Download Probability part 2

Probability
The Study of
Chance!
Part 2

In this
powerpoint we
will continue
calculating
probabilities
using
• Venn Diagrams
• Contingency
Tables
Let’s begin with Venn
Diagrams
A venn diagram
consists of
interlocking circles
that allows us to see
relationships
Learn about the history of the Venn
diagram at
http://www.pballew.net/arithme5.html
#venn
A great website created and maintained
by a great educator, Pat Ballew.
Relationships with Two Outcomes

Consider for a moment the following:
•At the Pueblo
Athletic Club,
72% of the
members use the
nautilus
machines, 51%
use the pool, and
30% use both.
•Create a venn
diagram that
shows this
relationship.
Nautilus
Swimming
Machines
.42
.21
Pool
.30
.07
•Step 1: Since there are 2 possibilities, we will use a venn diagram with 2
circles.
•Step 2: Now begin by putting the proportion of people who use both the
Nautilus machines and the swimming pool. P(M and S) = .30
•Step 3: Notice that the circle for the nautilus machines already has the 30%
of the people who use both. This means that we need to take the total of 72%
that use the nautilus machines and subtract those who use both (30%) to find
the percent that use the machines but not the pool. P(M and Sc) = .72-.30 = .42
•Step 4: We need to do the same process to find the percent of people who
use the pool but not the machines. P(Mc and S) = .51-.30 = .21
•Step 5: To complete the diagram we need to find the percent of people who
use neither the machines or the pool. P(Mc and Sc) = 1 – (.42+.30+.21) = .07
Using the Venn diagram to answer
questions.

Once we have created the diagram, we can answer
many different questions.
1. Find the probability
that a randomly
selected member
uses only the
swimming pool. .21
2. Find the probability
that a member uses
either the machines
or the swimming
pool, but not both
.42 + .21 = .63
Nautilus
Swimming
Machines
.42
.21
Pool
.30
.07
3. Find the probability that a member
uses either the machines or the swimming
pool. In statistics this type of “or” is an
inclusive or, which means one or the other
or both: so…. .42 + .30 + .21 = .93
Using a Contingency Table



We can represent the same
situation in a contingency
(two-way) table.
Step 1: Fill in the
information we have been
given.
Step 2: Since we are
dealing with proportions we
know the “grand total” is
equal to 1.

Step 3: Now using
subtraction, we can find the
missing proportions.
Nautilus Machines
S
W
I
M
M
I
N
G
Yes No
P
O
O
L
Yes
No
Total
.30
.42
.72
.21
Total
.51
.07
.49
.28
1
Using the table to answer
questions
Since our table is already in
proportions we can answer many
questions without any calculations
•What proportion of the
membership does not use the
Nautilus Machine: P(Mc) = .28
•What proportion of the
membership uses neither the
Nautilus Machines nor the pool?
P(Mc and Sc) = .07
Nautilus Machines
S
W
I
M
M
I
N
G
Yes No
P
O
O
L
Yes
No
Total
.30
.42
.72
.21
Total
.51
.07
.49
.28
1
Calculating “or” probabilities
Remember that the use of “or” in
statistics is inclusive. The easiest
way to find this probability is to
draw some lines on the table.
•First, a line across the row where
the member uses the swimming
pool.
•Then, a line down the column
where the member uses the
Nautilus Machines.
•Now add the row total and the
column total, but notice that the
value where the lines intersect has
been added in twice. So we must
subtract one of them
.51 + .72 - .30 = .93
Nautilus Machines
S
W
I
M
M
I
N
G
Yes No
P
O
O
L
Yes
No
Total
.30
.42
.72
.21
.07
.28
Total
.51
.49
1
Using Counts



What if our data is in counts rather than
proportions?
How does this change our calculations?
Well let’s see.
• In a large shopping mall, a marketing agency
conducted a survey on credit cards. Respondents
were asked to identify if they were employed or
unemployed and whether or not they owned a
credit card. The agency found the following. Of
the 109 respondents 46 owned a credit card, 62
were unemployed and 18 both owned a credit card
and were employed.
Counts instead of proportions
•The agency found the following.
Of the 109 respondents 46 owned
a credit card, 62 were unemployed
and 18 both owned a credit card
and were employed.
•Now, using subtraction complete
the table.
•Find the probability that a
randomly selected respondent was
employed: P(E) = 47/109
•Find the probability that a
randomly selected respondent was
both employed and did not have a
credit card. P(E and Cc)=29/109
Credit Card
Yes No
J
O
B
Yes
No
Total
18
28
46
Total
29
47
34
62
63
109
•Find the probability that a randomly selected respondent either has a
job or does not have a credit card. Again for this question, let’s use
lines on the table.
•So, P(E or Cc) = (47 + 63 – 29) / 109
Three Events???

What if we have three events?

How can we show the relationships?

Once again, we will use Venn diagrams to
show the relationships.
Three Events

In a survey carried out
in a school snack shop,
the following results
were obtained. Of 100
boys questioned, 78
liked sweets, 74 liked
ice cream, 53 liked cake,
57 liked both sweets
and ice cream, 46 liked
both sweets and cake,
10 liked only ice cream,
and only 31 liked all
three. We’ll create a
Venn diagram to model
the relationship.
Creating the Venn diagram
•Start with the intersection of all
three circles:
Sweets
Ice Cream
•31 liked all three
•Then move to the intersection two
events.
26
26
10
•57 liked both sweets and ice
cream. So 57-31=26
31
•46 liked both sweets and cake.
So 46-31 = 15
•Now look at the single events
1515
7
•10 liked only ice cream
•74 liked ice cream total,
20
74-(31+26+10) = 7
•53 liked cake,
53 – (15+31+7)= 20
•78 liked sweets
78 – (15+31+26) = 26
•Notice that this information tells us that
all boys like at least one of the three.
Cake
0
Additional Resources

The Practice of Statistics—YMM
• Pg 324 -331

The Practice of Statistics—YMS
• Pg 328-355