POD #49
11/29/2011
2003B #2
What is the marginal distribution of age?
What proportion of the sample made over $50,000?
Stats: Modeling the World
Chapter 14
From Randomness to Probability
Sample Space
The set of all possible outcomes
Try it!
Albert/Barbara
Albert/Carl
Albert/Dianna
Barbara/Carl
Barbara/Dianna
Carl/Dianna
AND VICE VERSA!!!!
P(Both Girls) = 2/12
Basic Probability Rules
1.
For any event A, 0 ≤ P(A) ≤ 1.
2.
The sum of the probabilities for all outcomes
in the sample space must equal 1.
P(S) = 1
Try it!
Is this a possible sample space? Why or why not?
a) {0.25, 0.25, 0.25, 0.25}
Yes
b) {0.1, 0.2, 0.3, 0.4}
Yes
c) {0.2, 0.3, 0.4, 0.5}
No
d) {0, 0, 1, 0}
Yes
e) {0.1, 0.2, 1.2, -1.5}
No
Addition Rule
P(A or B) = P(A) + P(B)
provided that A and B are disjoint.
Disjoint (Mutually Exclusive)
The two events, A and B, cannot occur at the
same time.
Complement Rule
The set of outcomes that are not in the event A is
called the complement of A, denoted NOT A.
The probability of an event occurring is
1 minus the probability that it doesn’t
occur:
P(A) = 1 – P(NOT A)
Try it!
2
4
1
4
1 1 2
 
4 4 4
2 1 3
 
4 4 4
1 3
1 
4 4
0
4
Multiplication Rule
P(A and B) = P(A) x P(B)
provided that A and B are independent.
Independent Events
Knowledge of event A does not affect the
probability of event B
Try it!
5 5
25
 
12 12 144
4 4
16
 
12 12 144
4 5
20
 
12 12 144
5 3
15
 
12 12 144
3 3
9
 
12 12 144
25
9
16
50



144 144 144 144
5 4
20
 
12 12 144
4 0
0
 
12 12 144
Example
The Masterfoods company says that before the introduction of
purple, yellow candies made up 20% of their plain M&Ms, red
another 20%, and orange, blue, and green each made up 10%.
The rest were brown.
a) If you pick an M&M at random, what is the probability that:
- it is brown? 30%
- it is yellow or orange?
20%+10% = 30%
- it is not green? 100%
- 10% = 90%
- it is striped?
0%
Example
The Masterfoods company says that before the introduction of
purple, yellow candies made up 20% of their plain M&Ms, red
another 20%, and orange, blue, and green each made up 10%.
The rest were brown.
b) If you pick three M&Ms in a row, what is the probability that:
- they are all brown? (.3)(.3)(.3) = 0.027
- the third one is the first one that’s red?
(.8)(.8)(.2) = 0.128
- none are yellow?
(.8)(.8)(.8) = 0.512
- at least one is green?
1-No Greens = 1 - (.9)(.9)(.9) = 0.271
Example
A consumer organization estimates that over a 1-year period 17% of
cars will need to be repaired once, 7% will need repairs twice, and 4%
will require three or more repairs.
a) What is the probability that a car chosen at random will need:
1. no repairs?
P(0) = 1 – 0.28 = 0.72
2. no more than one repair?
P(0 or 1) = 0.72 + 0.17 = 0.89
3. some repairs?
P(1 or more) = 1 – P(none) = 0.28
Example
A consumer organization estimates that over a 1-year period 17% of
cars will need to be repaired once, 7% will need repairs twice, and
4% will require three or more repairs.
b) If you own two cars, what is the probability that:
1. neither will need repair?
P(none and none) = (0.72)(0.72) = 0.5184
2.
both will need repair?
P(some and some) = (0.28)(0.28) = 0.0784
3.
at least one will need repair?
1 - P(none) = 1 – 0.5184 = 0.4816
Example
A certain bowler can bowl a strike 70% of the time. What is the
probability that she
a) goes three consecutive frames without a strike?
P(no/no/no) = (0.3)(0.3)(0.3) = 0.027
b) makes her first strike in the third frame?
P(no/no/yes) = (0.3)(0.3)(0.7) = 0.063
c) has at least one strike in the first three frames
1 - P(no/no/no) = 1 – 0.027 = 0.973
d) bowls a perfect game (12 consecutive strikes)
P(12 strikes) = (0.7)^12 = 0.0138
Example
Suppose that in your city 37% of voters are registered Democrats,
29% Republicans, and 11% other parties. Voters not aligned with
any party are termed “independent”. You are conducting a poll by
calling registered voters at random. In your first three calls, what
is the probability you talk to:
a. All Republicans?
P(RRR) = (0.29)(0.29)(0.29) = 0.024
b.
No Democrats?
P(ND/ND/ND) = (0.63)(0.63)(0.63) = 0.250
c.
At least one Independent?
1 – P(no Indy)= 1 - (0.89)(0.89)(0.89) = 0.295
P(40 or 100) = 3/12 = 0.25
P(10/10) = (0.5)(0.5) = 0.25
P(20/20/20) = (0.25)(0.25)(0.25) = 0.0156
P(20 or less 4 times) = (0.75)(0.75)(0.75)(0.75) = 0.316
P(No/No/No/No/WIN) = (11/12)^4 * (1/12) = 0.0588
1 - P(no gold in 6) = 1 – (11/12)^6 = 0.407
Common Errors

Don’t confuse disjoint and independent!!
Disjoint events CANNOT be independent!!
Example
In a previous WAP problem, we calculated the probabilities of
getting various M&Ms.
a) If you draw one M&M, are the events of getting a red one
and getting an orange one disjoint or independent or neither?
b) If you draw two M&Ms one after the other, are the events of
getting a red on the first and a red on the second disjoint or
independent or neither?
c) Can disjoint events ever be independent? Explain.