Download Chapter 10 - Hampden

Introducing probability
BPS chapter 10
© 2006 W. H. Freeman and Company
Brigitte Baldi, University of California, Irvine
Ellen Gundlach, Purdue University
Marcus Pendergrass, Hampden-Sydney College
Objectives (BPS chapter 10)
Introducing probability

The idea of probability

Probability models

Probability rules

Discrete sample space

Continuous sample space

Random variables

Personal probability
Randomness and probability
A phenomenon is random if individual
outcomes are uncertain, but there is
nonetheless a regular distribution of
outcomes in a large number of
repetitions.
The probability of any outcome of a random phenomenon can be
defined as the ideal theoretical proportion of times the outcome would
occur in a very long series of repetitions.
Coin toss
The result of any single coin toss is
random. But the result over many tosses
is predictable, as long as the trials are
independent (i.e., the outcome of a new
coin toss is not influenced by the result of
the previous toss).
The probability of
heads is 0.5 =
the proportion of
times you get
heads in many
repeated trials.
First series of tosses
Second series
Two events are independent if the probability that one event occurs
on any given trial of an experiment is not affected or changed by the
occurrence of the other event.
When are trials not independent?
Imagine that these coins were spread out so that half were heads up and half
were tails up. Close your eyes and pick one. The probability of it being heads is
0.5. However, if you don’t put it back in the pile, the probability of picking up
another coin and having it be heads is now less than 0.5.
The trials are independent only when
you put the coin back each time. It is
called sampling with replacement.
Probability models
Probability models mathematically describe the outcome of random
processes. They consist of two parts:
1) S = Sample Space: This is a set, or list, of all possible outcomes
of a random process. An event is a subset of the sample space.
2) A probability for each possible event in the sample space S.
Example: Probability Model for a Coin Toss
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
Sample space
Important: It’s the question that determines the sample space.
A. A basketball player shoots
three free throws. What are
the possible sequences of
hits (H) and misses (M)?
H
H -
HHH
M -
HHM
H
M
M…
H -
HMH
M -
HMM
…
B. A basketball player shoots
three free throws. What is the
number of baskets made?
S = {HHH, HHM,
HMH, HMM, MHH,
MHM, MMH, MMM }
Note: 8 elements, 23
S = {0, 1, 2, 3}
C. A person tosses a coin repeatedly. How long in seconds does it take
before the first head appears?
S = [0, ∞) = (all numbers ≥ 0)
Probability rules
1) Probabilities range from 0
(no chance of the event) to
1 (the event has to happen).
For any event A, 0 ≤ P(A) ≤ 1
Coin Toss Example:
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
Probability of getting a head = 0.5
We write this as: P(head) = 0.5
P(neither head nor tail) = 0
P(getting either a head or a tail) = 1
2) The probability of the
complete sample space must
equal 1.
3) The probability of an event
not occurring is 1 minus the
probability that it does occur.
P(sample space) = 1
P(A) = 1 – P(not A)
P(head) + P(tail) = 0.5 + 0.5 = 1
P(tail) = 1 – P(head) = 0.5
Probability rules (cont'd)
A and B disjoint
4) Two events A and B are disjoint if they have
no outcomes in common and can never happen
together. The probability that A or B occurs is
the sum of their individual probabilities.
P(A or B) = “P(A U B)” = P(A) + P(B)
This is the addition rule for disjoint events.
A and B not disjoint
Example: If you flip two fair coins and the first flip does not affect the second
flip, S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or
0.25.
The probability that you obtain “only heads or only tails” is:
P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50
Probability rules (cont'd)
5) Two events A and B are independent knowing that one event
occurred doesn’t affect the probability that the other occurs. If A and B
are independent, the probability that both A and B occurs is the product
of their individual probabilities.
P(A and B) = “P(A B)” = P(A) P(B)
This is the multiplication rule for independent events.
Example: Flip a fair coin twice. The coin has no “memory”, so the outcome of
the first flip can’t affect the probabilities for the second flip. The first and
second flips are independent.
Let A = {toss 1 is heads}, B = {toss 2 is tails}. Then
P(A and B) = P(HT) = P(H) P(T) = 0.5 + 0.5 = 0.25
Discrete sample space
Discrete sample spaces deal with data that can take on only certain
values. These values are often integers or whole numbers.
Dice are good examples of finite sample
spaces. Finite means that there is a limited
number of outcomes.
Throwing 1 die:
S = {1, 2, 3, 4, 5, 6},
and the probability of each event = 1/6.
Note: Discrete data contrast with continuous data that can take on any one of
an infinite number of possible values over an interval.
Example

Random phenomenon: toss a fair coin three times, and record the
outcomes.

What is the probability model?


Probability Model = Sample Space + Probability assignment

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

All outcomes equally likely (since the coin is fair)

So each outcome has probability 1/8.
We can represent the probability model as a table:
outcome
probability
In some situations, we define an event as a combination of outcomes.
In that case, the probabilities need to be calculated from our
knowledge of the probabilities of the simpler events.
Example: You toss two dice. What is the probability of the outcomes summing
to five?
This is the
sample space:
{(1,1), (1,2), (1,3),
……(6,6).}
There are 36 possible outcomes in S, all equally likely (given fair dice).
Thus, the probability of any one of them is 1/36.
Example: P(the roll of two dice sums to 5) = P(4,1) + P(3,2) + P(2,3) + P(1,4)
= 4 * 1/36
= 1/9 = 0.111
Example

A 6-sided die has been weighted so that 1, 4, and 6 are more likely
to appear than other digits. The probability model for the die is
given in the table:
outcome
1
2
3
4
5
6
probability
1/5
1/6
1/6
1/5
1/6
1/5

Is this a legitimate (valid) probability model?

If we wanted the probabilities for 1, 4, and 6 to all be 1/5, and the
other probabilities to all be equal to each other, what would they
have to be?
outcome
1
2
3
4
5
6
probability
1/5
??
??
1/5
??
1/5
Random Variables


A random variable is a variable whose value is a numerical
outcome of a random phenomenon.

Example: flip a coin three times; let X = number of heads obtained

Example: take an SRS of 1000 Virginia voters; let Y = number who
support the current economic stimulus plan.

Example: roll a pair of fair dice; let S = the sum of the spots that appear.

Throw a dart at a dartboard; let D = distance of dart from center.
A random variable describes a numerical outcome of a random
phenomenon. It’s values are random, so they are described by a
probability distribution.

Example; flip a fair coin three times, let X = number of heads obtained

P(X = 0) = ???

P(X = 1) = ???

P(X = 2) = ???

P(X = 3) = ???
Example


Toss a pair of fair dice, and record the outcomes
Random variable: let S = sum of the spots that appear
S=2
S=3
S=4
S=5
S=6
S=7

S=8
S=9
S=10
S=11
S=12
Probability Model for S:
outcome
S=2
S=3
S=4
S=5
S=6
S=7
S=8
S=9
S=10
S=11
S=12
probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
Give the sample space and probabilities of each event in the following cases:
A couple wants three children. What are the arrangements of boys (B) and girls
(G)?

•
Genetics tells us that the probability that a baby is a boy or a girl is the same,
0.5.
→ Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG}
→ All eight outcomes in the sample space are equally likely.
→ The probability of each is thus 1/8.
A couple wants three children. What are the numbers of girls (X) they could
have?

•
The same genetic laws apply. We can use the probabilities above to calculate
the probability for each possible number of girls.
→ Sample space {0, 1, 2, 3}
→ P(X = 0) = P(BBB) = 1/8
→ P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) =
3/8
→ etc.
Probability Model:
More problems


An on-campus club has 10 members: 3 freshmen, 2 sophomores, 1
junior, and 4 seniors.

Random phenomenon: pick a club member at random (i.e. choose an
SRS of size n = 1)

Random variable: let X = class rank of the chosen individual.

What is the probability model for X?
A club has 5 members: 3 men and 2 women.

Random phenomenon: choose an SRS of size 2 from the club.

Random variable: let N = number of women chosen.

What is the probability model for N?
Continuous sample space
Continuous sample spaces contain an infinite number of events. They
typically are intervals of possible, continuously-distributed outcomes.
Example: Pick a real number at random between 0 and 1 (e.g., 0.001, 0.4, 0.0063876).
S = {interval containing all numbers between 0 and 1}
How do we assign probabilities to events in an infinite sample space?
We use density curves and compute probabilities for intervals.
This is a uniform density curve.
There are a lot of other types of density curves.
The probability of the uniformly-distributed
variable Y to be within 0.3 and 0.7 is the
area under the density curve corresponding
to that interval. Thus:
y
P(0.3 ≤ y ≤ 0.7) = (0.7 − 0.3)*1 = 0.4
Probability distribution for a continuous random variable
% individuals with X
such that x1 < X < x2
The shaded area under the density
curve shows the proportion, or
percent, of individuals in the
population with values of X
between x1 and x2.
Because the probability of drawing
one individual at random
depends on the frequency of this
type of individual in the population,
the probability is also the shaded
area under the curve.
Intervals versus single outcomes
The probability of a single outcome is meaningless for a
continuous sample space. Only intervals can have a non-zero
probability, represented by the area under the density curve for that
interval.
Suppose X is uniformly distributed on [0, 6]
P(1  X  3) = (3 – 1)*(1/6) = 1/3
P( X = 3) = P(3  X  3)
= (3 – 3)*(1/6) = 0
Density curve for X
height = 1/6
0
1
3
6
0
1
3
6
height = 1/6
P(1  X  3) = P(1 < X  3) = P(1  X < 3) = P(1 < X < 3) = 1/3
We generate two independent random variables X and Y uniformly distributed
between 0 and 1. Let Z to be their sum. Z can take any value between 0 and 2.
The density curve for Z is:
What is the height of the triangle?
Height = 1. We know this because the
base = 2, and the area under the
density curve must be 1.
Y
0
1
2
What is the probability that Y is < 1?
1/2
What is the probability that Y < 0.5?
1/8
What is the probability that 0.5 < Y < 1.5?
The area of a this triangle is
½ (base*height).
Normal probability distribution
The probability distribution of many random variables is the normal
distribution. This is the same distribution we first encountered in Chapter
3. We’re just thinking about it in a slightly different way.
Example: Choose a woman at
random from a certain
population. Let X be the
chosen woman’s height. Then
X is a random variable.
Probabilities with the normal distribution are calculated in exactly
the same way as we computed proportions with the normal
distribution. (Use normalCDF on the TI-83 or 84.)
Previously, we wanted to calculate the proportion of individuals in the population
with a given characteristic.
Distribution of women’s heights ≈
N (µ, ) = N (64.5, 2.5)
Example: What's the proportion
of women with a height between
57" and 72"?
That’s within ± 3 standard
deviations  of the mean m, thus
that proportion is roughly 99.7%.
Since about 99.7% of all women have heights between 57" and 72", the chance
of picking one woman at random with a height in that range is also about 99.7%.
Example
Assume the distribution of women’s heights in the previous slide. What is the
probability, if we pick one woman at random, that her height is between 68” and
70” ?
Let X be the height of the woman who is selected.
Because the woman is selected at random, X is a random variable.
From the previous slide, we know X ~ N(m = 64.5,  = 2.5).
So the probability that the selected woman is between 68’’ and 70’’ in height is
P(68 < X < 70) = normalCDF(68, 70, 64.5, 2.5) = 0.06686 = about 6.7%
Reminder: standardizing N (m,)
We standardize normal data by calculating z-scores so that any Normal
curve N(m,) can be transformed into the standard Normal curve N(0,1).
N(64.5, 2.5)
N(0,1)
→
z
x

z
Standardized height (no units)
(x  m )

Inverse problem:
Your favorite chocolate bar is dark chocolate with whole hazelnuts.
The weight on the wrapping indicates 8 oz. Whole hazelnuts vary in weight, so
how can they guarantee you 8 oz. of your favorite treat? You are a bit skeptical...
To avoid customer complaints and
lawsuits, the manufacturer makes
sure that 98% of all chocolate bars
weight 8 oz. or more.
The manufacturing process is
roughly normal and has a known
variability  = 0.2 oz.
How should they calibrate the
machines to produce bars with a
mean m such that P(x < 8 oz.) =
2%?
 = 0.2 oz.
Lowest
2%
x = 8 oz.
m=?
How should they calibrate the machines to produce bars with a mean m such that
P(x < 8 oz.) = 2%?
 = 0.2 oz.
Lowest
2%
x = 8 oz.
m=?
x
z = ???
0
z = (x-m) / 
Standardize X
Lowest
2%
Find z:
z = invNorm(0.02, 0, 1)
= - 2.054
Plug in what we know:
Solve for m:
- 2.054 = (8 – m) / 0.2
m = 8.4107
How should they calibrate the machines to produce bars with a mean m such that
P(x < 8 oz.) = 2%?
 = 0.2 oz.
Lowest
2%
x = 8 oz.
m=?
x
z = ???
0
z = (x-m) / 
Lowest
2%
Thus, the chocolate bar should weigh μ = 8.41 oz on average.
Meaning of a probability
We have several ways of defining a probability, and this has consequences on
its intuitive meaning.

Theoretical probability
 From understanding the phenomenon and symmetries in the problem


Example: Six-sided fair die: Each side has the same chance of turning up;
therefore, each has a probability 1/6.

Example: Genetic laws of inheritance based on meiosis process.
Empirical probability
 From our knowledge of numerous similar past events

Mendel discovered the probabilities of inheritance of a given trait from
experiments on peas, without knowing about genes or DNA.

Example: Predicting the weather: A 30% chance of rain today means that it rained
on 30% of all days with similar atmospheric conditions.

Personal probability
 From subjective considerations, typically about unique events

Example: Probability of a large meteorite hitting the Earth. Probability of
life on Mars. These do not make sense in terms of frequency.
A personal probability represents an individual’s personal degree of
belief based on prior knowledge. It is also called Baysian probability for
the mathematician who developed the concept.


We may say “there is a 40% chance of life on Mars.” In fact, either there
is or there isn’t life on Mars. The 40% probability is our degree of belief,
how confident we are about the presence of life on Mars based on what
we know about life requirements, pictures of Mars, and probes we sent.
Our brains effortlessly (?) assess risks (probabilities) of all sorts, and
businesses try to formalize this process for decision-making.