# Statistics Questions I need help with these 4 statistic questions

Thanks MTH 245 Lesson 10 Notes

Theoretical Probability

The theoretical prob

Statistics Questions I need help with these 4 statistic questions

Thanks MTH 245 Lesson 10 Notes

Theoretical Probability

The theoretical probability approach is used when consists of a reasonably

small number of countable outcomes, each of which is equally likely to

occur.

There are three steps to calculating a theoretical probability:

1. Define the experiment, the sample space , and the event space .

2. For both and , count the number of outcomes in each.

3. Calculate the probability:

( ) =

number of outcomes in

number of outcomes in

Warning! This approach assumes that each outcome in is equally likely. If that

isn’t the case, then the above formula does not produce the correct results.

Example 1: To continue Example 1 in Lesson 9, suppose we roll a single six-

sided die once. What is the probability of rolling greater than a 4?

We have already defined the sample space as = {1, 2, 3, 4, 5, 6} and the

event space as = {5, 6}. It follows that

( ) =

= 2

6

= 1

3

= 0.333 .

Example 2: Consider an experiment where a fair coin is flipped three times

in a row and the result of each set of flips is recorded as a single sequence.

What is the probability that the next sequence of flips will contain two or

more “tails?”

There are eight possible sequences of flips: = { , , , ,

, , , }. Of these, four contain two or more “tails”:

= { , , , }. Therefore, ( ) = 4

8

= 1

2

= 0.500.

The Addition Rule

A compound event is an event that combines two or more simple events.

The compound event ∪ (read “A

union B” or “A or B”) is the outcome

where, in a single trial, either

occurs or occurs or they both occur.

If A and B can both occur, then their

intersection ∩ (read “A intersect

B” or “A and B”) is not empty and

contains outcomes that are in both A

and B.

To calculate the probability of ∪ occurring, count the number of ways

can occur and the number of ways can occur, and add those totals

together in such a way that no individual outcome is counted twice. Then

divide that sum by the total number of possible outcomes. Formally, this

process is represented by the Addition Rule:

For any two events A and B,

( ∪ ) = ( ) + ( )− ( ∩ )

Subtracting the term P( ∩ ) ensures we don’t double count the outcomes

that appear in both A and B when calculating P( ∪ ).

Example 4: Continuing Example 1 of this section, what is the probability

rolling greater than a 4 or an even number?

We already know that = {1, 2, 3, 4, 5, 6} and = {5, 6}. In addition,

define = { } = {2, 4, 6}. Then ∩ = {6} and

( ∪ ) = ( ) + ( ) − ( ∩ ) = 2

6

+ 3

6

− 1

6

= 4

6

= 2

3

= 0.333 .

Example 4: A sample of 200 tennis rackets – 100 graphite and 100 wood – is

taken from the warehouse. Suppose 6 of the wood rackets and 9 of the

graphite rackets are defective. If one racket is randomly selected from the

sample of 200, find the probability that the racket is either wood or

defective.

Define = {100 }, = {15 }, and

∩ = {6 ℎ ℎ }. Then

( ∪ ) = ( ) + ( )− ( ∩ ) = 100

200

+ 15

200

− 6

200

= 109

200

= 0.545.

Example 5: Refer to the table below. If 1 subject is selected at random from

among 1,000 subjects given a drug test, what is the probability that the

subject had a positive drug test, actually uses drugs, or both?

Positive

Test Result

Negative

Test Result

Total

Subject uses drugs 44 6 50

Subject doesn’t

use drugs

90 860 950

Total 134 866 1,000

Define = {134 ℎ }, = {50 ℎ }, and

∩ = {44 ℎ }. Then

( ∪ ) = ( ) + ( )− ( ∩ ) = 134

1,000

+ 50

1,000

− 44

1,000

= 140

1,000

= 0.140

Disjoint Events

The events and are said to be disjoint (or

mutually exclusive) if they have no outcomes in

common and therefore cannot occur

simultaneously. In other words, ∩ = ∅.

If A and B are disjoint, then ( ∩ ) = 0, and

( ∪ ) = ( ) + ( )

Example 6: Suppose we draw a single card from a standard 52-card poker

deck. What is the probability that the card is either a face card or a “5”?

(Refer to the “Resources” page in the “Start Here” module for a link with

more information on card decks.)

Define = {12 }, = {4 ” ” }. A card can’t be both

A face card and a “five,” so ∩ = ∅. Then

( ∪ ) = ( ) + ( ) = 12

52

+ 4

52

= 16

52

= 4

13

= 0.308

Example 7: In a certain lottery, winning numbers are determined by

selecting from a set of 10 plastic balls numbered 0 through 9. If a single ball

is selected at random, what is the probability of selecting either an even-

numbered ball or the ball marked “3”? (Note: assume 0 is even.)

Define = {0, 2, 4, 6, 8}, = {3}. A ball can’t display a number that is

both even and a 3, so ∩ = ∅. Then

( ∪ ) = ( ) + ( ) = 5

10

+ 1

10

= 6

10

= 3

5

= 0.600

Complementary Events

The complement of event A is the subset of all outcomes in that are not in

. There is no standard notation for the complement of a subset; we will

use to stay consistent with the (optional) textbook.

For any properly constructed experiment, an

individual outcome cannot be in both and

. In other words, and are disjoint, and

since they contain all the outcomes in

between them, it follows that ∪ = , and

further that

( ) + ( ) = 1

This equation can also be written in one of the following two equivalent

forms:

( ) = 1 − ( )

( ) = 1 − ( )

Example 8: Suppose we draw a single card from a standard 52-card poker

deck (assume a “2” is the lowest card and an ace is the highest). What is the

probability of drawing a “5” or higher?

If we define = {” ” ℎ ℎ }, then = {2, 3, 4 ℎ }

(note that has 12 total elements—three cards × four suits). Then

( ) = 1 − ( ) = 1 − 12

52

= 40

52

= 10

13

= 0.769.