Ch. 13 Notes
MATH 2400
Mr. J. Dustin Tench
Recap Ch. 12
- P(A or B) =
- P(A) + P(B) – P(A∩B)
- If A and B are disjoint, P(A∩B) = 0, so we get
P(A) + P(B).
- P(A and B) =
- P(A|B) • P(B)
- If A and B are independent, P(A|B) = P(A), so we
get P(A) • P(B)
When are A and B disjoint?
When the sample spaces for A and B contain no
common elements, they are considered
disjoint.
Ex: Rolling a 2 or a 4 on a die.
Ex: Drawing a black ace or a red ace out of a
standard deck of cards.
Ex: Selecting a senior citizen or a teenager at
random for a prize from a collection of
surveys.
Venn Diagram of Disjoint Events
P(A∩B)=0
When are A and B Non-Disjoint?
When A and B contain common elements in
their sample space.
Ex: Drawing a red card or a king from a standard
deck of cards.
Ex: Rolling a 4 or an even number.
Ex: Buying a used car or buying a Toyota.
Venn Diagram of Disjoint Events
P(A∩B) ≠ 0
When are A and B Independent?
Events A and B are independent when the
outcome of A cannot affect the outcome of B.
Ex: Rolling a 6 and flipping Heads.
Ex: Having twins and the first being a boy and
the second being a girl.
Conditional Probability
Conditional probability exists when we are only
considering a subset of the original sample
space. Generally, the word “given” is used,
but not always.
Ex: Of the freshmen, what’s the probability they
are a science major?
Ex: Given a student has a job, what’s the
probability they have above a 3.5 GPA?
Example
The Clemson University Fact Book for 2007
shows that 123 of the university’s 338
assistant professors were women, along with
76 of the 263 associate professors and 73 of
the 375 full professors.
1) What is the probability that a randomly
chosen Clemson professor is a woman?
2) Given a selected individual is a full professor,
what is the probability that the person is a
woman?
3) Are the rank and gender of Clemson
professors independent?
Example
A recent study has shown that 75% of teenagers
own cell phones. The study also indicates that
65% of teenagers own a phone and text on
their phones.
a) What is the probability that a randomly
selected teen is a “texter” if we know they
own a phone.
b) Of the teens that own phones, 15% send
more than 6,000 texts a month. What
percent of all teens own a phone, are texters,
and send more than 6,000 texts a month?
Ch. 13 Notes – FINALLY!!!
Binomial Distributions are simply situations in
which only two different outcomes are
possible for each trial.
Ex: Shooting free throws in a basketball game.
Ex: Having a baby.
Ex: Computerized telephone survey successfully
calling a residence.
Combinatorics
Permutations & Combinations
A permutation is an arrangement of objects in
which the order they are arranged can be
counted differently than a different
arrangement of the same objects
ABC is a different arrangement than ACB.
A combination is a grouping of objects in which
order does not matter and the group is
counted as the same group if it contains that
same elements
ABC is the same group as ACB.
Permutations
Consider assigning three people 3 different roles
in an organization.
A – President
B – Vice-President
C – Secretary
We can arrange these same 3 people 6 different
ways.
ABC, ACB, BAC, BCA, CAB, CBA
Permutations
Consider choosing the president, we have 3
choices.
Then, consider choosing the vice-resident, we
have 2 choices.
Finally, consider choosing the secretary, we have
1 choice. So, the number of arrangements of
3 people is 3*2*1 = 3! = 6.
Factorial: n! = n*(n-1)*(n-2)*(n-3)*…*3*2*1
Permutations
What if we don’t use all the objects in a group.
Consider the case where a President and VicePresident are to be chosen from a group of 4
people. How many different ways can we
assign these two positions?
For the president, we have 4 choices. For the
vice-president, we have 3 choices. So, the
number of ways we can assign these positions
is 4*3 = 12.
Permutations – A Formula
There is a formula, but it is
much easier just to think
about the problem and
count the number of ways
each part can happen.
Combinations
Consider forming a team where each person has
the same role. If 3 people are to be used from
a pool of…3……people…..then there is only 1
group that can be formed.
But, what if the pool was larger?
Consider a terrible situation in which there are 4
people and a team of 3 is to be formed. How
many ways can this happen?
ABC ACD
ABD BCD
There are 4 ways.
Combinations
Combinations, unlike Permutations, have an
accepted notation used.
For the situation we just looked at, n=4 and r=3,
so
= 4.
On The Ti-84
Permutations: MATH, PRB tab, 2: nPr
Combinations: MATH, PRB tab, 3: nCr
Type in what you want to calculate like…
4
3 and hit ENTER.
Okay, Back to Binomial Distributions
We use combinations to count the number of
ways an event can occur.
For example, if Jordan makes 70% of his free
throws, what is the probability he makes 4 of
the 6 free throws he attempts in a game?
The number of ways he can make 4 of the 6 is
= 15.
Jordan
There are 15 different ways Jordan can make 4 of
his 6 free throws. The probability Jordan hits any
one of the free throws is .7. The probability
Jordan misses is .3.
Lets consider the case where Jordan makes the first
four free throws, but misses the next two.
The probability of this happening is
(.7)(.7)(.7)(.7)(.3)(.3) = .021609
Consider the case where Jordan makes the first
three, misses two, but makes the last.
(.7)(.7)(.7)(.3)(.3)(.7) = .021609
They are exactly the same!
Jordan continued…
So, there are 15 ways of making 4 out of 6 free
throws, so the probability that Jordan makes
exactly 4 out of the 6 free throws is.
15*(.7)(.7)(.7)(.7)(.3)(.3) =
15* (.7)4(.3)2 = .324135
More Jordan
1. Calculate the probability that Jordan makes
at least 2 of the 6 free throws.
P(X≥2) =
2. Calculate the probability that Jordan makes
at most 4 of the 6 free throws.
P(X≤4) =
3. Calculate the probability that Jordan makes 2,
3, or 4 of the 6 free throws.
P(2≤X≤4) =
On the Calculator
2nd, DISTR
- 0:binompdf(n,p,r), for P(x = r)
- A:binomcdf(n,p,r), for P(x <,≤,>, or ≥ r)
So, for Jordan…
P(X=4) = binompdf(6,0.7,4)
P(X≥2) = 1 - binomcdf(6,0.7,4)
P(X≤4) = binomcdf(6,0.7,4)
P(2≤X≤4) = binomcdf(6,0.7,4) – binomcdf (6,0.7,2)
μ and σ for Binomial Distributions
1) μ = np where μ is the mean, n is the number
of trials, and p is the probability of success
for each trial.
2) σ = 𝒏𝒑(𝟏 − 𝒑) where σ is the standard
deviation.
An Application
Delta has been collecting data regarding its
flights. Data indicates that 82% of the flights
from Atlanta to Las Vegas depart on time.
1) If there are 227 flights from Atlanta to Las
Vegas this year, what is the expected number
(mean) of flights that should depart on time?
2) What is the standard deviation using this
data?
HW
13.1, 13.2, 13.3, 13.4, 13.5, 13.24, 13.26