UNIT SIX
COMBINATORICS AND PROBABILITY MATH 621B
12 HOURS
Grade 7
- identify situations for which the probability would be near 0 1 4
1 2
3
4 1
, , , , and
- solve probability problems, using simulations and by conducting experiments
- identify all possible outcomes of two independent events, using tree diagrams and area models - create and solve problems, using the numerical definition of probability
- compare experimental results with theoretical results
- use fractions, decimals and percents as numerical expressions to describe probability
Grade 8
- conduct experiments and simulations to find probabilities of single and complementary events - determine theoretical probabilities of single and complementary events
- compare experimental and theoretical probabilities
- demonstrate an understanding of how data is used to establish broad probability patterns
Grade 9
- make predictions of probabilities involving dependent and independent events by designing and conducting experiments and simulations
- determine theoretical probabilities of independent and dependent events
- demonstrate an understanding of how experimental and theoretical probabilities are related - recognize and explain why decisions based on probabilities may be combinations of theoretical calculations, experimental results and subjective judgements
Combinatorics (from NCTM Standards) is the mathematics of systematic counting.
G20 demonstrate an understanding of the Fundamental Counting Principle
1. Count the number of ways that customers at Kent’s Deli can construct a sandwich. They have a choice of 3 types of meat - ham, chicken or beef. They also have a choice of 2 types of bread - white or rye.
a) construct a tree diagram by choosing bread first then meat second.
b) construct a tree diagram by choosing meat first then bread second.
c) how do the answers from parts (a) and (b) compare?
2. At Bonnie’s Deli customers have a wider selection. They can choose from 4 types of meat - ham, chicken, beef or salami and 3 choices of bread - white, rye or brown. Use a tree diagram to illustrate all possible types of sandwiches.
3. From the above problems make a conjecture as to the patten developing. Write a conjecture that explains how to solve counting problems where there are different levels of choices to be made.
This conjecture is called the Fundamental Counting Principle.
12, students will be expected to:
G20 demonstrate an understanding of the FCP
Fundamental Counting Principle (7.1)
Invite student groups to do the Explore and Inquire on p.334.
Student groups should read and discuss p.335-6. Students should understand how tree diagrams can be streamlined into the FCP greatly simplifies the solving of problems.
Ex. How many license plates can be generated containing 4 letters of the alphabet?
a) where repetitions are not allowed.
b) where repetitions are allowed.
a) 26 × 25 × 24 × 23 = 358,800
in the first blank any of the 26 letters can be placed
in the second blank any of the remaining 25 letters can be placed in the third blank any of the remaining 24 letters can be placed in the fourth blank any of the remaining 23 letters can be placed Note to Teachers: This is a permutation problem 26 P4 .
On the TI-83 press 26 math < < < PRB 2: nPr 4.
b) 26 × 26 × 26 × 26 = 456,976
in the first blank any of the 26 letters can be placed in the second blank any of the 26 letters can be placed in the third blank any of the 26 letters can be placed in the fourth blank any of the 26 letters can be placed
Fundamental Counting Principle (7.1) Group Discussion
A series of recipes is written on index cards. The recipes are either for appetizer, entree or dessert. There are 5 appetizer cards, 7 for entrees and 9 for desserts. In how many ways can a three course meal be selected from the menu?
Journal
What is the major advantage of using the FCP instead of tree diagrams to solve counting problems?
Activity
A store carries 8 different sizes of shirts, each size is available in men’s and women’s designs. Each is made with either cotton, silk or micro-fiber. Each shirt is either a solid colour or a print design. How many different kinds of shirts does the store carry?
Group Activity
The block diagram shown represents nine city blocks, with the lines indicating streets. A fire truck at point A wishes to travel to point B by travelling only North or East. Use a tree diagram to find the number of different routes this truck could take.
(See tree diagram at the end of the unit)
Fundamental Counting Principle (7.1)
Math Power 12 p.336 #1-7 odd, 8-14 even
12, students will be expected to:
A19 describe what a permutation is, how to evaluate a permutation and how to use permutations to solve problems
Permutations (7.2)
Definition: The process or result of making or becoming different.
Synonyms: change, variation, modification, mutation.
Factorial notation should be introduced to students and its calculation on scientific calculators (and the TI-83) demonstrated.
5 math < < < PRB 4: ! enter then press enter again to evaluate 5 !
Invite students to read p.338-341 and do the Explore and Inquire.
A key concept is that in permutations “order is important”. An example will illustrate what this means:
Five people are to get their picture taken. If the people are moved to different positions then these are considered to be mutations or
permutations. There are considered to be 5! or 120 different ways( 5P5) (permutations) of arranging these 5 people for the photo.
In the next section on combinations, “order is not important” and no matter how the people are re-positioned the same 5 people are in the picture and this situation is said to have only 1 combination.
There are two mutations of permutations:
1) no repetitions allowed:
a) n distinct objects taken n at a time (essentially this is a FCP problem) n! or nPn
b) n distinct objects taken r at a time.
n rP n n r
= −
! ( ) !
Note to Teachers: The familiar permutation formula only works for no repetitions.
n rP n
n r
= −
! ( ) !
2) repetitions allowed
n distinct objects taken n at a time, where a objects are alike, b objects are alike...
n a b
!
! !. . .
Permutations (7.2) Activity
Express 7 × 6 × 5 × 4 × 3 × 2 × 1 in factorial notation.
Pencil/Paper/Technology Determine the value of:
a) 4!
b) 9P6 c) 8P3 Activity
A baseball team has 9 players on the field. In how many ways can the 9 players on the field be re-positioned?
Group Discussion
The block diagram shown represents nine city blocks, with the lines indicating streets. A fire truck at point A wishes to travel to point B by travelling only North or East. Use permutation theory to find the number of different routes this person could take.
Note to Teachers: This is the same problem as in section 7.1 on the FCP. It is a case 2 permutation problem (repetitions are allowed). A person must travel 3 blocks East and 3 blocks North ( for example EENENN). So one must make a total of 6 moves where there are 3 East moves (3 repetitions of East) and 3 North moves (3 repetitions of North).
Permutations (7.2)
Do Permutation Worksheet at end of unit.
Math Power 12 p.342 #1-19 odd Applications p.342 # 21,30
Note to Teachers: Why is 0! = 1?
The FCP predicts that if 6 objects are chosen from a list of 6 objects then they can be chosen in 6 × 5 × 4 × 3
× 2 × 1 = 6! Ways.
This is also a permutations problem and can be solved using:
6 6
6
6 6
6 0
P n
n r
= −
= −
=
!
( ) !
!
( ) !
!
!
Which means that the denominator 0!
must be equivalent to 1.
12, students will be expected to:
B24 analyze and count the number of permutations and combinations possible in various situations
G14 describe what a combination is and how to evaluate a combination and use them to solve problems
G15 distinguish between combinations and permutations
Combinations (7.3)
Challenge student groups to do the Explore and Inquire on p.345.
Invite student groups to read and discuss p.345-348.
In the Explore on p.345 the tree diagram looks like this:
From 5 numbers we pick 2 numbers with no repetitions and order is not important.
Therefore we can do 5C2 = 10
A combination is a selection of objects where order is not important.
nCr n rP r
n n r r
= =
−
!
! ( ) ! !
For a real world example illustrating the difference between
permutations and combinations see the explanation sheet at the end of the unit.
Ex. A five person committee is to be selected from a class with 10 women and 8 men. In how many ways can a committee be formed if it must contain 3 women?
Solution:
10C3 × 8C2 = 3360 ways to form the committee
Combinations (7.3) Group Discussion
There are 2000 CD players in a company warehouse. The shipper is going to select 40 of these using their serial numbers to check for defects. In how many ways can the shipper choose the 40 players?
Group Activity
A salesperson handles 25 brands of golf clubs. At a trade show, he has, however, space to display only 18 of the brands.
In how many ways can the display be set up?
Group Activity
In a class of 35 students, in how many ways can 8 people be chosen to work on the decorations committee for the
graduation dance? If 13 people say that they don’t want to be on that committee, how many ways can the 8 people be chosen?
Journal
Mary says “A permutation is the same as a combination.”
Write to explain that her statement is incorrect.
Group Discussion
The block diagram shown below represents nine city blocks, with the lines indicating streets. A fire truck at point A wishes to travel to point B by going only North or East. How many different ways can the truck go from A to B?
Combinations (7.3)
Math Power 12 p.348 #1,3,5,20,19
Applications p.348 #15-18,21,22,23
12, students will be expected to:
G17 connect Pascal’s Triangle with combinational coefficients
Pascal’s Triangle (7.4)
Invite student groups to read and discuss p.350-352 and do the Explore and Inquire.
See the connections between Combinations and Pascal’s Triangle sheet at the end of the unit.
The block problems we have been looking at using FCP, Permutations and Combinations can be done as well using Pascal’s Triangle.
Consider the nine city block problem that we having been looking at:
Looking at this block diagram superimposed on Pascal’s Triangle we immediately see the solution.
Encourage student groups to realize that Pascal’s Triangle is generated from combinations as shown on the top of p.351.
The pathway problems like Ex. 1 and 2 on p.351 can be solved using Pascal’s Triangle. The solution is the sum of the numbers in the correct row of Pascal’s Triangle.
Pascal’s Triangle (7.4) Group Presentation
Determine the number of ways to go from A to B by travelling only East or North if the street pattern is a 4 × 4 square pattern using:
a) FCP
b) permutations c) combinations d) Pascal’s Triangle
Activity/Discussion
Julie has invited 8 people to a birthday party at her house. On the invitation sent out she did not ask for a RSVP, so she is not sure of how many will come to the party. How many combinations of guests could come? How does the answer of (a) relate to Pascal’s Triangle?
Journal
Explain how Pascal’s Triangle could have been used to solve Julie’s problem.
Research
Write a short essay on the life and contributions of Blaise Pascal.
Pascal’s Triangle (7.4)
Math Power 12 p.352 # 1-3,5-8
Pascal’s Triangle Worksheet (7.4)
12, students will be expected to:
G16 demonstrate an understanding of binomial expansion and its connection to combinations
The Binomial Theorem (7.5)
Student groups should do the Explore and Inquire on p.354. Challenge student groups to read and discuss p.355-356. Students should
understand that combinations are used to determine the coefficients of an expanded binomial.
The Binomial Theorem states:
(a b) C a b
where r is the number of the term
n n
r n
r n r r
+ =
=
∑ − 0
The general term can be obtained using: tn+1=nC ar n r r− b
Ex. Find the 7th term of (x ! 3)11. Solution:
To find the 7th term; r = 6, a = x, b = !3, n = 11
nC ar n rbr
C x x
x
−
− −
×
11 6 11 6 6
5
5
3
462 729
336798
( )
Note to Teachers: Most texts stop at very simple and abstract examples of the Binomial Theorem. At the end of the unit are worksheets on basic probability and applied Binomial Theorem problems. It assumes students have a basic knowledge of probability; which students should have obtained in Junior High and Grade 10 Data Management For binomial events: ie. either a success or a failure occurs.
The formula for binomial application problems is:
let F = the probability of failure let S = the probability of success
(F + S)n=nC Fs n s− Ss
The Binomial Theorem (7.5) Pencil/Paper
Use Pascal’s Triangle and the pattern of decreasing powers of a and increasing powers of b to expand:
a) (x ! 3)6 b) (2x + 5y)5
Group Activity
Use the Binomial Theorem to expand:
a) (x + 2)6 b) (3x ! 2y)4
Group Activity
Use the general term of the Binomial Theorem to determine:
a) the 4th term of (x + 5)8 b) the 9th term of (x ! 2y)13
Group Presentation
A true-false test has 12 questions. Suppose all questions are guessed. Determine the probability of guessing 6 correct answers.
Group Activity
About 9% of women will contract breast cancer sometime in their lifetimes. There are 100 women at a sales meeting. Find the probability that exactly four of them will contract breast cancer in their lifetimes.
A company manufactures computer chips. About 2% of the chips are defective. A quality control inspector pulls 50 chips at random off an assembly line. Find the probability that exactly 10 of them are defective.
The Binomial Theorem (7.5)
Math Power 12 p.356#1-21odd, 25,27,30
Binomial Theorem Worksheet
12, students will be expected to:
G11 apply area and tree diagrams to interpret and determine probabilities
G12 find probabilities for sequences of 2 or more events
G21 apply the FCP to probability
Probability & Sample Space (8.1)
As an introduction to probability, students should do the Probability Worksheet at the end of the unit on probability of single events.
Challenge student groups to do the Explore and Inquire on p.368.
Invite student groups to read and discuss p.369-371. Students should be able to express probabilities as common fractions, decimals and
percents.
Students should be able to combine the concepts of probability and tree diagrams to form a probability tree diagram.
Invite students to discuss the concepts of:
< outcomes - things that may happen in an experiment
< event - a collection of outcomes (a subset of the sample space) < sample space - the set of all possible outcomes of an experiment < complementary events - two events whose sum is 1.
Note to the Teachers: Some of the problems in this section are involved with making more than one selection. These are called compound events and will be dealt with in more depth in section 8.2.
For a simple example, see Ex. 1 , p.369. A second example is below:
Ex. 10 p.372
The probability that both trades are with the US is 0.494.
The probability that neither trade is with the US is 0.084.
The probability that the export is to the US but the import is not from the US is 0.266.
Note that the sum of all probabilities in the sample space must equal 1.
Probability & Sample Space (8.1) Pencil/Paper
Use a probability tree to determine the probability of the following events:
a) obtaining a tail and a queen of hearts. (1/2 × 1/52 = 1/104) b) rolling a 5 and obtaining a head. (1/6 × ½ = 1/12)
c) picking a day of the week and an hour of the day. (1/7 × 1/24 = 1/168)
Research
Do research to determine the percentage of Islanders that are of Acadian descent. Statistics Canada claims that about 12%
of Canadian adults have a university degree. What is the probability that:
a) an Island Acadian has a university degree.
b) an Island non-Acadian has a university degree.
c) an Island Acadian hasn’t a university degree.
d) an Island non-Acadian hasn’t a university degree.
e) What is the sum of the probabilities for parts (a) to (d)? Is this the entire sample space?
Journal
Write to explain the differences between an outcome, an event and a sample space.
Group Activity
In your class, what is the probability that:
a) A person selected is a female and has brown hair?
b) A person selected is male and has blue eyes?
Probability & Sample Space (8.1) Probability Worksheet
Math Power 12 p.371 #1-7,9,10
12, students will be expected to:
G8 demonstrate an understanding of “and” “or” and “complementary” as they relate to probability
G9 demonstrate an understanding and apply complementary probability
G12 find probabilities for sequences of 2 or more events
Classifying Events (8.2)
Note to Teachers: This section has to do with making more than one selection (compound events). They are classified below. At the end of the unit is a more detailed explanation of the classifications. This topic should be dealt with carefully and kept to an introductory level.
Invite student groups to read and discuss p.375-379 and do the Explore and Inquire on p.375.
Classifications for discussion are:
< independent vs dependent events
< mutually exclusive and non-mutually exclusive events < complementary events
both selections must be made
For independent: P(A & B) = P(A) × P(B) For dependent: P(A & B) = P(A) × P(B *A)
Note: P(B*A) deals with conditional probability and will be dealt with later in the unit.
one selection or the other
must be made
For mutually exclusive: P(A or B) = P(A) + P(B)
For non-mutually exclusive: P(A or B) = P(A) + P(B) ! P(A & B)
Classifying Events (8.2) Group Activity/Presentation
Classify the events as either dependent or independent:
a) selecting 2 blue socks, one at a time, from a drawer that contains 6 black socks and 4 blue socks.
b) selecting 2 drinks from a cooler, one at a time, that contains 2 bottles of pepsi and 4 bottles of juice.
c) select a card from a deck of cards, replace the card, then make a second selection.
d) roll two dice and get a sum of 7, then roll a second time and get a sum of 4.
Group Activity/Presentation
Classify the events as either mutually exclusive or not mutually exclusive:
a) selecting a penny or a dime from 4 pennies, 3 nickels and 6 dimes.
b) selecting a boy or a brown-haired person from 12 girls, 5 of whom have brown hair , and 15 boys, 6 of whom have brown hair.
c) selecting a king or queen from a deck of cards.
d) selecting a card from a deck and it being either red or a face card.
Classifying Events (8.2)
Note to Teachers: At the end of the unit is a more detailed explanation of section 8.2 and accompanying
exercises to help point out the differences.
Math Power 12 p.380 # 1-13
12, students will be expected to:
G12 find probabilities for sequences of 2 or more events
Probability & Combinatorics (8.3)
Student groups should do the Explore and Inquire on p.382.
Invite student groups to read and discuss p.382-385. When tree diagrams become too complicated, combinatorics can be used to
determine the size of the sample space and the size of the event set. The problems in section 8.2 that were dependent employed conditional probability. These problems can be solved using combinatorics as well.
Students should do the Combinatorics Worksheet at the end of the unit.
Ex.1 Find the probability that 2 cards drawn, without replacement, are both face cards.
Solution:
The phrase: “without replacement” makes this a dependent situation.
Method 1:
P(A) = 12/52 = 3/13; P(B*A) = 11/51 P(A & B) = P(A) × P(B*A) = .0498 Method 2:
Event set: From the 12 face cards we need to choose, in any order, 2 face cards: 12C2.
Sample space: From the total 52 cards we choose 2 cards: 52C2
P face cards C
(2 ) 12C2 .0498
52 2
= =
Ex. 2 Four door prizes are awarded at a party. 26 people hold one ticket each. a) What is the probability that Sara, Jane and Mary win the first, second and third prizes in that order? b) What is the probability that the same 3 people win the prizes but not necessarily in that order?
a) P Sara Jane Mary
( , , ) = 1P = 1
15600
26 3
b) P in any order
( ) P!
3 3 6
15600
26 3
= =
There are 3! = 6 ways of arranging Sara, Jane and Mary and therefore 6 times as much of a chance of these 3 people winning if they don’t have to win in a specific order.
Probability & Combinatorics (8.3) Pencil/Paper
A committee of 5 people is to be selected from your class.
Determine the number of women and men in your class and find the probability that there are exactly 3 women on the committee.
Discussion
There are 7 women and 5 men in a room. A committee of 4 is to be randomly selected. What is the probability of there being:
a) 4 women on the committee.
b) 3 women on the committee.
c) 2 women on the committee.
d) 4 men on the committee.
e) 3 men on the committee.
f) 2 men on the committee.
Group Activity
Three balls are randomly drawn from a bag containing 6 white balls and 4 black balls. What is the probability of drawing:
a) 3 white balls.
b) 2 white balls and 1 black ball.
c) 1 white ball and 2 black balls d) 3 black balls
Group Discussion
Four people are to be randomly selected from a group of 9 boys and 7 girls. What is the probability of each event?
a) exactly 3 being girls.
b) all 4 being boys.
c) 2 being girls.
Group Activity
Four balls are randomly drawn from a bag containing 3 white balls and 5 black balls. What is the probability that there are exactly 2 white balls?
Presentation
Probability & Combinatorics (8.3) Math Power 12 p.385 #1-3, 10-12 Combinatorics Worksheet at end of unit
12, students will be expected to:
G12 find probabilities for sequences of 2 or more events
Applications of Conditional Probability (8.4) Invite student groups to read and discuss p.388-391.
Conditional Probability is the probability that an event will occur given that another event has occurred.
For two events A and B:
a) if they are independent, then the joint probability, P(A & B), of them both happening is:
P(A & B) = P(A) × P(B)
Note that if the events are independent, the second event in unaffected by the first event happening. The expression on the left above, P(A & B) could have been written as P(B & A). A particular event does not have to happen before the other. Hence conditional probability does not come into play here.
b) if they are dependent, then the joint probability, P(A & B), of them both happening is:
P(A & B) = P(A) × P(B*A) or P(B & A) = P(B) × P(A*B)
The joint probabilities P(A & B) and P(B & A) are equal. Thus from the above two equations we can get:
P B A P A B P A
P B P A B ( ) ( & ) P A
( )
( ) ( )
= = ×( )
or P A B P A B P B
P A P B A ( ) ( & ) P B
( )
( ) ( )
= = ×( )
This is known as BAYES’ Law or Theorem.
Note to Teachers: This may be rather abstract for students. Conditional probability problems can be solved using a joint probability table. The conditional probability worksheet has solutions done both ways.
Students may very well find that using the table is a simpler and more understandable procedure.
Applications of Conditional Probability (8.4) Activity
A household is considered prosperous if its income exceeds
$75,000. A household is considered educated if the
householder has completed college. A household is selected at random and event A is: the selected household is
prosperous, and event B is: the household is educated.
According to Statistics Canada P(A) = 0.15, P(B) = 0.25 and P(A & B) = 0.09. What is the probability of each event?
a) a household is educated, given that it is prosperous.
b) a householder is prosperous, given that it is educated.
Group Discussion
In the table below are the numbers (in hundreds) of earned degrees in Canada in a recent year, classified by level and by gender of the degree recipient.
a) If a degree recipient is chosen at random, what is the probability that the person chosen is a women?
b) What is the probability that a woman was chosen, given that the person chosen had received a professional degree?
c) What is the probability that the person selected had received a Bachelor’s degree, given that the person chosen was male?
Applications of Conditional Probability (8.4)
Math Power 12 p.391 #1-3
Applications of Conditional
Probability Worksheet at the end of the unit.
This represents only the top half (10 paths) in the tree diagram. Students should quickly see that more efficient methods must be found to solve problems like this. This problem will be revisited in
Permutations (7.2) where students will write the solution as:
# !
paths = 6! ! 3 3
no repetitions allowed
1. In how many ways can 6 people be arranged in a row for a photo shoot? 6P6 = 720
2. In how many ways can 5 students be selected from a class containing 30 students? 30P5 = 17,100,720 repetitions allowed
3. In how many ways can 6 people be arranged in a row for a photo shoot if there are 2 identical twins in the group? 6!/2! = 360
4. How many distinctly different arrangements can be made from the following words:
a) selected 8!/3! = 6720 b) license 7!/2! = 2520 c) initials 8!/3! = 6720 d) letters 7!/2!2! = 1260
e) arrangements 12!/2!2!2!2! = 29,937,600 f) mississippi 11!/4!4!2! = 34650
5. Rearrange the letters to form as many words as possible a) mslie
b) pto c) iter d) opst
Note to Teachers: One method is to write all possible arrangements(permutations) and see which ones make words. Ex: pto 6 pto, opt, top, otp, tpo, pot
6. For a four letter word (no repetitions) there are 4! or 24 different arrangements. Write out all 24 arrangements for the letters opst.
7. Write to explain what is meant by “ order is important”.
For a group of 10 people, a 3 person committee is to be chosen having a president, a vice president and a secretary. In how many different ways can the committee be chosen if:
a) we will pick the 3 people to be on committee now and at some later time decide which will be president and so on.
b) the first person chosen is to be president, the second is to be VP and the third is to be secretary.
a) Here order is not important (combinations) because if from the 10 people we pick Jane, Bob and Mary it doesn’t make any difference who is chosen first, second or third. We don’t have the added restriction that the first has to be president and so on. Usually adding extra conditions reduces the number of answers but just the opposite happens here. Adding restrictions makes different groupings of the same three people distinguishably different creating more possible answers. Thus the number of ways (combinations) is:
10C3 = 120
b) Here order is important (permutations) and the number of ways to pick the committee is : From the FCP we get : 10 × 9 × 8 = 720
or from permutations we get 10P3 = 720 In this scenario if we had picked:
1st pick < Jane < she would be president 2nd pick < Mary < she would be VP 3rd pick < Bob < he would be secretary on the other hand if we had picked:
1st pick < Mary < she would now be president 2nd pick < Bob < he would be VP
3rd pick < Jane < she would be secretary
The same three people were picked from the 10 to choose from but each of the two scenarios yields different committees. Thus the order in which the people are picked is important.
How many different ways can 3 people be arranged for a picture or a committee:
FCP = 3 × 2 × 1 = 6 or 3P3 = 6
So in part (a) there is only one combination that has Jane, Mary and Bob but this one combination yields
The 20 in Pascal’s Triangle is found using 6C3,
Join the following series of points with as many different line segments as possible (including diagonals).
Complete the accompanying table.
Predict the number of line segments that can be drawn for an octagon. Verify your answer by drawing an octagon and all possible line segments.
Predict the number of line segments that can be drawn for a dodecagon (12 sided figure).
Predict the number of diagonals that can be drawn for an octagon. Verify your answer by drawing an octagon and all possible diagonals.
a) a face card. 12/52
b) a black, non-face card. 20/52 c) an ace 4/52
2. A box contains 5 yellow, 6 blue and 4 black marbles. What is the probability that the marble selected will be:
a) black 4/15 b) not be blue 9/15 c) green 0
d) yellow 5/15
3. One flower is selected at random from a vase containing 5 red flowers, 2 white flowers and 3 orange flowers. Find the probability of each:
a) P(red) 5/10 b) P(not orange) 7/10 c) P(white) 2/10 d) P(not white) 8/10
1. A pitcher throw 4 strikes for every 7 pitches. What is the probability that 3 of the next four pitches will be strikes?
2. A baseball player’s batting average is .200 . Find the probability that for the next 5 times at bat:
a) he has 1 hit.
b) he has 3 hits.
3. A true-false test has 12 questions. Suppose all questions are guessed. Determine the probability of guessing 5 correct answers.
4. About 9% of women will contract breast cancer sometime in their lifetimes. There are 100 women at a sales meeting. Find the probability that exactly four of them will contract breast cancer in their lifetimes.
5. A company manufactures computer chips. About 2% of the chips are defective. A quality control inspector pulls 50 chips at random off an assembly line. Find the probability that exactly 10 of them are defective.
6. A multiple-choice test has 12 questions. Each question has 5 choices, only one of which is correct. All questions are guessed at. What is the probability of getting 3 correct answers? 7 correct answers?
7. An archer can hit the bullseye 80% of the time. What is the probability that the archer can hit the bullseye 9 times out of 10.
8. The Twins and the Orioles are rival baseball teams. In the past few years, the Twins have win 60% of the games between the two teams. What is the probability that the Twins will win 5 of the next 7 games?
9. The Flames and the Avalanche are in the Stanley Cup Finals. Each team hopes to win the best-of- seven series. The probability of an Avalanche win is 70%. What is the probability that the Avalanche will win in 5 games?
10. Eight out of every 10 persons who contract a certain viral infection can recover. If a group of 7 people become infected, what is the probability that 3 will recover?
11. During the Gulf War in 1990-1991, SCUD missiles hit 20% of their targets. In one attack six missiles were fired at a fuel storage installation. What is the probability that 4 of them hit the target?
1. P(F) = 4/7; P(S) = 3/7;
( ) ( )
.
F S
C
+ n = +
=
F
H GIKJFH GIKJ
= 3 7
4 7 3 7
4 7 32
4
4 3
1 3
2. P(F) = .8; P(S) = .2;
a)
(. . )
( ) (. ) (. ) .
8 2
1 8 2
4096
5
5 1 4 1
+
=
= P hit C
b)
(. . )
( ) (. ) (. ) .
8 2
3 8 2
0512
5
5 3 2 3
+
=
= P hits C
3. P(F) = .5; P(S) = .5 (. . )
( ) (. ) (. )
. 5 5
5 5 5
1934
12
12 5 7 5
+
=
=
P correct C
4. P(F) = .91; P(S) = .09
(. . )
( ) (. ) (. )
. 91 09
4 91 09
03
100
100 4 96 4
+
=
= P will get cancer C
(. . )
( ) (. ) (. )
. 98 02
10 98 02
4 688 10
50
50 10 40 10
8
+
=
= × −
P defective C
6. P(F) = .8; P(S) = .2
P( correct) C (. ) (. ) .
3 8 2
2362
12 3 9 3
=
=
P( correct) C (. ) (. ) .
7 8 2
0033
12 7 5 7
=
=
This problem an all the others can be solved with the TI-83.
The formula for the general term of a binomial expression is nC ar n r r− b
For this problem; n = 12; r = x; a = .8 and b = .2 . The window must have the horizontal dimension be 94 divided by some integral value (I used 10 here). From the table the answer can be see to be .00332 and similarly from the graph we can see the same answer.
Note to Teachers: The graph shows the shape of this particular binomial distribution. If you are also teaching Math 621A these, along with normal distributions, will be done in more detail in Unit 6.
(. . )
( ) (. ) (. )
. 2 8
9 2 8
2684
10
10 9 1 9
+
=
= P bullseyes C
8. P(F) = .4; P(S)== .6
(. . )
( ) (. ) (. ) .
4 6
5 4 6
2613
7
7 5 2 5
+
=
= P wins C
9. P(F) = .3; P(S) = .7
(. . )
( ) (. ) (. ) .
3 7
5 3 7
3177
7
7 5 2 5
+
=
= P wins C
10. P(F) = .2; P(S) = .8
(. . )
( ) (. ) (. ) .
2 8
3 2 8
0287
7
7 3 4 3
+
=
= P live C
11. P(F) = .8; P(S) = .2
(. . )
( ) (. ) (. ) .
8 2
4 8 2
0154
6
6 4 2 4
+
=
= P hits C
12. P(F) = .9; P(S) = .1
(. . )
( ) (. ) (. )
. 9 1
8 9 1
3558 10
20
20 8 12 8
4
+
=
= × −
P with disease C
Independent vs Dependent Events
Independent events are when the first event does not affect the second event.
Dependent events are when the first event does affect the second event.
Ex. Find the probability of selecting a face card and then selecting an ace:
a) when the first card is replaced in the deck.
b) when the first card is not replaced in the deck.
Solution:
a) Event A does not affect event B and these events are therefore independent. P(A & B) = P(A) × P(B).
For event A: P(A) = 12/52 = 3/13 For event B: P(B) = 4/52 = 1/13 Therefore P(A & B) = 3/13 × 1/13 = 3/169 = .0178
b) Event A does affect event B and these events are thus dependent. P(A & B) = P(A) × P(B*A).
P(B*A) is known as the conditional probability of event B happening, given that event A has already happened. Here we will deal with only simple conditional probability problems. This topic will be dealt with in more detail in section 8.4.
For event A: P(A) = 12/52 = 3/13 For event B: P(B*A) = 4/51 Therefore P(A & B) = 3/13 × 4/51 = 4/221 = .0181
Independent/Dependent exercises
Determine if each event is independent or dependent. Then determine the probability.
1. Statistics collected in a particular area show that the probability that a person will develop diabetes is 5/11.As well, the probability that a person in that same area will develop arthritis is 1/5. If one health problem does not affect the other, what is the probability that a randomly selected person from that area of Canada will not develop diabetes but will develop arthritis.
2. Determine the probability of rolling a sum of 7 on the first toss of two dice and a sum of 4 on the second toss.
6. Determine the probability of selecting a blue marble, without replacement, then a yellow marble from a box that contains 5 blue marbles and 4 yellow marbles.
7. Determine the probability of randomly selecting 2 oranges, one at a time without replacement, from a bowl of 5 oranges and 4 tangerines, if the first selection is replaced.
8. A green number cube and a red number cube is tossed. What is the probability that a 4 is shown on the green cube and a 5 is shown on the red cube?
9. Find the probability of randomly taking 2 blue notebooks, one at a time without replacement, from a shelf which has 4 blue and 3 black notebooks.
10. A bag contains 4 nickels, 4 dimes and 7 quarters. Three coins are removed in sequence without replacement. What is the probability of selecting a nickel, a dime and a quarter in that order?
Note to teachers: If the phrase “in that order” is not there, then there are 3! = 6 different orders of choosing a nickel, a dime and a quarter. The probability would then be 6 × as great. This idea applies once you start working with different objects/people.
11. What is the probability of removing 13 cards from a deck of cards, one at a time without replacement, and have all of them red?
12. What is the probability of randomly selecting a knife, a fork and a spoon in that order, without replacement, from a kitchen drawer containing 8 spoons, 8 forks and 12 knives? (Note that the order of selecting is given).
13. Determine the probability of selecting 3 different coloured crayons from a box containing 5 red, 4 black and 7 blue crayons, assuming each crayon is replaced?
14. Find the probability of drawing a black marble, replacing it, then drawing a yellow marble from a bag with 3 yellow and 5 black marbles.
15. Determine the probability of rolling a sum of 2 and the first toss of two dice and a sum of 6 on the second toss. (Note order is given).
16. Determine the probability of randomly selecting 2 yellow markers from a box, one at a time without replacement, that contains 4 yellow and 6 pink markers.
17. A box contains 3 black and 9 white balls. A ball is drawn and not replaced. A second ball is drawn from the box. What is the probability that both balls drawn are black?
18. A box contains 3 black and 9 white balls. A ball is drawn and replaced. A second ball is drawn from the box. What is the probability that both balls drawn are black?
1. independent. P(A) = 1 ! 5/11; P(B) = 1/5; P(A & B) = 6/55 2. independent. P(A) = 6/36; P(B) = 3/36; P(A & B) = 1/72 3. dependent. P(A) = 3/9; P(B*A) = 2/8; P(A & B) = 1/12 4. dependent. P(A) = 5/12; P(B*A) = 4/11; P(A & B) = 5/33 5. dependent. P(A) = 4/6; P(B*A) = 3/5; P(A & B) = 2/5 6. dependent. P(A) = 5/9; P(B*A) = 4/8; P(A & B) = 5/18 7. independent. P(A) = 5/9; P(B) = 5/9; P(A & B) = 25/81 8. independent. P(A) = 1/6; P(B) = 1/6; P(A & B) = 1/36 9. dependent. P(A) = 4/7; P(B*A) = 3/6; P(A & B) = 2/7
10. dependent. P(A) = 4/15; P(B*A) = 4/14; P(C*A&B) = 7/13; P(A & B & C) = 8/195 11. dependent. P = 19/1,160,054
12. dependent. P(A) = 12/28; P(B*A) = 8/27; P(C*A&B) = 8/26; P(A & B & C) = 32/819 13. independent. P = 35/1024
14. independent. P(A) = 5/8; P(B) = 3/8; P(A & B) = 15/64 15. independent. P(A) = 1/36; P(B) = 5/36; P(A & B) = 5/1296 16. dependent. P(A) = 2/10; P(B*A) = 1/9; P(A & B) = 1/45 17. dependent. P(A) = 3/12; P(B*A) = 2/11; P(A & B) = 1/22 18. independent. P(a) = 3/12; P(B) = 3/12; P(A & B) = 1/16
Mutually Exclusive vs Non-Mutually Exclusive Events
Mutually Exclusive events are two events that share no common outcomes. That is they can’t happen at the same time.
P(A or B) = P(A) + P(B)
Non-Mutually Exclusive events are two events that may share common outcomes. That is the two sets of outcomes may have an intersection set.
P(A or B) = P(A) + P(B) ! P(A & B) Example of Mutually Exclusive:
In a contest, Jim can win a car if he selects a blue ball or a red ball. He must select the ball from a box containing 2 blue, 3 red, 9 yellow and 11 green balls. What is the probability of his winning the car?
Solution:
Theses are mutually exclusive events. He is allowed to make only one draw and he is permitted some flexibility in what he can draw to win.
Event A: P(A) = 2/25; Event B: P(B) = 3/25 P(A or B) = 2/25 + 3/15 = 1/5.
The probability of winning the car is 1/5.
Example of Non-Mutually Exclusive:
What is the probability of rolling a 2 dice where the first die shows a 2 or the sum of the two dice is 6 or 7?
Solution:
It is possible to have a 2 on the first die and have the sum of the two dice being 6 or 7. Therefore these events are non-mutually exclusive.
Event A: P(A) = 6/36 First die showing a 2 Event B: P(B) = 11/36 Sum of the dice is 6 or 7
Event A & B: 2/36 First die showing a 2 and the sum is 6 or 7 P(A) + P(B) ! P(A & B) = 6/36 + 11/36 ! 2/36 = 15/36
The probability of the first die showing a 2 or the sum of the dice being 6 or 7 is 15/36 = 5/12.
Event A & B
P(A or B) = 15/36
Determine if the events are mutually exclusive or non-mutually exclusive. Then determine the probability of each.
1. Find the probability of choosing a penny or a dime from 4 pennies, 3 nickels and 6 dimes.
2. Find the probability of selecting a boy or a blond-haired person from 12 girls, 5 of whom have blond hair, and 15 boys, 6 of whom have blond hair.
3. Find the probability of drawing a king or queen from a standard deck of cards.
4. The probability for a driver’s license applicant to pass the road test the first time is 5/6. The probability of passing the written test on the first attempt is 9/10. The probability of passing both test the first time is 4/5. Are the events mutually exclusive? What is the probability of passing either test on the first attempt?
5.Find the probability of tossing two dice and either on showing a 4.
6. Find the probability of selecting an ace or a red card from a deck of cards.
7. Determine the probability that a card drawn from a deck is red or a face card.
8. Find the probability of two dice being tossed and showing a sum of 6 or a sum of 9.
9. A weather forecaster states that the probability of rain is 3/5, the probability of lightning is 2/5, and the probability of both is 1/5. What is the probability of a sporting event being cancelled due to rain or lightning?
10. A bag contains cards numbered from 1 to 14. One card is drawn at random. Find the probability of:
a) selecting a prime number or a multiple of four.
b) selecting a multiple of two or a multiple of three.
c) selecting a 3 or a 4.
d) selecting an 8 or a number less than 8.
1. mutually exclusive; P(A) = 4/13; P(B) = 6/13; P(A or B) = 4/13 + 6/13 = 10/13 2. non- mutually exclusive; P(A) = 15/27; P(B) 11/27; P(A & B) = 6/27
P(A or B) = 15/27 + 11/27 ! 6/27 = 20/27
3. mutually exclusive; P(A) = 4/52; P(B) = 4/52; P(A or B) = 8/52 = 2/13 4. non mutually exclusive; P(A) = 5/6; P(B) = 9/10; P(A & B) = 4/5 P(A or B) = 5/6 + 9/10 ! 4/5 = 14/15
5. non -mutually exclusive; P(A) = 1/6; P(B) = 1/6; P(A & B) = 1/36 P(A or B) = 1/6 + 1/6 ! 1/36 = 11/36
6. non-mutually exclusive; P(A) = 4/52; P(B) = 26/52; P(A & B) = 2/52 P(A or B) = 4/52 + 26/52 ! 2/52 = 28/52 = 7/13
7. non-mutually exclusive; P(A) = 26/52; P(B) = 12/52; P(A & B) = 6/52 P(A or B) = 26/52 + 12/52 ! 6/52 = 32/52 = 8/13
8. mutually exclusive; P(A) = 5/36; P(B) = 4/36 P(A or B) = 9/36 = 1/4
9. non- mutually exclusive; P(A) = 3/5; P(B) = 2/5; P(A & B) = 1/5 P(A or B) = 3/5 + 2/5 ! 1/5 = 4/5
10.
a) mutually exclusive; P(A) = 6/14; P(B) = 3/14 P(A or B) = 9/14
b) non-mutually exclusive; P(A) = 7/14; P(B) = 4/14; P(A & B) = 2/14 P(A or B) = 7/14 + 4/14 ! 2/14 = 9/14
c) mutually exclusive; P(A) = 1/14; P(B) = 1/14 P(A or B) = 2/14 = 1/7
d) mutually exclusive; P(A) = 1/14; P(B) = 7/14 P(A or B) = 8/14 = 4/7