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

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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

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Possible Outcomes: 1, 2, 3, 4, 5, 6

So there are 6 total possible outcomes.

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Even Outcomes: 2, 4, 6

There are 3 even outcomes.

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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.

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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.

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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.

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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.”)

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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


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.

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We already calculated the number of ways to get 7 out of 10 questions correct:

C(10, 7) = 10 nCr 7 = 120.

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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.

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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.

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The total possible number of ways to answer the quiz remains 1,024, as we

already calculated in Example 4.

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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


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)

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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)

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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?


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.



Football Band

35 455



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


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?



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


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.



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



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.



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



We can get the same result by using the formula:





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


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)


Two events are independent if

P(B from A) = P(B)

or

P(A from B) = P(A)


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)?



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)



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.


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



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.



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.


Example 10: What is the probability that a student chosen at random from the entire school population will be a female sophomore?


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.

Probability, Part 1

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