# Statistics Questions I need help with these 4 statistic questions Thanks MTH 245 Lesson 10 Notes Theoretical Probability The theoretical prob

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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.

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
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

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.