Review of
Probability and
Binomial
Distributions
The Mathematics of Chance
• How many possible outcomes are
there with a single 6-sided die?
• What are your “chances” of rolling a
6?
• Can we generalize what you just
did?
The gambler’s dispute (1654)
• This famous dispute led to the
formal development of the
mathematical theory of probability
"A gambler's dispute … a game consisted
in throwing a pair of dice 24 times; the
problem was to decide whether or not to
bet even money on the occurrence of at
least one "double six" during the 24
throws.
Let’s simulate this…
• How many possible
outcomes are there?
• What fraction of these is
a “double-six”?
• How can we quantify the
odds?
• How many times would
expect to get 6-6 in 24
tries?
• How likely would it be to
play this game 36 times
and NOT get 6-6?
You have a 36% chance of not
getting 6-6 in 36 throws (1:2 odds)
Link to Excel simulation
Defining Probability
• We define probability by comparing
an outcome or set of outcomes with
the set of all possible outcomes for
an event.
• This will lead us to an “intuitive”
definition of probability
Examples…
• A coin toss:
– Two possible outcomes H or T
– Probability for H is 1 of the 2 or ½ = 0.5 = 50%
• You win the “Stats 300 Lottery”
– 39 possible outcomes
– Only 1 of you! Probability is 1/39 = 2.5%
• Odds of a full-house in Poker
– There are 2,598,960 possible poker hands
– There are 3,744 ways to get a full house or
3744/ 2,598,960 = 0.024% (1 in 4165 hands!)
Independent Events
• When events are independent – the
outcome (or probability) of the one does
not change the probability of the other.
• Example:
– You flip a coin and get heads – what is the
probability that you heads on the next flip?
– NOTE – this is not the same as asking what is
the probability of flipping two heads in
succession
Probability of HH is
(1/2)(1/2) = 1/4
Four Possible Outcomes
Probability Rules
(for events)…
• A probability of 0 means an event
never happens
• A probability of 1 means an event
always happens
P
• Probability
is a number always
between 0 and 1
Probability Rules
(for events)…
• If the probability of an event A is
P(A) then the probability that the
event does not occur is 1-P(A)
• This is also called the compliment of
A and is denoted AC
• Example: what is the probability of
not rolling a 6 when using an honest
die?
Solution: P6 = 1/6, PC6 = 1 - 1/6 = 5/6
Probability Rules
(in pictures)…
• If events A and B are
completely independent of
each other (disjoint) then
the probability of A or B
happening is just:
P( A B)  P( A)  P(B)
Sample Questions…
• What is the probability of flipping 5
successive heads?
• What is the probability of flipping 3 heads
in 5 tries?
• From your text: 4.8, 4.13,4.14
Probability Rules
• If events A and B are
independent of each
other (but not disjoint)
then the probability of
A and B happening is
just:
P( A B)  P( A)P(B)
(in pictures)…
The Binomial Distribution
A motivating example…
• 35% of Canadian university students work
more than 20 hours/week in jobs not related
to their studies. This can have a serious
impact on their grades. What is the
probability that I have at least 5 such
students in this class?
Answer: There is better than
a 99% chance!
What is a Binomial Distribution?
•Any random statistic that can be cast in a
“yes/no” format where:
•N successive choices are independent
•“yes” has probability p and “no” has probability 1-p
fits a binomial distribution.
Suggest 3 other examples of data sets that
can be modeled as binomial distributions
Looking a bit deeper…
• Suppose someone offered you the following
“game”:
Toss a coin 5 times. If you get 3 heads
I pay you a dollar, otherwise you pay me 50 cents.
• Should you accept the bet?
• What is your expected return on this bet?
• How can we calculate the odds?
Pascal to the rescue!
There are exactly 10
ways to get 3 heads
What is the probability
of flipping 6 tails in 8 trials?
How to generate Pascal’s Triangle
•Pascal’s triangle “unlocks” the mystery of binomial
distributions
•The cells in the triangle represent binomial coefficients
which also represent all possible “yes/no” combinations
•In “math-speak” we use the following notation to
calculate the number of ways “k” events can occur in “n”
choices:
n
n!
 k   k !(n  k )!
 
Factorial notation
5! = 5x4x3x2x1 = 120
How many ways can 3 people be selected from a class of 39?
Math detail (FYI)
• The general binomial probability is:
n k
P(k )    p (1  p ) n k
k 
Example: B(9,0.4),what is P(5)?
• The Binomial Table is built from these terms
How to use the binomial distribution
• Assign “yes” and “no” and
their respective
probabilities to the
instances in your problem
•Assign “n” and “k” and
either use the formula, look
up in a table or use a stats
package (Excel works well)
•Example: 5.5
Look up in table
3 ways:
Use formula
Use Excel
15 
P(3)    (0.3)3 (0.7)12  0.1700
3
From Binomial to Normal Distributions
• Binomial is a discrete probability
distribution
• Normal is a continuous distribution
• When n becomes very large we can often
approximate by using a N(m,s) dist.
m X  np
s X  np(1  p)
• How large is “large”?
Rule of Thumb: when np >= 10 and n(1-p) >= 10 we can use the
Normal Distribution approximation
Sample Proportions…
• We often are interested in knowing the
proportion of a population that exhibits a
specific property (statistic). We denote this
the following way:
count of successes X
pˆ 

size of sample
n
• p is a proportion (often interpreted as a
probability) and is therefore a number
between 0 and 1
Mean and Standard Deviation of a Sample
Proportion
• If p is the proportion of “successes” in a
large SRS of n samples, then:
m pˆ  p
s pˆ 
p(1  p)
n
Look at Example 5.7
Working through some examples…
• 5.19: ESP
• A) ¼ = 0.25
• B) p(10)+p(11)+…+p(20) or… 1- [p(0)+…p(9)], this
can be read from Table C or done in EXCEL
• C) use m X  np
s X  np(1  p)
• You would expect 5 correct choices with a standard
deviation of 1.936
• D) Since the subject knows that all 5 of the shapes are on the card
the choices are no longer random and hence a binomial model is
not appropriate – this was not the case in parts a-c
• 5.21
• A) just use m X  np
• B) now use:
m pˆ  p
s pˆ 
• C)
z ( pˆ  0.24) 
s X  np(1  p)
p(1  p)
n
0.24  0.2
 3.16
0.01265
• D) p = 0.01  z = 2.33, use z 
X m
s
; X  m s z
• 5.24
• Identify relevant statistics: n = 1500, p = 0.7
•
•
•
•
A) X = np = (1500)(0.70) = 1050
B) z = (1000-1050)/17.748,  better than 99% chance
C) z = (1200-1050)/17.748,  NO CHANCE!!!!!
D) X = np = 1190 and s = 18.89, chance that more than
1200 accept is now pretty good (p = 0.2892)