9. Introducing probability
The Practice of Statistics in the Life Sciences
Third Edition
© 2014 W. H. Freeman and Company
Objectives (PSLS Chapter 9)
Introducing probability

Randomness and probability

Probability models

Discrete vs. continuous sample spaces

Probability rules

Continuous random variables

Risk and odds
Randomness and probability
In a random event, outcomes are uncertain, but there is nonetheless a
regular distribution of outcomes in a large number of repetitions.
We define the probability of any outcome of a random phenomenon
as the proportion of times the outcome would occur in a very long
series of repetitions.
The sex of a newborn is random
P(Male) ≈ 0.512, the proportion of
times a newborn is male in many,
many births.
Year
Males
Females
Proportion
of males
2002
2,058
1,964
2004
2,105
2,007
2006 2008
2,184 2,173
2,081 2,074
0.5117 0.5118 0.5121 0.5117
Probability models
Probability models mathematically describe the outcome of random
processes. They consist of two parts:
1) S = Sample Space: This is a list or description of all possible
outcomes of a random process. An event is a subset of the
sample space.
2) A probability assigned for each possible simple event in the
sample space S.
Probability model for the sex of a newborn
S = {Male, Female}
P(Male) = 0.512; P(Female) = 0.488
Year
Males
Females
Proportion
of males
2002
2,058
1,964
2004
2,105
2,007
0.5117
0.5118
2006
2,184
2,081
2008
2,173
2,074
0.5121 0.5117
Discrete vs. continuous sample spaces
Discrete sample space:
Discrete variables that can take on
only certain values (a whole
number or a descriptor).
Blood types
For a random person:
S = {O+, O-, A+, A-, B+, B-, AB+, AB-}
Continuous sample space:
Continuous variables that can take
on any one of an infinite number of
possible values over an interval.
Cholesterol level
For a random person:
S = any reasonable positive value
(or the interval zero to infinite)
A. A couple wants three
children. What are the
possible sequences of
boys (B) and girls (G)?
B
B -
BBB
G -
BBG
B
G
G…
B -
BGB
G -
BGG
S = { BBB, BBG,
BGB, BGG, GBB,
GBG, GGB, GGG }
…
B. What is the number of days last
week that a randomly selected teen
exercised for at least one hour?
S = { 0, 1, 2, 3, 4, 5, 6, 7 }
C. A researcher designs a new maze for lab rats. What are the possible
outcomes for the time to finish the maze (in minutes)?
S = interval zero to infinity (all positive values)
Probability rules
Probabilities range from 0 (no chance) to 1 (event has to happen):
For any event A, 0 ≤ P(A) ≤ 1
The probability of the complete sample space S must equal 1:
P( sample space ) = 1
The probability that an event A does not occur (not A) equals 1 minus
the probability that is does occur:
P( not A ) = 1 – P(A)
The 2011 National Youth Risk Behavior Survey provides insight on the physical
activity of U.S. high school students. Physical activity was defined as any activity
that increases heart rate. Here is the probability model obtained by asking,
“During the past 7 days, on how many days were you physically active for a total
of at least 60 minutes per day?”
What is the probability that a randomly selected U.S. high school student
exercised at least 1 day in the past 7 days?
A) 0.08
B) 0.23
C) 0.77
D) 0.85
E) 0.92
Disjoint events
Two events are disjoint, or
mutually exclusive, if they can
never happen together (have no
outcome in common).
Events A
and B are
disjoint.
Events A and
B are NOT
disjoint.
• “male” and “pregnant”  disjoint events
• “male” and “Caucasian”  not disjoint events
Addition rules
Addition rule for disjoint events:
When two events A and B are disjoint:
P(A or B) = P(A) + P(B)
General addition rule for any two
events A and B:
P(A or B) = P(A) + P(B) – P(A and B)
Probability that a random person is type O+
P(O+) = 0.38
P(all blood types) = .38 + .07 + .34 + .06
+ .09 + .02 + .03 + .01 = 1
Probability that a random person is not type A+
P(not A+) = 1 – P(A+) = 1 – 0.34 = 0. 66
Probability that a random person is “blood group O”
P(O) = P(O+ or O-) = P(O+) + P(O-) = .38 + .07 = .45
Probability that a random person is “rhesus neg”
P(O- or A- or B- or AB-) = .07 + .02 + .06 + .01 = 0.16
Probability that a random person is either “blood group O” or “rhesus neg”
P(O or -) = P(O) + P(neg) – P(“O-”) = .45 + .16 – .07 = .54
Probabilities of hearing impairment and blue eyes among Dalmatian dogs.
HI
B
= Dalmatian is hearing impaired
= Dalmatian is blue eyed
Neither HI nor B
0.66
HI and not B
0.23
B and not HI
0.06
HI and B
0.05
Are the traits HI and B disjoint? Find the following:
P (HI) =
P (B) =
P (HI or B) =
Continuous random variables
Continuous sample spaces contain an infinite number of events.
We use density curves
to model continuous
probability distributions.
They assign probabilities
over the range of values
making up the sample space.
Continuous probabilities are assigned for intervals
Events are defined over intervals of values.
The total area under a density curve represents the whole population
(sample space) and equals 1 (100%).
Probabilities are computed as areas under the corresponding portion of
the density curve for the chosen interval.
The probability of an event being
equal to a single numerical value
is zero when the sample space
is continuous.
Let Y be a continuous random variable with a uniform distribution as illustrated
here. We compute various probabilities as the corresponding areas under this
uniform distribution.
P(0 ≤ y ≤ 0.5) = 0.5
P(0 < y < 0.5) = 0.5
P(0 ≤ y < 0.5) = 0.5
Height
=1
P(y = 0.5) = 0
y
P(y ≤ 0.5) = 0.5
P(y > 0.8) = 0.2
P(y ≤ 0.5 or y > 0.8) = P(y ≤ 0.5) + P(y > 0.8) = 0.7
Risk and odds
In the health sciences, probability concepts are often expressed in terms
of risk and odds.

The risk of an undesirable outcome of a random phenomenon is the
probability of that undesirable outcome.
risk(event A) = P(event A)

The odds of any outcome of a random phenomenon is the ratio of
the probability of that outcome over the probability of that outcome
not occurring.
odds(event A) = P(event A) / [1 − P(event A)]
Sickle-cell anemia is a serious, inherited blood disease affecting the shape of
red blood cells. Individuals carrying only one copy of the defective gene (“sicklecell trait”) are generally healthy but may pass on the gene to their offspring.
If a couple learns from prenatal tests that they both carry the sickle-cell trait,
genetic laws of inheritance tell us that there is a 25% chance that they could
conceive a child who will suffer from sickle-cell anemia. What are the
corresponding risk and odds?
The risk of conceiving a child who will suffer from sickle-cell anemia is the
probability, so
risk of sickle-cell anemia = 0.25.
The odds is the ratio of two probabilities, so
odds of sickle-cell anemia = 0.25/(1 − 0.25) = 0.333,
which can also be written as odds of 1 to 3 (1:3).