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6.2 Introduction to Probability Summary Table, Study notes of Probability and Statistics

values that could occur, and the probability for each occurring. Probability Distributions. The probability distribution for the results of tossing two coins.

Typology: Study notes

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STAT1010 – Intro to probability part 3
1
1
6.2 Introduction to Probability
part 2
!From Tables to Probability
"Computing relative frequency probabilities from a
table.
!Probability Distributions
2
!The table below provides information on the
hair color and eye color of 592 statistics
students (The American Statistician, 1974).
Summary Table
Eye Color
Row
Brown
Blue
Hazel
Green
Tota l
Black
68 20 15 5
108
Hair
Brown
119 84 54 29
286
Color
Red
26 17 14 14
71
Blond
7 94 10 16
127
592$
3
!If a student is drawn at random from this
population
"What is the probability that they are blond?
!ANS: 127/592 = 0.2145
"What is the probability that they are NOT blond?
!ANS: P(NOT blond) = 1-P(blond) =1-127/592 = 0.7855
!ANS: (71 +286 + 108)/592 = 465/592 = 0.7855
Eye Color
Row
Brown
Blue
Hazel
Green
Tota l
Black
68 20 15 5
108
Hair
Brown
119 84 54 29
286
Color
Red
26 17 14 14
71
Blond
7 94 10 16
127
592$
pf3
pf4
pf5

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Download 6.2 Introduction to Probability Summary Table and more Study notes Probability and Statistics in PDF only on Docsity!

6.2 Introduction to Probability

part 2

! From Tables to Probability

" Computing relative frequency probabilities from a

table.

! Probability Distributions

! The table below provides information on the

hair color and eye color of 592 statistics

students (The American Statistician, 1974).

Summary Table

Eye Color Row Brown Blue Hazel Green Total Black 68 20 15 5 108 Hair Brown 119 84 54 29 286 Color Red 26 17 14 14 71 Blond 7 94 10 16 127 592 3

! If a student is drawn at random from this

population…

" What is the probability that they are blond?

! ANS: 127/592 = 0.

" What is the probability that they are NOT blond?

! ANS: P(NOT blond) = 1-P(blond) =1-127/592 = 0.

! ANS: (71 +286 + 108)/592 = 465/592 = 0.

Eye Color Row Brown Blue Hazel Green Total Black 68 20 15 5 108 Hair Brown 119 84 54 29 286 Color Red 26 17 14 14 71 Blond 7 94 10 16 127 592

! If a student is drawn at random from this

population…

" What is the probability that they are blond

and have brown eyes?

! ANS: 7/592 = 0.

" What is the probability that they have red

hair and have green eyes?

! ANS: 14/592 = 0.

Eye Color Row Brown Blue Hazel Green Total Black 68 20 15 5 108 Hair Brown 119 84 54 29 286 Color Red 26 17 14 14 71 Blond 7 94 10 16 127 592 5

! Based on this table…

" What is the probability that a blond student

has brown eyes?

! ANS: 7/127 = 0.

" What is the probability that a brown-haired

student has hazel eyes?

! ANS: 54/286 = 0.

Eye Color Row Brown Blue Hazel Green Total Black 68 20 15 5 108 Hair Brown 119 84 54 29 286 Color Red 26 17 14 14 71 Blond 7 94 10 16 127 592

We’ve subsetted to a smaller

population from which we are

drawing a person

Relative Frequency Method

! Step 1. Repeat or observe a process

many times and count the number of times

the event of interest, A , occurs.

! Step 2. Estimate P ( A ) by

! P ( A ) =

number of times A occurred

total number of observations

! A probability distribution shows the possible

values that could occur, and the probability

for each occurring.

Probability Distributions

The probability of each value is

shown by the height of the bar.

! Find the probability distribution for the ‘sum

of two dice’.

Example: Roll two 6-sided dice

Probability (^2 3) Sum of two dice 4 5 6 7 8 9 10 11 12

Most likely sum

is a sum of 7.

Only one way

to get sum of

12 (double 6’s).

P(sum of 12) =

! Find the probability distribution for the ‘sum

of two dice’.

Example: Roll two 6-sided dice

Probability (^2 3) Sum of two dice 4 5 6 7 8 9 10 11 12

Recall,

P(sum of 11) =

And that is the

height of the bar

above the number

! Find the probability distribution for ‘the

number that turns up’.

Example: Roll one 6-sided dice

Each of 6 values is

equally likely, at 1/

value on die probability 1 2 3 4 5 6

0.^ 0.^ 0.^ 0.^ 0.^

This is a uniform distribution.

! How to make a probability distribution.

1) List all possible outcomes. Use a table or

figure it is helpful.

2) Identify outcomes that represent the same

event. Find the probability of each event.

3) Make a table in which one column lists

each event and another column lists each

probability. The sum of all the probabilities

must be 1.

Probability Distributions

! Random experiment: Toss Three Coins

! Make a probability distribution for the

number of heads that occurs.

1) List the possible outcomes (equally likely):

HHH,HHT,HTH,THH,HTT,THT,TTH,TTT

2) Identify outcomes that represent the same

event. There are four possible events:

0, 1, 2, or 3 heads.

Example: Making a probability

Probability Distribution