Download - Probability, Part 1
Probability, Part 1
We’ve looked at several counting techniques to determine the number
of possible arrangements or groupings in a given scenario. These techniques include the multiplication
principle, the addition principle, permutations, and combinations.
Probability, Part 1
Together these techniques provide us with a powerful set of tools for
calculating probabilities.
Probability, Part 1
Probability tells us the likelihood of a given event occurring.
We normally represent it as a decimal or fraction between 0 and 1, or as a percentage between 0% and 100%.
Probability, Part 1
For example, a weather reporter might say there is a 70% chance, or
probability, of rain tomorrow.
We could also say
P(rain) = 0.7
Probability, Part 1
A probability of 1, or 100%, indicates certainty that a given even will occur.
A probability of 0, or 0%, indicates certainty that a given even will not
occur.
Probability, Part 1
Example 1: Say you roll a die. What is the probability that you will get a
number between 1 and 6? What is the probability you will get a 7?
P(1-6) = 1, because we are guaranteed to get one of the numbers
from 1 to 6.
Probability, Part 1
Example 1: Say you roll a die. What is the probability that you will get a
number between 1 and 6? What is the probability you will get a 7?
P(7) = 0, because it is impossible to get a 7 with only one die.
Probability, Part 1
This example was rather simple, but the problems can become much more complex. We
need to have a more systematic approach to handle those complex problems.
In general, we will want to determine the number of possible outcomes that fit a set
of conditions, and compare that to the total possible number of outcomes.
Probability, Part 1
In general, the probability of a given event occurring is equal to the ratio:
__Number of Outcomes That Work__Total Possible Number of Outcomes
Probability, Part 1
Example 2: If you roll a die, what is the probability of getting an even number?
____Number of Even Outcomes____Total Possible Number of Outcomes
Probability, Part 1
Example 2: If you roll a die, what is the probability of getting an even number?
Possible Outcomes: 1, 2, 3, 4, 5, 6
So there are 6 total possible outcomes.
Probability, Part 1
Example 2: If you roll a die, what is the probability of getting an even number?
Even Outcomes: 2, 4, 6
There are 3 even outcomes.
Probability, Part 1
Example 2: If you roll a die, what is the probability of getting an even number?
3 16 2
P(even) = =
Probability, Part 1
Again, this example was rather simple. Let's look at a couple other examples that
require more thought.
Probability, Part 1
Example 3: You are taking a 5-question matching quiz; you are asked to match 5
vocabulary terms with 5 definitions. Alas, you are ill-prepared for the quiz, and end up totally guessing on the whole thing. What is the probability that by guessing
you will score 100%?
Probability, Part 1
Example 3: We can view this as a permutation problem to determine the
total possible number of ways to answer the quiz. In a sense, you are being asked
to arrange 5 definitions in order. The number of ways to do this is
P(5, 5) = 5 nPr 5 = 120.
Probability, Part 1
Example 3: The outcomes that “work” in this case are the outcomes that will give
you a score of 100%. There is only 1 such outcome.
Hence, the probability of scoring 100% is
1 = 0.0083, or about 0.83%.
120
Probability, Part 1
Example 4: You are taking a 10 question True/False quiz. You are so unsure of the questions that you decided to flip a coin to
help you answer each one. (If you get heads, you answer true; tails you answer false.) In other words, you're answering
completely at random. What is the probability you will score exactly 70%?
Probability, Part 1
Example 4: 10 question True/False Quiz
First, let's figure out the total number of ways to answer the quiz. The first question has two options: true or false. So does the second question, and the third, ... So the total possible number of outcomes is
2*2*2*2*2*2*2*2*2*2 = 210 = 1,024.
Probability, Part 1
Example 4: 10 question True/False Quiz
Now let's figure out the number of outcomes that “work,” or in this case that give you a score of 70%. Assuming there is no partial credit, this means getting 7 out of the 10 questions correct. It doesn't matter which 7 questions you get correct, though—it could be 1-7, 4-10, 1-5 and 8-9, etc.
Probability, Part 1
Example 4: 10 question True/False Quiz
In other words, we need the number of combinations of 7 correct answers out of 10 problems. (Why combinations rather than permutations?) The number of ways to score 70% is
C(10, 7) = 10 nCr 7 = 120.
Probability, Part 1
Example 4: 10 question True/False Quiz
The probability of getting 70% is then
__120__
1,024 = 0.117, or about 11.7%.
Probability, Part 1
In the previous example we used the multiplication principle together with
combinations. In some instances we will use the addition principle in concert with
combinations or permutations.
(Recall that a key word for the addition principle is frequently “or.”)
Probability, Part 1
Example 5: You are again completely guessing on a 10 question True/False quiz. What is the probability that you will score
at least 70%?
Probability, Part 1
Example 5: 10 question True/False Quiz
Again assume that no partial credit is given. In this case, the outcomes that
“work” are scores of 70% or 80% or 90% or 100%. We need to determine these
possibilities and then add them together.
Probability, Part 1
Example 5: 10 question True/False Quiz
We already calculated the number of ways to get 7 out of 10 questions correct:
C(10, 7) = 10 nCr 7 = 120.
Probability, Part 1
Example 5: 10 question True/False Quiz
Consider also the number of ways to get 8, 9, or 10 correct:
C(10, 8) = 10 nCr 8 = 45.
C(10, 9) = 10 nCr 9 = 10.
C(10, 10) = 10 nCr 10 = 1.
Probability, Part 1
Example 5: 10 question True/False Quiz
Now add together the number of ways to get 7, 8, 9, or 10 questions correct.
120 + 45 + 10 + 1 = 176
This gives us a total of 176 outcomes that work.
Probability, Part 1
Example 5: 10 question True/False Quiz
The total possible number of ways to answer the quiz remains 1,024, as we
already calculated in Example 4.
Probability, Part 1
Example 5: 10 question True/False Quiz
This gives us a probability of scoring at least 70% on the quiz as:
176__
1,024 = 0.172, or about 17.2%.
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
When considering 2 events, A and B, we can find the probability of one or the other occurring
by simply adding together their individual probabilities, with one adjustment.
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
When considering 2 events, A and B, we can find the probability of one or the other occurring
by simply adding together their individual probabilities, with one adjustment.
We need to subtract the overlap between the 2 events.
P(A or B) = P(A) + P(B) – P(A and B)
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
In a case where A and B are mutually exclusive, both cannot occur at the same time. As a result, P(A and B) = 0, and our equation simplifies to
P(A or B) = P(A) + P(B)
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Confused yet?
Let's look at some examples to help illustrate.
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 6: A car is being given away at a high school graduation. What is the probability that a football player or band member will win the car, if among the 400 graduates there are 40 football players and 50 band members, including 5 that participated in both football and band?
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Out of 40 football players, 5 were also in band, so there were 35 who played football but were not in band.
Out of 50 band members, 5 also played football, so there were 45 who were in band but did not play football.
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Football Band
35 455
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Football Band
35 455
P(Football or Band) = P(Football) + P(Band) – P(Football and Band)
= 40_ + 50_ - 5 = 85 = 17 400 400 400 400 80
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 7: At the same high school graduation, what is the probability that a football player or cross country runner will win the car, if among the 400 graduates there are 40 football players and 50 cross country runners, and no one participated in both?
Probability, Part 1The Addition Principle for Mutually Exclusive Events
Example 7: Car Giveaway
Football
40
CC
50
P(Football or CC) = P(Football) + P(CC)
= 40_ + 50_ = 90 = 9 400 400 400 40
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
We can use the multiplication principle to help determine the probability of two events both
occurring (and rather than or):
P(A and B) = P(A) * P(B from A)
Once again, this may be confusing without an example.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: A certain high school has the following student population:
What is the probability that a student selected at random will be a male junior?
Female Male Total
Fr. 123 105 228
Soph. 168 132 300
Jr. 145 110 255
Sr. 124 93 217
Total 560 440 1000
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
Consider being a junior as Event A, and being male as Event B. We’re trying to determine
P(A and B).
We could do this by constructing a tree diagram.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
Fr
Sr
Jr
So
Label the probabilities for each grade level. These come from the number of students in each grade divided by the total number of students.
E.g., P(So) = 300/1000 = .3
.3
.228
.255
.217
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
Fr
Sr
Jr
So.3
.228
.255
.217
M
M
M
M
F
F
F
F
.461
.539
.560
.440
.431
.569
.423
.577
Label the probabilities of being male or female within each grade level.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
Fr
Sr
Jr
So.3
.228
.255
.217
M
M
M
M
F
F
F
F
.461
.539
.560
.440
.431
.569
.423
.577
Follow the path for Juniors and then Males, and multiply the probabilities.
.110
.255*.431 = .110
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
We can get the same result by using the formula:
P(A and B) = P(A) * P(B from A)
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
P(A and B) = P(A) * P(B from A)
P(Junior and Male) = P(Junior) * P(Male from Juniors)
P(Junior) = 255/1000 = .255
P(Male from Junior) = 110/255 = .431
P(Junior and Male) = .255 * .431 = .110
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Our formula for conditional probabilities becomes a small bit simpler when the events are independent. In this case:
P(A and B) = P(A) * P(B)
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Two events are independent if
P(B from A) = P(B)
or
P(A from B) = P(A)
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event A is being a Junior, and Event B is being a male.
Does P(B from A) = P(B)?
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event A is being a Junior, and Event B is being a male.
P(B) = P(Male) = 440/1000 = .440
P(B from A) = P(Male from Juniors) = 110/255 = .431
So P(B from A) P(B)
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event A is being a Junior, and Event B is being a male.
P(A) = P(Junior) = 255/1000 = .255
P(A from B) = P(Junior from Males) = 110/440 = .250
So P(A from B) P(A)
Being a junior and being male are therefore not independent.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female independent in the school of Example 8?
Female Male Total
Fr. 123 105 228
Soph. 168 132 300
Jr. 145 110 255
Sr. 124 93 217
Total 560 440 1000
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female independent in the school of Example 8?
P(B) = P(Female) = 560/1000 = .560
P(B from A) = P(Female from Sophomores) = 168/300 = .560
Because P(B from A) = P(B), being a sophomore and being female are independent in this school.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female independent in the school of Example 8?
In other words, the probability of picking a female from among the sophomores is the same as the probability of picking a female from the school population at large.
In contrast, the probability of picking a male from among the juniors is slightly less than the probability of picking a male from the school population at large.
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 10: What is the probability that a student chosen at random from the entire school population will be a female sophomore?
Probability, Part 1Conditional Probability:Using the Multiplication Principle
Example 10: What is the probability that a student chosen at random from the entire school population will be a female sophomore?
We already established that being female and being a sophomore are independent events in this case, so we can use our simplified formula:
P(Female and Sophomore) = P(F) * P(Soph)
= .560 * .300 = .168This makes sense as there are 168 female sophomores out of 1,000 total students in the school.