A)·P(B|A) = P(B)·P(A|B)
Note that if A and B are independent, then this reduces to P(A ∩ B) = P(A)·P(B).
We sometimes illustrate the intersection of two events with a picture called a Venn Diagram. In these diagrams, outcomes are represented by points in a plane. The plane is typically drawn as a rectangle, and labeled either U or S. Events in the universe are drawn as geometric shapes (circles, primarily). The intersection of events A and B is illustrated below.
Examples:
[9.] What is the probability that when two coins are tossed, both land heads up?
Well, since we're asked about both landing heads up, that means the first is heads AND the second is heads—this is an intersection problem. So: are the events independent? If you know how the first coin landed, will that change the probability that the second coin lands heads? No! Since they are independent, we can simply multiply the probabilities to get the intersection. ½·½ = ¼.
[10.] A bag contains 3 red, 4 blue and 6 white marbles. A marble is taken out, its color noted, and replaced in the bag. Then a second marble is taken out and its color noted. What is the probability that both marbles are blue?
Both marbles are blue means blue AND blue—an intersection. Are the events independent? Yes—knowing how the first marble came out doesn't change the probability for the second one. So we can multiply! .
[11.] A bag contains 3 red, 4 blue and 6 white marbles. A marble is taken out, its color noted, and then discarded. Another marble is withdrawn and its color noted. What is the probability that both marbles are blue?
Still an intersection problem, but are the events independent? If you know that the first marble was blue, does that change the probability that the second one is blue? Yes! These events are independent. So P(blue AND blue) = P(first blue)·P(second blue, given that the first was blue).
P(first blue) = . For P(second blue, given that the first was blue), we know that the first marble was blue—thus, there are now only 12 marbles, 3 of which are blue. So P(second blue, given that the first was blue) = = ¼. Finally, the answer is .
What if we want the probability for one event or the other? This is the probability of either A or B. Now, this isn't like the word "or" in English—it doesn't mean "one or the other (but not both)." The math word for that is XOR (exclusive or). When we say "or" in math, we mean "one event, or the other, or both."
This type of combination of events is called a union, and the key word to look for is (you guessed it!) or.
P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Note the subtraction there—when you add P(A) and P(B), you might be adding some outcomes twice (perhaps there are outcomes that belong to both A and B!); thus, you must subtract the intersection. If there is no intersection, then the events are called mutually exclusive, or disjoint.
Here's a Venn Diagram of the union:
Examples:
[12.] In a group of 500 college students, 275 are female, 30 are statistics majors, and 4 are female and statistics majors. A member of this group is randomly selected. What is the probability that the person selected is either a female or a statistics major?
"female or a statistics major" means that this is a union problem. P(female) = , P(stats major) = , and P(female and stats major) = 0.008. So P(female or stats major) = 0.55 + 0.06 - 0.008 = 0.602.
[13.] Roll two dice and look at the numbers on top. What is the probability that a 2 is showing on at least one of the dice?
If a two is showing, then it's showing on the first die, or the second die—this is a union problem. For any single die, P(2) = . The probability of getting a 2 on both dice is P(2 and 2) = . So P(a 2 is showing) = P(2 or 2) = P(2) + P(2) - P(2 and 2) = .
Note the summary on page 187!
Knowing that probability involves counting outcomes, techniques that allow us to count outcomes quickly are a good idea.
Tree diagrams are useful when the experiment can be broken down into a series of steps, and we know the number of outcomes that can result from each step.
I can't explain (in writing) how to make a tree diagram. Here are some examples.
Examples:
[14.] A box contains three coins—a penny, a nickel, and a quarter. A coin is randomly drawn, its value noted, and then set aside. Another coin is now randomly drawn, and its value noted. In how many different ways can this be done?
So there are six ways to do this—the probability of any particular outcome is .
[15.] A bag contains 2 red, 3 blue and 4 white marbles. A marble is taken out, its color noted, and then discarded. Another marble is withdrawn and its color noted. What is the probability that both marbles are blue?
If we do this the regular way, we'd need LOTS of arrows!
And that's just the first step! We could make this better if we write in counts (how many branches), rather than making so many lines.
Multiply along the lines to see how many branches there ought to be.
So there are 6 ways to draw two blue…add the number in parentheses to find the total number of ways to complete the experiment (72). So the probability is .
Well, if you can just count and multiply branches, why draw the branches at all? Why not just multiply the numbers?
The Multiplication Rule: if the experiment can be broken down into two smaller events, and the first event can occur in m different ways, and the second event can occur in n different ways, then the experiment can occur in m × n ways.
This can be extended for three, four…any number of events.
Examples:
[16.] I have four different colors of pants, three different colors of shirt, and five different types of ties. If I don't care about whether they match or not (and some accuse me of this!), and I make my selections at random, in how many different ways can I get dressed?
So—the experiment (getting dressed) can be broken down into pants, shirt, and tie. These smaller events can happen in four, three, and five different ways (respectively). The multiplication rule tells me that I can get dressed in 4 × 3 × 5 = 60 different ways.
[17.] A phone number uses seven digits. How many phone numbers can be made if there are no restrictions on which digits can be used? How many phone numbers are possible if the first digit cannot be a 9 or a 0?
With no restrictions, there are 10 possible ways to select each digit. Thus, 10×10×10×10×10×10×10 = 107 = 10,000,000.
If the first digit cannot be 9 or 0, then we get 8×10×10×10×10×10×10 = 8×106 = 8,000,000.
[18.] When playing the Carolina Four lottery game, players must select a four digit number. How many number are possible if there are no restrictions on the digits? How many numbers are possible if all four digits must be different?
With no restrictions, we're back to the telephone problem—10×10×10×10 = 104 = 10,000.
If each number can only be used once, then we get something different. 10×9×8×7 = 5,040.
Factorials are useful for determining how many ways that a number of things can be arranged. For example, if I want to line up the three students in my class and march them down to the office, I can pick any one of the three to start the line; then either of the remaining two for the middle. There is only one choice for the remaining student. So there are 3×2×1 = 6 ways to arrange them in a line. This product (when it starts at a number and reduced by one each time) is called a factorial. It is denoted with the ! symbol. Here's the definition:
n! = n×(n - 1)×(n - 2)×…×(2)×(1).
Examples:
[19.] In how many ways can I arrange the 14 statistics textbooks on my bookshelf?
14! = 14×13×12×…2×1= 87,178,291,200.
[20.] In how many ways can I arrange the letters IRMO?
4! = 4×3×2×1 = 24.
Extra! If some of the items to be arranged are identical, then you need to divide by the number of repetitions (to take care of arrangements that are really the same). Some examples might make things more clear…
Examples:
[21.] In how many ways can the letters in the word SCHOOL be arranged?
Normally, we'd just use the length of the word: 6! However, one letter shows up more than once—O. Since it shows up 2 times, we use a 2! in the denominator. The answer is .
[22.] In how many ways can we arrange the letters in the word MASSACHUSETTS?
The length is 13, but some letters are duplicated: A 2 times, S 4 times, and T twice. The answer is .
Permutations are like factorials—they count the number of ways to arrange things—but they differ in that you're not arranging everything; you pick some of the items and arrange them (but leave the rest alone). In the calculation, this means that you don't keep lowering the numbers until you get to 1; you stop at some point. For example: I've got 20 students in the Math Club, and I've decided to ignore the voting and simply select three officers (President, Vice-President, and Secretary) myself. Putting students into different offices like this is a way of arranging them; but I'm not using all of the students. Thus, this is a permutation of three out of 20 students—there will be only three terms when applying the multiplication rule: 20×19×18 = 6,840. Notice that this is the same as —which leads me to the formula.
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Examples:
[23.] Having won a large sum in the lottery, I now want to give it to a few of my closest friends. I've narrowed the field to 10 lucky friends. I want to pick four people—the first will get $1000; the second, $500; the third, $250, and the fourth will get $125. In how many ways can I do this?
This is a permutation of 4 out of 10: .
[24.] 100 seniors have applied for 5 parking spaces that have become available because of some unfortunate expulsions. The spaces are all at different distances from the building (so some are more desirable than others). The administration decides to hold a raffle—the first ticket drawn gets first pick of the spaces; the second ticket drawn gets a pick of those that remain; etc. In how many ways can the spaces be assigned?
This is a permutation of 5 out of 100: .
Combinations are like permutations, except that there is no arranging involved—you're just selecting some of the possible items (but probably not all of them). The formula is very similar: .
For example—there are ten different pieces of candy in a bag, and I want to east three of them. In how many different ways can I select three of the ten? Since I'm only selecting (not selecting and arranging), this is a combination problem. The answer is .
Examples:
[25.] The first five numbers on a Powerball ticket must be different numbers between 1 and 53. In how many different ways can I pick the first five numbers on a Powerball ticket?
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[26.] There are 20 students in a classroom, and I want to select 3 of them to receive an extra credit opportunity. In how many ways can I select the three students?
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Page last validated 2010-08-15