Probability and Statistics
Chapter 3 Notes
Section 3-3
Mutually exclusive events are events that
both happen at the
time.
The Addition Rule (For “OR” probabilities)
“Or” can mean one of three things:
A occurs and B does not occur.
B occurs and A does not occur.
Both A and B occur.
If A and B are mutually exclusive, simply
their probabilities of occurring together.
If A and B are NOT mutually exclusive, add their probabilities of occurring together and then subtract the probability
that
.
Example 1: Determine whether these two events are mutually exclusive or not.
1) Roll 3 on a die AND roll 4 on a die.
2) Randomly select a male student AND randomly select a nursing major.
3) Randomly select a blood donor with type O blood AND randomly select a female blood donor.
4) Randomly select a jack from a deck of cards AND randomly select a face card from a deck of cards.
5) Randomly select a 20-year-old student AND randomly select a student with blue eyes.
6) Randomly select a vehicle that is a Ford AND randomly select a vehicle that is a Toyota.
SOLUTIONS
1) You
roll and 3 AND a 4 on the same roll. These are
2) You
be both a male and a nursing major. These are
3) You
be both a female and have type 0 blood. These are
4) Jacks are face cards, so any jack is also a face card. These are
5) You
be a 20-year-old student and have blue eyes. These are
6) A vehicle
be a Ford and a Toyota at the same time. These are
Example 2:
1) You select a card from a standard deck. Find the probability that the card is a 4 or an ace.
2) You select a card from a standard deck. Find the probability that the card is a queen or a red card.
SOLUTIONS
1) “Ace” and “4” are
.
the probability of getting an ace to the probability of getting a 4.
The probability of getting an ace is
; the probability of getting a four is also
.
4
4
8
2
The probability of getting an ace or a four is + = = ≈
.
52
52
52
13
2) The events are
(there are
red queens).
the probability of getting a
to the probability of getting a
, then
the probability of getting a
.
P(queen) =
; P(red) =
;
P(red queen) =
4
26
2
30
2
.
.
.
.
.
.
28
(52 + 52) − 52 = 52 − 52 = 52 ≈
Example 3
The frequency distribution shows the volume of sales (in dollars) and the number of months a sales representative
reached each sales level during the past three years. If this sales pattern continues, what is the probability that the
sales representative will sell between $75,000 and $124,999 next month?
Sales Volume
Months
0-24,999
3
25,000-49,999
5
50,000-74,999
6
SOLUTION:
Define Event A as monthly sales between
Define Event B as monthly sales between
Because events A and B are
$75,000 and $124,999 next month is:
𝑃(𝐴 𝑜𝑟 𝐵) =
≈ 0.444
75,000-99,999
7
100,000-124,999
9
125,000-149,999
2
150,000-174,999
3
175,000-199,999
1
and
and
, the probability that the sales representative will sell between
Example 4
A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last
five days. The number of donors who gave each blood type is shown in the table. A donor is selected at random.
1. Find the probability that the donor has type O or type A blood.
2. Find the probability that the donor has type B or is Rh-negative.
Rh-Factor
Positive
Negative
Total
O
156
28
184
A
139
25
164
Blood Type
B
37
8
45
AB
12
4
16
Total
344
65
409
SOLUTIONS:
1. These events are
.
𝑃(𝑂 𝑜𝑟 𝐴) = 𝑃(𝑂) + 𝑃(𝐴)
184
164
348
+
=
≈ 0.851
409
409
409
2. These events are
.
𝑃(𝐵 𝑜𝑟 𝑅ℎ − 𝑛𝑒𝑔) = 𝑃(𝐵) + 𝑃(𝑅ℎ − 𝑛𝑒𝑔) − 𝑃(𝐵 𝑎𝑛𝑑 𝑅ℎ − 𝑛𝑒𝑔)
45
65
8
102
+ 409 − 409 = 409 ≈ 0.249
409
Example 5
Use the graph to find the probability that a randomly selected draft pick is not a running back or a wide receiver.
SOLUTION:
Define Events A and B
Event A: Draft pick is a running back.
Event B: Draft pick is a wide receiver.
These events are
, so the probability that the draft pick is a running pick or a wide receiver is:
𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
25
34
59
+ 255 = 255 ≈ 0.231
255
To find the probability that the draft is NOT a running back or wide receiver, simply subtract the probability that
he was one of those positions from 1 (
rule)
1 − 0.231 = 0.769
Section 3-4
Permutations
A permutation is an
arrangement of objects. The number of different permutations of n distinct object is n!
The ! Symbol means “factorial” and indicates that you start with the number given and multiply that times every
number between that number and zero.
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880
The calculator will do factorials for you;
Math – Prb – 4 accesses that feature.
To find the number of ways that a permutation can occur, use the TI-84.
Math – PRB – 2
Enter the total number of items in your list.
Press Math - PRB – 2 and Enter
Enter the number of items you wish to order and press enter.
Distinguishable Permutations
A distinguishable permutation must be done by hand.
𝑛!
𝑛1 !𝑛2 !𝑛3 !…𝑛𝑘 !
Alpha and y= on your calculator will allow you to enter this as a fraction, making it easier.
Again, the calculator will also let you enter the factorial symbol, saving time and reducing the chance of making a
simple error in entering the numbers.
Combinations
A combination is a
of r objects from a group of n objects
regard to
.
To find the number of ways that a combination can occur, use the TI-84.
Math – PRB – 3
Enter the total number of items in your list.
Press Math - PRB – 3 and Enter
Enter the number of items you wish to select and press enter.
Example 1
The objective of a 9 x 9 Sudoku number puzzle is to fill the grid so that each row, each column, and each 3 x 3 grid
contains the digits 1 to 9, with no repeats. How many different ways can the first row of a blank 9 x 9 Sudoku grid be
filled.
SOLUTION:
There are
digits that could fill the first spot,
for the second spot, and so forth.
This can be expressed as either
, or as
.
9! =
.
9P9 also equals
So, there are
ways to fill the first row of a 9 x 9 Sudoku number puzzle.
Example 2
Find the number of ways of forming three-digit codes in which no digit is repeated.
SOLUTION:
To form a three-digit code with no repeating digits, you need to select
digits from a group of
, so n =
and r =
.
.
10P3 =
You could also simply say that you had
choices for the first digit,
choices for the second digit, and
choices for the third digit.
*
*
=
.
Example 3
Forty-three race cars started the 2007 Daytona 500. How many ways can the cars finish first, second, and third?
SOLUTION:
=
*
*
=
cars could win the race. Once the winner has won, there are
left to fight for 2nd. Once 2nd has been
rd
determined, there are
left to battle it out for 3 .
Example 4
A building contractor is planning to develop a sub-division. The sub-division is to consist of 6 one-story houses, 4 twostory houses and 2 split-level houses. In how many distinguishable ways can the houses be arranged?
SOLUTION:
12!
=
different distinguishable ways to arrange the houses.
6!4!2!
Example 5
A state’s department of transportation plans to develop a new section of interstate highway and receives 16 bids for
the project. The state plans to hire four of the bidding companies. How many different combinations of four companies
can be selected from the 16 bidding companies?
SOLUTION:
=
.
There are
different combinations of 4 companies that can be selected from the 16 bidding companies.
Example 6
A student advisory board consists of 17 members. Three members serve as the board’s chair, secretary, and
webmaster. Each member is equally likely to serve any of the positions. What is the probability of selecting at random
the three members that hold each position?
SOLUTION:
There are
different ways that the three positions can be filled.
There are
different ways that the three positions can be filled.
1
The chance that you randomly pick the one correct outcome is 4080 ≈
Example 7
You have 11 letters consisting of one M, four I’s, four S’s and two P’s. If the letters are randomly arranged in order,
what is the probability that the arrangement spells the word Mississippi?
SOLUTION:
This is a distinguishable permutation question.
11!
There are 1!4!4!2! different distinguishable ways to arrange those 11 letters.
There are
different distinguishable ways to arrange those 11 letters.
1
The probability of randomly picking the one arrangement that spells Mississippi is 34650 ≈
.
Example 8
Find the probability of being dealt five diamonds from a standard deck of playing cards.
SOLUTION:
You need to determine how many ways you can get 5 diamonds, and then how many different 5 card hands are
possible.
Once you have these values,
the number of ways to get 5 diamonds by the number of possible hands
to find out how likely it is that you get 5 diamonds.
and 52C5 =
13C5 =
1287
P(5 diamonds) =
≈
.
2,598,960
The probability of getting a diamond flush is approximately
.
Example 9
A food manufacturer is analyzing a sample of 400 corn kernels for the presence of a toxin. In this sample, three kernels
have dangerously high levels of the toxin. If four kernels are randomly selected from the sample, what is the probability
that exactly one kernel contains a dangerously high level of the toxin?
SOLUTION:
Find the number of possible ways to choose one toxic kernel and three non-toxic kernels,
those
together and
the answer by the number of ways to choose 4 kernels. This tells you how likely it is that
you will randomly choose exactly one toxic kernel out of four.
=
and
=
3 * 10,349,790 =
There are
different ways to select one toxic kernel out of 400.
=
There are
different ways to select 4 kernels out of 400.
31,049,370
P(exactly 1 toxic kernel) = 1,050,739,900 ≈
.
The probability of getting exactly one toxic kernel is approximately
.