MDM4UUnit 1 Master
Overview of Unit 1: Introduction to Probability (Chapter # 4 – Nelson)
Lesson 1: / Sec.4.1 An Introduction to SimulationsLesson 2: / Sec.4.2 Theoretical Probability
Lesson 3: / Sec.4.3a. Finding Probability Using Sets
Lesson 4: / Sec.4.3b. Finding Probability Using Venn Diagrams
Lesson 5: / Sec.4.4 Conditional Probability
Lesson 6: / Sec.4.5 Finding Probability Using Tree Diagrams
Lesson 7: / Sec.4.6a Permutations
Lesson 8: / Sec.4.6b Probability & Permutations
Lesson 9: / Sec.4.7a Combinations
Lesson 10: / Sec.4.7b Probability & Odds
Lesson 11: / Sec.4.7c Problem Solving with Combinations
Lesson 12: / Unit Review Day # 1
Lesson 13: / Unit Review Day # 2
Unit 1: Homework
Topic / HomeworkSimulations / p.210 # 4, 5, 8, 10, 11, 14, 19
Theoretical Probability / p.218 # 1 , 2b , 3 – 6 , 8 , 10 , 11, 14 , 18
Probability Using Sets / p.228 # 1abcdf , 6a , 8a
Probability Using Venn Diagrams / p.228 # 2 , 3 , 4 , 6b , 7 , 8b , 9 , 12 , 13 , 14
Conditional Probability / p.236 # 1 , 2 , 5 , 8 , 9 , 14ab
Tree Diagrams / p.245 # 3 , 4ab , 5abc (ii,iii,iv) , 6ab , 9ab
Permutations / p.255 # 1bde , 2b , 3c , 4abc , 5 , 6ab , 9a , 11 , 13 , 15 - 17
Probability & Permutations / YES
Combinations / p.262 # 1 - 3 , 6 , 7 omit g
Odds / YES
Problem Solving with Combinations / YES
Probability & Odds
Homework:
1.a)Given:Find:odds in favour of A
b)Given:Find:odds in favour of A
c)Given:(odds against A) = 5 : 4Find:P(A)
d)Given:(odds in favour of A) = 10 : 8
(odds against B) = 5 : 1
A & B are mutually exclusive
Find:(i)
(ii)odds in favour of either A or B occurring
e)Given:P(A) = 0.5
P(B) = 0.3
A & B are independent
Find:odds in favour that both A & B occur
[ 7:29 , 2:1 , 4/9 , 13/18 , 13:5 , 3:17 ]
2.Show that if (odds in favour of A) = then
3.What are the odds in favour of rolling a pair, followed by two even numbers in two successive rolls of a pair of dice?
[ 1:23 ]
4.Over time, Natalie has experienced that if she washer her car, it rains the next day 80% of the time. What are the odds in favour of rain tomorrow if Natalie is washing her car today?
[ 4:1 ]
- Statistics show that in 75% of the fatal accidents involving two cars, at least one of the drivers is impaired. If you have the misfortune to witness such an accident, what are the odds in favour of one of the drivers being impaired?
[ 3:1 ]
Probability & Permutations
Homework:
1.Match each expression on the left with an equivalent expression on the right.
a)i)10 100
b)ii)6 !
c)iii)52
d)iv)10!
e)90 x 8!v)14
f)30 x 4!vi)20!
2.Find the value for each expression.
a)b)c)d)
e)f)[ 336 , 19535040 , 5985 , 36 , 3652110 , 2919735 ]
3.Evaluate each of the following expressions, using the definition of the factorial operation to simplify the work.
a)b)c)d)
e)f)g)h)
[ 30240 , 586051200 , 84 , 495 , 56 , 8008 , 20 , 990 ]
4.Simplify the following expressions
a)b)c)
d)e)f)
[ n! , (n+1)! , (n+1)! , (n+2)! , n2-n , (n+2)(n+1)n ]
5.Solve for n,
a)b)
c)d)
[ 8 , 5 , 6 , 8 ]
6.In how many ways can the letters of each word be arranged?
a)MAXIMUMb)CANADAc)SASKATCHEWAN
d)INTERESTINGe)UNINTERESTINGf)MISSISSAUGA
[ 840 , 120 , 39916800 , 2494800 , 129729600 , 415800 ]
7.List all five-digit numbers than can be formed by using two 4’s and three 6’s.
[ 10 ]
8.How many seven-digit integers, are there which include
a)two 3’s, three 2’s and two 8’s?
b)four 3’s and three 4’s?
[ 210 , 35 ]
9.A man bought two vanilla ice cream cones, three chocolate cones, four strawberry cones, and one
pistachio cone for his ten children. In how many ways can he distribute the flavours among the children?
[ 12600 ]
10.A coin is tossed nine times. In how many ways could the results be six heads and three tails?
[ 84 ]
11.Anya is starting out on her evening run. Her route always takes her eight blocks east and five blocks north to her grandmother’s apartment building. But she likes to vary the path she follows. How many different possibilities does she have? Hint: Consider the permutations of 13 letters, 8E’s and 5 N’s
[ 1287 ]
12.How many numbers greater than 300000 are there using only the digits 1 , 1 , 1 , 2 , 2 , 3 ?
[ 10 ]
13.Yurak is shelving books in a display in the school library. He has four different books with three copies of each. In how many ways can he arrange the books on the shelf for display?
[ 369600 ]
14.A developer will build 12 houses on the same side of Costly Court in a new subdivision. If he has room for two houses modeled on Plan A, four modeled on Plan B, and six modeled on Plan C, in how many different ways can he arrange the houses on the street?
[ 13860 ]
15.Emily’s minor soccer team played a total of 14 games in the season. Their record was eight wins, four losses, and two ties. In how many orders could this have happened?
[ 45045 ]
16.a)How many permutations are there of the letters of the word BASKETBALL?
b)How many of the arrangements begin with K?
c)How many of the arrangements start with a B?
d)In how many of the arrangements would the two L’s be together?
[ 453600 , 45360 , 90720 , 90720 ]
Problem Solving in Combinations
Homework:
1.In how many ways can a committee of at least one person be formed from seven club members?
[ 127 ]
2.Chang arrives at the giant auction sale late in the afternoon. There are only five items left to be sold. How many different purchases could he make?
[ 31 ]
3.In how many different ways could a team of three students be chosen from Lin’s Finite Mathematics class of 25 students to complete in the Country Mathematics Contest?
[ 2300 ]
4.a)How many different sums of money can be made from a $2-bill, a $5-bill, and a $10-bill?
[ 7 ]
b)How many different sums of money can be made from the bills in (a) as well as one more $10-bill?
[ 11 ]
c)Why does the situation become much more complicated if another $5-bill is added?
5.A committee of students and teachers is being formed to study the issue of student parking privileges. Fifteen staff members and 18 students have expressed an interest in serving on the committee. In how many different ways could a five person committee be formed if it must include at least one student and one teacher?
[ 225765 ]
6.In the binary number system which is used in computer operations, there are only two digits allowed: 0 and 1.
a)How many different binary numbers can be formed using at most four binary digits (ie: 0110) ?
[ 16 ]
b)If eight binary digits are used (ie: 11001101), how many different finery numbers can be formed?
[ 256 ]
Sec.4.1 An Introduction to Simulationsand Experimental Probability
A simulation is an experiment, model or activity that imitates real or hypothetical conditions. It is used to estimate quantities that are difficult to calculate and for verifying theoretical calculations.
ie: biological experiments, drug testing, IKEA furniture testing, multiple choice guessing, births
Example # 1: Simulating Birth Sequences
Suppose a family plans to have four children. Use a simulation to estimate the likelihood that the family will have three girls in a row and then a boy.
Solution # 1a)
Toss a single coin four times to simulate a single trial. Let Heads represent the birth of a girl and let Tails represent the birth of a boy. Repeat this experiment for 50 trials to get a fair representation of the likelihood of having three girls followed by a boy.
Solution # 1b)
Toss four different coins to simulate the births of all four children at once. Let Heads represent the birth of a girl and let Tails represent the birth of a boy. The Penny will represent the 1st birth, the Nickel will represent the 2nd birth, the Dime will represent the 3rd birth and the Quarter will represent the 4th birth. Repeat this experiment for 50 trials to get a fair representation of the likelihood of having three girls followed by a boy.
Example # 2: Batting Averages
Suppose that Nicholas has a batting average of 0.320. This means that he will hit the ball 320 times out of 1000 at bats. This reduces to 32 out of 100 (or 8 out of 25). Use a simulation to estimate the likelihood that this player has no hits in one game (assuming 3 hits @ bat/game).
Solution # 2
Fill a container with 25 slips of paper, 8 of which have the word HIT written on them. All the other papers will have the word MISS written on them, or you could leave them blank for saving time. Draw a sheet of paper out of the container and record “hit or miss”, return the paper to the container and draw again. You will draw 3 papers to represent Nicholas’s likelihood of hitting in a game. Repeat this experiment for 50 trials to get a fair representation of the likelihood that Nicholas will have no hits in a game.
Using Graphing Calculators for Simulations
The graphing calculator can be used for simulations. In Example # 1 for the births, you can generate a list of random numbers to represent the births of girls and boys.
On the TI-83+ select the MATH button, use the right arrow 3 times so that PRB (for probability) is selected. Either scroll down or select #5 for RANDINT( , you now have to set a lower bound, upper bound and number of integers. In the case of example # 1, we will let our lower bound be 1 (which will represent the birth of a girl) and let our upper bound be 2 (which will represent the birth of a boy) and choose 4 because we wish to represent the births of four children in total.
This function will generate a list of four numbers at random to simulate the birth of the four children. Repeat this experiment for 50 trials to get a fair representation of the birth sequences.
ie: 2 2 1 2 (would represent a boy, boy, girl, boy for the four births)
In the case of example # 2 we would need a random integer between 1 & 1000 with 3 numbers each time. The numbers 1 – 320 would represent HITS and the numbers 321 – 1000 would represent MISS. Repeat this experiment for 50 trials to find the likelihood that Nicholas will have no hits in a game.
ie: 978 279 276 (would represent a miss, hit, hit during one game)
Using Excel Spreadsheets for Simulations
For Example # 1:
Trial / Birth 1 / Birth 2 / Birth 3 / Birth 41 / 2 / 1 / 1 / 2
2 / 1 / 1 / 2 / 2
3 / 1 / 1 / 1 / 2
4 / 1 / 1 / 2 / 1
5 / 2 / 1 / 1 / 1
6 / 2 / 2 / 2 / 1
7 / 1 / 2 / 2 / 1
8 / 1 / 1 / 1 / 1
9 / 1 / 1 / 2 / 1
10 / 2 / 1 / 1 / 1
Under the Birth 1 column you would enter “=int(2*rand( ) + 1). You then use the fill handle on this cell to drag this same formula into all other columns and rows necessary.
Sec.4.2 Theoretical Probability
Simple Event
- an event that consists of exactly one outcome.
Experimental Probability
- only estimates the likelihood that an event occurs in a given experiment.
Theoretical Probability
- is deduced from analysis of all possible outcomes.
Sample Space
- is represented by “S” and is the collection of all possible outcomes.
Event Space
- is represented by “A” and is the collection of all those outcomes that we want.
We will look at the probability of experiments where the outcomes are easily determined.
Example #1:
A die is rolled once. What is the probability of rolling a four?
Solution # 1:
Let S represent the collection of all possible outcomes. S = { 1 , 2 , 3 , 4 , 5 , 6 }
Let A represent the successful outcomes. A = { 4 }
Example # 2:
A bag contains 6 red smarties, 2 blue smarties, and 7 green smarties. If you take one, what is the probability that it is green?
Solution # 2:
Let S represent the collection of all smarties. S = { 15 smarties all together }
Let A represent the green smarties. A = { 7 green }
Example # 3:
If a single die is rolled what is the probability of;
a) rolling an odd number
b) rolling a number less than 3
c) rolling either a three or a five
Solution # 3:
a) rolling an odd number
b) rolling a number less than 3
c) rolling either a three or a five
Example # 4:
Find the probability that a number picked at random between 1 and 10 is a perfect square?
Solution # 4:
n { S } = 10 elements in total
n { A } = 3 elements that are perfect squares between 1 & 10
Example # 5:
When drawing a single card from a standard deck of cards, what is the probability that;
a) a queen is drawn
b) a heart is drawn
c) a black card is drawn
Solution # 5:
a) a queen is drawn
b) a heart is drawn
c) a black card is drawn
One immediate consequence of the definition of probabilities is the relationship between an event and it’s compliment, that is the set of outcomes in which the event does not occur.
In General:
Example # 6:
What is the probability of not selecting a prime number in a random selection of a number from 1 to 20?
Solution # 6:
Let A represent the prime numbers between 1 and 20.
A = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 }
n ( A ) = 8 & n ( S ) = 20
Therefore,
Example # 7:
My music collection at home consists of eight rock CD’s, twelve blues CD’s and 4 classic CD’s. If a CD is chosen at random, find the probability that the one chosen is;
a) rock
b) not classical
c) neither blues nor classical
Solution # 7:
a) rock
b) not classical
c) neither blues nor classical
Sec.4.3a Finding Probability Using Sets
A Set
is a collection or group of objects witch have a specific relationship.
ie:a set of dishes (all have the same pattern)
the set of natural numbers (no fractions or decimals, 1 and up)
An Element
is all the members of a set and are usually listed in curly brackets { } and each entry is separated by comma’s.
ie:D = { tea cup, saucer, bowl } this is called a finite set. n ( D ) = 3
N = { 1 , 2 , 3 , 4 , 5 , ………. } this is an infinite set n ( N ) = undefined
Null Set
is the set with no members in it. It is denoted by , or empty { } but not by{ 0 }.
Disjoint Sets
two sets that have no elements in common
ie:C = { 4 , 6 , 7 , 9 , 10 }
A = { 1 , 5 , 11 , 13 }C & A are disjoint sets.
Subset
if all elements of one set are contained (also found in) a second set, then the first is a subset of the second.
ie:C = { 1 , 3 , 5 , 7 } G = { 1 , 3 , 5 , 7 , 9 } C is called a subset of G
Compliment
C ` is the compliment of C and contains all elements not in C but in the sample space S.
Let our sample space be the odd #’s between 1 & 10
If C is defined as C = { 1 , 3 , 5 , 7 } the same as above, then C ` = { 9 }
Example #1:
Describe the relationship between the sets described below using the termsequal, disjoint and subset.
a) N = { the natural numbers } & W = { the whole numbers }
b) R = { colours in a rainbow } & B = { grey, black, brown }
c) D = { prime divisors of 34 } & E = { 2 , 17 }
Solution # 1:
a) N = { the natural numbers } & W = { the whole numbers }
the natural numbers are contained within the whole numbers
b) R = { colours in a rainbow } & B = { grey, black, brown }
are disjoint, no element is the same in the two sets.
c) D = { prime divisors of 34 } & E = { 2 , 17 }
are equal because all the elements in D are also in E.
All sets are subsets of a universal set for that particular situation
ie: the vowelsV = { A , E , I , O , U } are a subset of the universal setcalled the alphabet where U = { A , B , C , D , E , F , G , …………, X , Y , Z }
Let's say that our universe contains the numbers 1, 2, 3, and 4.
Let Abe the set containing the numbers 1 and 2. That is, A= {1, 2}.
Let Bbe the set containing the numbers 2and 3; that is, B = {2, 3}.
Then we have the following relationships:
The Union
contains all the elements from both sets without repeats
ie:
The Intersection
contains all the elements that are common to both sets
ie:
Recall the Compliment
ie: A ` means everything in the universe not in A
Try This?
a) Find b) Find
Example # 2:
Determine the elements in the union & intersection of M & N.
M = { 1 , 3 , 6 , 9 , 12 }N = { 2 , 4 , 6 , 8 , 10 , 12 }
Solution # 2:
Union of M & N is: { 1 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 }
Intersection of M & N is : { 6 , 12 }
Example # 3:
Determine the elements in the following, given;
S = { K , E , Y , B , O , A , R , D }
A = { E , O , A }
B = { B , O , A , R , D }
Find;
a) A ` b) B `c) d) e)
Solution # 3:
a) A `= { K , Y , B , R , D }
b) B ` = { K , E , Y }
c) { O , A }
d) { E , B , O , A , R , D }
e) { K , E ,Y , B , R , D }
Sec.4.3b Finding Probability Using Venn Diagrams
Try This?
Example #1:
Find the probability of rolling an even number or a number greater than 3 when rolling a die. These two events are dependent on each other. ie: there is overlap between the two events.
Solution # 1:
S = { 1 , 2 , 3 , 4 , 5 , 6 }n(S) = 6
Let A be the event of rolling an even number.A = { 2 , 4 , 6 }n(A) = 3
Let B be the event of rolling a number greater than 3B = { 4 , 5 , 6 }n(B) = 3
(AB) = { 4 , 6 }n(AB) = 2
*** you must take away the intersection ***
Now let’s look at a situation where the two events are not dependent on each other. They are independent or disjointed events, sometimes also referred to as mutually exclusive.
Example # 2:
If two dice are rolled, one red and one green (so that you can tell the difference between them), find the probability that a sum of;
a)3 or 12 will occur
Solution # 2a
Let A be the event of rolling a sum of 3 Let B be the event of rolling a sum of 12
** These two events are mutually exclusive, independent **
b)6 or a pair will occur
Solution # 2b
Let A be the event of rolling a sum of 6 Let B be the event of rolling a pair
** These two events are not mutually exclusive, they are dependent because there is 6 in both sets! **
** account for 6 in both **
Mutually exclusive events occur when 2 events can not happen at the same time.
** there is nothing to take away when the two sets are disjoint **
Example # 3:
Jody attends a fundraiser at which T-shirts are given away as door prizes. Door prize winners are randomly given a shirt from a stock of 2 black shirts, 4 blue shirts and 9 white shirts. Assuming that Jody wins the first door prize what is the probability that she will get a black or white shirt?
Solution # 3:
Let A be the event of drawing a black shirt Let B be the event of drawing a white shirt
** these are mutually exclusive, independent events **
Example # 4:
A card is randomly selected from a standard deck of cards. What is the probability that either a heart or a face card (JQK) is selected.
Solution # 4:
Let A be the event of selecting a heart Let B be the event of selecting a face card
** these are not mutually exclusive, they are dependent events **
** J of heart, Q or heart, K of hearts **
Example # 5:
As a result of a recent survey it was estimated that 85% of the grade twelve population at a local high school enjoys an alcoholic beverage once a week, 35% smoke at least one cigarette a day and 25% indulge in both habits. What is the probability that an individual chosen at random from the targeted population either smokes or drinks.
Solution # 5:
Let A be the event that the individual drinks Let B be the event that the individual smokes
** these are not mutually exclusive, they are dependent events **
P(A) = 0.85 P(B) = 0.35 P(AB) = 0.25
Therefore the probability that an individual either smokes or drinks when chosen at random from the targeted population is 95 %.
Sec. 4.4 Conditional Probability
When calculating probabilities, additional information may become available that will affect the calculation of the outcome.
ie: If you are interested in the probability of your passing the next test, the probability will depend on whether you studied, if you worked late the night before, if you had a good breakfast, have you been doing your homework regularly, etc…
When additional information is known that will affect the probability of an outcome, we calculate what is called the conditional probability represented by P(A | B), read as “the probability of event A, given that the event B has occurred”, or, “the probability of A given B”.