MATH 125 NOTES ABOUT HOMEWORK, ANNOUNCEMENTS, ETC.
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Assigned on: |
After introducing examples of a new section, if I assign a range of problems from the current section for you to select work from, you should try the odds in that range, check the answers in back, go to tutoring, etc., and bring questions to class. BE SURE TO CLICK ON RELOAD/REFRESH
ON YOUR COMPUTER OR THE CURRENT ADDITIONS TO THE PAGE MAY NOT APPEAR! You may also not see current pages if your computer does not
have an up-to date browser… download a
new version or use a library/lab computer. I see how far we get in lecture before I come up with a homework assignment, and will generally post that new assignment by 12pm the day of lecture! |
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W 01/21 |
Read section 2.2 about graphing by translation families of functions (parabolas, cubics, squareroots, absolute values, etc.). We will do more with them on Friday, along with exponentials from 2.4. |
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F 01/23 |
We did more examples of translation of the families of functions above. Read section 2.4 about exponentials (they prepare you for interest problems in Ch. 3).
Homework due M 01/26: Graph by translation (you do not have to plot extra points besides the “vertex” for each, but show the basic family member for each and how the new one relates to it in size and placement, as I did in the class). 1. y= – 1/2*(x cubed) – 5 2. y= 3*(absolute value of (x +7)) – 2 3. y= 2/3*((x-3) squared) + 2 4. y= – (squareroot of (x+4)) – 5 5. y=3/2*(x squared) +4 |
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M 01/26 |
Homework due W 01/28: 2.4 # 4, 5, 8, 22, 26 3.1 # 22, 24
Quiz on graphing by translation on Weds. |
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W 01/28 |
Homework due F 01/30: 3.1 p133 #8, 36, 40, 44, 48 Read section 3.2 for next time
One more quiz on graphing by translation on Fri. |
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F 01/30 |
We are moving from simple interest (linear functions) in 3.1 to compound interest (exponential functions) in 3.2. We have not talked about continuous compounding yet, so the 3.2 homework below only uses the relationship on p136. It helps to think of the relationship in the following way:
where m is the number of compounding periods per year, so m=1 for yearly compounding m=2 for semiannual compounding m=4 for quarterly compounding m=12 for monthly compounding m=360 (in this section) for daily compounding
Homework due M 02/02: 3.1 p134 # 52 3.2 146 # 4, 18, 20, 22, 32, 62, 68
Quiz on Monday on hmk assigned M 01/26 above |
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M 02/02 |
Homework due on W 02/04: 3.2 p146 # 12, 64, 66, 67, 76
Quiz on hmk assigned 1/28 next time. |
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W 02/04 |
I talked about how the key to doing annuity problems in both section 3.3 and 3.4 is to draw an abbreviated chart as on p149 (if there are 60 periods in the compounding, don’t write them all down, just write enough to get onto the pattern) and use the sum of a geometric series on page 150 to put them all together. Then see that this is the same as using the formulas. It is important to understand the series of compoundings instead of just plugging numbers into a formula!
Try 3.3 #21 in the way done on p149 and see if you can get the correct answer. It is not going to be collected, but is important that you try it before we go over it in class on Friday!
Quiz on hmk. assigned on 01/30 next time. |
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F 02/06 |
We looked at more annuity examples from 3.3 and 3.4 from the standpoint of using the sum of the geometric series instead of just plugging numbers into the formulas: see pg. 150 a+ar+ar2+… arn-1 = a[(rn-1)/(r-1)]
Sometimes we only have time to go over the set-ups for problems, so here are the answers to the chapter 3.1/3.2 even problems you have had for homework so far:
3.1 #8 209% and 1/6 of a yr #22 0.22 #24 ¼ of a yr #36 25.05 #40 0.078 #44 0.2124 #48 976.07 #52 0.22112
3.2 #4 100626.57 #12 12935.03 #18 0.035 #20 0.0003 #22 0.0175 #32 0.13 #62 16961.66 #64 20683.10 #66 173319.50 #68 38 yrs #76 17 quarters or about 4 ¼ yrs.
Some set-ups for 3.3 (find the future value of a series of
regular deposits):
#21 p156 mt=(12)(10)=120 (after 1 month, the 1st deposit is made)
future value = 500+500(1+0.0665/12)1+500(1+0.0665/12)2 +…+
500(1+0.0665/12)118+500(1+0.0665/12)119 use sum of geometric series a+ar+ar2+… arn-1 = a[(rn-1)/(r-1)] from pg. 150 with a=500 and r=(1+0.0665/12)1 answer: 84895.40
#23 p156 mt=(12)(5)=60 (after 1 month, the 1st deposit is made)
future value = 300+300(1+0.06/12)1+300(1+0.06/12)2
+…+
300(1+0.06/12)58+300(1+0.06/12)59 use sum of geometric series with a=300 and r=(1+0.06/12)1 answer: 20931.01
#25 p156 mt=(12)(15)=180 (after 1 month, the 1st deposit is made)
future value = 2000000 = P
+ P(1+0.0635/12)1+ P(1+0.0635/12)2+…+ P(1+0.0635/12)178+ P(1+0.0635/12)179 use sum of geometric series with a=P and r= P(1+0.0635/12)1 answer: Solve for P and get 667.43
Some set-ups for 3.4 (find present value of a series of
regular withdrawals):
#21 p167 was done in class
#23 p167 paying down a debt is withdrawing from the loan until the balance of the debt is zero. mt=(12)(3)=36 months (after 1 month, the 1st withdrawal of debt is made and the present value of that payment has 1 month negative interest:
present value = 350(1+0.0756/12)-1+ 350(1+0.0756/12)
-2+ …+ 350(1+0.0756/12)-34 + 350(1+0.0756/12)-35 + 350(1+0.0756/12) -36 use sum of geometric series with a=350(1+0.0756/12)-1 and r= (1+0.0756/12)-1 answer: 11241.81
#25 p167 do as #25 p156 above, where you are solving for P, but using “negative interest” to find the present value.
HOMEWORK due Monday (show tables as above, show use of the sum of a geometric series instead of using the formulas in the book!): 3.3 p156 # 24 (answer 343214.73), 26 (answer 626.93) 3.4 p167 # 22 (answer 55135.98), 26 (answer 94.69)
Also, you have your first test on Weds. 2/11. Here is the format so you can start preparing!
Test #1 format: 8 problems (the first with 2 or 3 parts): 1. section 2.2 two or three functions to graph using the
method of transformations, described on p69 and the examples in the section. 2. section 2.4 one exponential function to graph using a table
of values, as #3-8 on p103. 3. section 3.1 one problem like #33-48. 4. section 3.1 one problem like #51-54. 5. section 3.2 one problem like #61-66. 6. section 3.2 one problem like #75-76. 7. section 3.2 one problem like #67-68. 8. section 3.3 one problem like #21-24.
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M 02/09 |
Homework is to study for Test #1 as above. |
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W 02/11 |
You have scheduled holidays for Friday and Monday: Presidents’ Days. Enjoy. No homework to turn in on Weds. 02/18, but you should take a look at the examples in section 4.2 about how to use a matrix to solve a system of equations. I did one brief example in class the session before the test, but you should read some more so that we can more smoothly do more examples on Weds. together. Section 4.1 will not be formally covered, because it contains review material that you should remember from Algebra. Perhaps you should take a read thru the examples in 4.1 just to reassure yourself that you know how to solve a system of 2 equations in 2 variables by the methods of graphing, substitution, and elimination.
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W 02/18 |
HOMEWORK due Friday: 4.2p 197 #42, 46, 50 (all 2x2!)
the rest of the solution to the 3x3 from in class is below: |
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note that (-41/13)(1)+(41/13)=0, (-41/13)(2)+(82/13)=0 and (17/13)(1)+(-17/13)=0, (17/13)(2)+(-60/13)=-2
so X1=0, X2=-2, and X3=2 (plug them back into the original equations)
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F 02/20 |
We looked at more matrix problems and I almost finished 4.3 #49 as another 3x3 example. At the very last steps of the problem, when you make a new row 1 by taking (-4)Row3+Row1 you get row elements 1 0 0 1 and when you make a new row 2 by taking (-6)Row3+Row2 you get row elements 0 1 0 -2. This means that X1=1, X2=-2, and X3=1.
A little more practice with matrix operations, then we will do some of the more exotic ones that involve no solution or infinite solutions, and matrices where the number of equations is not the same as the number of variables. But for right now, all of the problems for homework have solutions and it doesn’t hurt to give you the solutions to work towards for the last two—the first two problems are matrices that are almost done, but give you a little warm-up:
Homework due Monday: 4.3 p208 # 24 answer (7, -2), # 28 answer (4, -4, -1), # 34 answer (1, 4, -2), #50 answer (-1, 1, 3)
It’s usually good to try more problems, but beware that the exercises are loaded up with those that have infinite or no solution, so those are not the greatest to practice on!
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M 2/23 |
Note: I have been given a surprise in the schedule by administration. They have decided to make March 11 a so-called “flex day” where teachers come but students don’t! Unfortunately, that means that Test#2 which was scheduled for that day will have to be taken on Friday the 13th instead…how appropriate!
We worked on matrices today that do not have a nice solution. Homework due Weds.: (try looking at the examples in the sections first, then look at some odds based on their answers in the back of the book) 4.2 p197 # 48 answer: no solution, #52 answer: (2t-1, t) 4.3 p208/209 # 36 answer: (-t-1, 2t+3, t), #40 answer: no solution, #42 answer: (2t+3, -3t+5, t) Look also at ex. 5 and ex. 6 p204 and we will do more with this type next time.
Quiz on something like 4.3 p208 #27-30, where you show how to work with 3x3 matrices without having to do all of the work!
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W 02/25 |
We talked about how to deal with matrices that have fewer equations than variables. I briefly showed you an example like ex. 6 p205: #68 p209. I set it up and gave the answer to verify. You must follow the 4 requirements in the blue box on p198. And in all cases, have a plan of action: try to get the 1st one in the 1st column by switching rows or dividing thru by a number, then the other zeros in that column by adding multiples of the row containing the 1, then move to the next column and try to get the next 1 down. If this is not possible, move to the next column and so on. Read the examples such as ex. 5 p204 and ex. 6 p205 before trying homework problems.
Homework due ???(For the word problems, first look at # 67, 69 with answers in the back of the book and the supplemental notes below! Note that for problems 69 and 70, you are not doing another matrix, you are plugging in the values found in problems 67 and 68!): 4.3p208 #44, 46, 58, 67, 69
Example #67 (with answers in back, this is more clarification): Using the equations provided, divide the second equation thru by the GCF 8000 to make the matrix easier to work with:
So the first row says
X1-X3=-17 then X1=X3-17 and the second row says
X2+2X3=41 then X2=-2X3+41 So using X3 for t, X1
becomes t-17, and X2 becomes -2t+41 But there cannot be a
negative number of tank cars, so t-17 is greater than or equal to 0 so t is
greater than or equal to 17. Also, -2t+41 is greater than or equal to 0 so t
is less than or equal to 41/2=20.5. Therefore, the only possibilies for t are
17, 18, 19, 20. (X1, X2, X3) = (t-17,
-2t+41, t) For t=17 (X1, X2, X3) = (0, 7, 17) For t=18 (X1, X2, X3) = (1, 5, 18) For t=19 (X1, X2, X3) = (2, 3 19) For t=20 (X1, X2, X3) = (3, 1, 20) Problem #69 is connected to
the above: Let C=450X1+650X2+1150X3 For (0, 7, 17) then
C=450*0+650*7+1150*17=24100 For (1, 5, 18) then
C=450*1+650*5+1150*18=24400 For (2, 3, 19) then
C=450*2+650*3+1150*19=24700 For (3, 1, 20) then
C=450*3+650*1+1150*20=25000 So the first value minimizes
the cost! The end!
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F 02/27 |
Homework due Monday: I wrote the wrong due date on the last assignment, so now it is due on Monday along with the following: 4. 4 p220 # 6, 10, 14, 32, 34 |
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M 03/02 |
Homework due Weds. 4.3 p210 #71, 75 4.4 p222 # 61, 65 and maybe start reading section 4.5!
Note that since the school cancelled classes for Weds. 3/11, your Test #2 will now occur on Friday 3/13! |
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W 03/04 |
Homework due Friday 4.5 p234 # 32, 36, 42
Answers to last hmk: 4.3 71. If x1 is the fed tax, and state and local tax are removed from the income before the fed tax is computed, x1 = 0.5(7,650,000 – x2 –x3) distribute to get x1 + 0.5x2 + 0.5x3 = 3,825,000
If x2 is the state tax, and fed and local tax are removed from the income before the state tax is computed, x2 = 0.2(7,650,000 – x1 –x3) distribute to get 0.2x1 + x2 + 0.2x3 = 1,530,000
If x3 is the local tax, and fed and state tax are removed from the income before the state tax is computed, x3 = 0.1(7,650,000 – x1 –x2) distribute to get 0.1x1 + 0.1x2 + x3 = 765,000
75. (A)
switch rows to get 1
(B): no solution since matrix operations yield
(C):
so x1=8, x2+2x3=10 so if x3 is t, then x2=10-2t and since x2 must be greater than or equal to 0, t is less than or equal to 5.
4.4
61. A+B
gives
65. (A): Multiply the first row of M by the first column of N to get 0.6*17.30 + 0.6*12.22 + 0.2*10.63 = 10.38 + 7.332 + 2.126 = 19.838 = 19.84 (B): Multiply row 3 of M by column 2 of N to get 1.5*14.65 + 1.2*10.29 + 0.4*9.66 = 38.187 = 38.19 (C): MN yields labor costs for each plant, but NM is not defined (D): multiply the 3x3 matrix by the 3x2 matrix to get the 3x2 matrix in the back of the book |
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F 03/06 |
Start reading section 4.6 which we will cover on Monday. No hmk. to turn in on Monday, but you should use the time to start studying for Test #2 which will occur on Friday 03/13. Test format: 1. section 4.3 one problem like p208 #35, 36 2. section 4.3 one problem like p207 #17/18 where you have an already reduced matrix but you must interpret the solutions in terms of “t”, then assume t must be greater than zero to find actual possible point values. 3. section 4.3 one problem like p208 #27-30 where the 3x3 is partially done and you must finish it 4. section 4.3 one problem like p209 #67-70 where you set up the matrix but do not solve it. 5. section 4.4 one or two problems like p220 #1-14 6. section 4.4 one problem like p222 #65 (A) or (B) 7. section 4.5 one problem like p235 #29-44
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M 03/09 |
All classes on this campus are cancelled on Weds. 03/11. You will come on Friday 03/13 to take Test #2 as above. Homework is to study for your test. We went over some topics in section 4.6 and you will have homework in that section given to you after the test and due on Monday. |
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F 03/13 |
Homework due Monday 03/16 4.6 p242 #2, 8, 12, 16, 26 |
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M 03/16 |
Homework due Weds 03/18 4.6 p244 #49 (read ex. 4 p240 first to know what to do!) |
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M 03/20 |
Homework due Monday 5.2 p273 #8, 24, 40 and 5.3
p285 #14, 20 (and turn in problem 49
from last time if you didn’t today!) |
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M 03/23 |
Homework due Weds. 5.3 p289 #46 and 6.1 p298
#10 example 5.3 p298 #45 do as
in 5.2: graph, find corners, then evaluate P at each: Let x1=# mice and x2=#rats
and maximize P = x + y subject to 10x+20y< 800 (or equal
to) x-intercept (80, 0) and y-intercept (0, 40) 20x+10y<640 (or equal
to) x-intercept (32, 0) and y-intercept (0, 64) corner values (0, 0), (0,
32), (40, 0), (16, 32) and P is maximized here P=48 |
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W 03/25 |
We started work on section
6.2. It is a tough read, so I suggested mainly looking at the procedure boxes
in the text, then looking at the examples. I started 6.2 p312 #13 as an
example but will have to complete it on Friday. For homework, do the related
problem: 5.3 p285 #13 by graphing and get the corner points. Try plugging
them in to the P that must be maximized in section 6.2: P=30x1+40x2. We will
finish up the example from 6.2 next time and compare it to your answer! |
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F 03/27 |
In the final matrix from class,
there was a mistake at the end of the problem.The bottom row was found by
taking 10 times the pivot row and adding to the bottom row. In class, I wrote
the elements 0 20 0 0 30 1 360 but they should have read 0 0 0 20 10 1 260. Homework due Monday (try
some of the odds around them first!): 6.1 p297 #6, 8, 12 6.2 p312 #6, 9 (this is
shorter than the one in class--check the answer in back!) |
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M 03/01 |
We went over the homework
and did not have time for new material! Homework is to start studying for
your test. The format for Test #3 is as follows: 1. section 4.6 p243 one
problem like #23-26 (inverse will be supplied!) 2. section 5.2 p273 one
problem like #19-25 to shade
region only, not find corner points 3. section 5.2 p274 one
problem like #39/40 to write a system of inequalities, but not graph or solve
it 4. section 5.3 p285 #10/14
find corner points and find max/min value of given function given the shaded
graph already done from the given equations 5. section 6.1 p298 one
problem like #7/8 6. section 6.2 p312 one
problem like #9/10 to set up the simplex tableau only (do not solve) as in
this morning’s quiz 7. section 6.2 p312 one
problem like #5/6 to identify the pivot and perform one or two operations on
the pivot |
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F 04/03 |
In view of the lecture on
Weds, before the test, where we started a new segment on counting and
probability, try the following: Homework due M 04/13: 7.3 p372 #6 (make a tree
diagram), 7.4 p385 #12, 16, 32, 34 Look at p382: for work in
probability that we will do after break, you are responsible for knowing the
contents of a 52 card deck. Notice that it is a regular
deck of 52 cards with no jokers/wildcards. There are four suits:
hearts, diamonds , clubs (clovers) , and spades . Half of the deck is red
and half of the deck is black. Each suit has 13 kinds
of cards : K,Q,J,10, 9,8,7,6,5,4,3,2,A, where A=Ace counts as a one. We will use the
abbreviations: K for King, Q for Queen, J for Jack, A for Ace. There are four cards of
each kind in a deck (i.e., K hearts, K diamonds, K clubs, K spades) Each suit has 3 face
cards : King, Queen, and Jack (face cards are cards with people on
them!). Therefore, each suit has 10
non-face cards. (DON’T FORGET TO HAVE A
GREAT BREAK!!!) |
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M 04/13 |
Class
cancelled |
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W 04/15 |
We
are skipping section 7.1 (at least for now) and we will probably look at the
concepts from 7.2 either Friday or in conjunction with section 8.2. For now
(from 4/3 and today), we are looking at concepts in 7.3 and 7.4, especially: p369
how to make a tree diagram and use the multiplication rule to count the
branches of the tree p375
factorial notation p377
how to find the number of permutations nPr (they use notation Pn,r) p380
how to find the number of combinations nCr (they use notation Cn,r) p380
as in example 4, how to know which of nPr and nCr to use. Homework
due Friday 04/17: 1. Permutations: From the set of four
letters { A, B, C, D } list the number of ways that you can select three
letters at a time (such as ABC), where writing letters in a different order
counts as a different selection. (a tree diagram might be helpful). 2. Combinations: From the set of four
letters { A, B, C, D } list the number of ways that you can select three
letters at a time, where writing the letters in a different order counts as a
duplicate selection and is therefore eliminated from the list. 3. Compute 4P3 and 4C3 using the formulas
and compare these counts to your lists in problems 1 and 2. In the following counting problems
(decide whether “order matters” in each situation, and compute the value of
“nPr” or “nCr” as appropriate, showing factorial notation and cancellations
as in class or as in the exs at the top of p381!). 4. In how many different ways can four
of 12 basketball teams be ranked first, second, third and fourth by a panel
of coaches? 5. In how many different ways can a
committee of four be selected from the 72 staff members of a hospital? (Don’t
make any inferences on your own about the makeup of the committee when
deciding whether or not order matters!). 6. In how many different ways can 3
horses in a 10-horse race finish at the racetrack? 7. In the Illinois Lottery, in how many different ways can one
pick the set of six winning numbers from the 51 available to choose from? (
Never played a lottery game? Not sure whether order matters or not? Compute
both permutations and combinations for this one. Then think about the
probability of winning being 1 chance in the number you just found. For
example, if you found the number of different ways to pick lottery numbers
was 12,345 , then if you purchased one ticket, you would have a one chance in
12,345 of winning. That is, 1/12,345 = 0.000081004 chance of winning.
Consider that there are about 13 million people in Illinois, assume that
almost that many tickets are sold each time, and that the Lottery officials
would like someone to win at least every other drawing! Which looks more
appropriate now, 51C6 or 51P6?) |
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|
F 04/17 |
We worked on using the
counting technique of nCr to start forming probabilities with respect to a
deck of cards. In 7.4 p382 ex. 6 (and p383 ex.8 is related), you see how nCr
can be used to make a count of the number of ways cards can be chosen, but
then we put this method of counting to use in 8.1 p401 ex. 7 to form a
probability. Using the methods of 7.4, to find the
number of ways that you can draw a single card from a regular deck of 52, you
would write 52C1. Note that in general, nC1 = n, so 52C1=52. One might get
the idea that the nCr notation is not helpful from that example, but it
becomes more helpful as the counts become more complicated: 52C2= 52!/(2!)(50!)=
(52x51x50x49x….x2x1)/(2x1)(50x49x8x…x2x1)=1326. This represents how many
pairs can be taken from a deck of cards, where order doesn’t matter. So it is
not so easy to just count cards as the problems get harder and involve taking
more cards at a time! The above number is not one you can just come up with
from knowing how many cards are in the deck. There is no nice shortcut for
computing 52C2, either! Now from 8.1, a restatement of the
theorem on p400 is as follows: the probability that an event occurs can be formed by using
the classical method, that is, by forming the following quotient (fraction): (the # of ways to get a “success”) /
(the total # of equally likely things that could occur). We will form the top and the bottom of the above
fraction with nCr counts, because in the probabilities we will focus on,
order does not matter. For instance, in the case of being dealt cards from a
deck, once they have been given to you, the order in which you shuffle them
in your hand does not change the set of cards you have. Permutations come into play with more difficult
probability situations that we probably will not have time for. The probability that you would get two
Kings in two cards drawn from a deck of cards is 4C2/52C2
since on the bottom of the fraction, you need to count all of the ways you
could get any 2 cards from a deck, and on the top of the fraction, you need
to count all of the ways you could get 2 “successful” cards (Kings), of which
there are 4 available. Then, 4C2=(4!)/(2!)(2!)=(4*3*2*1)/(2*1)(2*1)=(4*3)/(2*1)=12/2=6
and 52C2=(52!)/(2!)(50!)=(52*51*50***1)/(2*1)(50*49***1)=(52*51)/(2*1)=2652/2=1326
so 4C2/52C2=6/1326=about 0.005 (a rather
rare event!). The following are some more “simple” examples of writing probabilities using
the above notation (not simple as in easy, but meaning not involving compound
events where more than one thing is happening!). Count the cards involved in
being a “success” and you will see where most of the numbers are coming from
(there are 4 aces, 26 red cards, 13 hearts, etc.): Probability of an ace in 1 draw
=4C1/52C1=4/52=0.08 Probability of a red card in 1 draw
=26C1/52C1=26/52=0.50 Probability of a heart in 1 draw
=13C1/52C1=13/52=0.25 Probability of a face card in 1 draw
=12C1/52C1=12/52=0.23 Probability of a black four in 1 draw
=2C1/52C1=2/52=0.04 Probability
of a red face card in 1 draw =6C1/52C1=6/52=0.12 Probability of a non-face card in 1 draw
=40C1/52C1=40/52=0.77 Probability of 2 aces in 2 drawn
=4C2/52C2=6/1326=0.005 Probability of 2 sevens in 2 drawn =4C2/52C2=6/1326=0.005
Probability of 2 hearts in 2 drawn
=13C2/52C2=78/1326=0.06 Probability of 2 face cards in 2 drawn
=12C2/52C2=66/1326=0.05 Probability of 3 Kings in 3 cards drawn
=4C3/52C3=4/22100=0.0002 The following “compound” probabilities are more difficult than
the “simple” ones above, because we must worry about more than one kind of
event involved. This involves more
difficult questions involving intersections (multiplication rule) and unions
(addition rule) of sets. We will talk about these rules more next week. But
for now, we can apply the rules to less complex compound problems in the
following way: When you see the key word “and” in a
sentence, use multiplication to connect the two counts together. When
you see the key word “or” in a sentence, use addition to connect the
two counts together: 1. What is the probability of getting a
3 and a 4 in two
cards drawn? Answer:
(4C1*4C1)/52C2=(4C1*4C1)/52C2=(4*4)/1326=0.01 2. What is the probability of a jack and an ace in 2 drawn Answer: same as the previous one, since
they are both still “kinds” of cards! 3. What is the probability of a jack or an ace in 1 card drawn Answer: (4C1+4C1)/52C1=(4+4)/52=0.15 4. What is the probability of getting a
diamond and a heart
in two cards drawn? Answer: (13C1*13C1)/52C2=(13*13)/1326
=0.13 5. What is the probability of getting a
heart and a club in
two cards drawn? Answer: same as the previous one since
they are both still “suits”! 6. What is the probability of a heart or a club in 1 card drawn Answer: (4C1+4C1)/52C1=(4+4)/52=0.15 ********** HOMEWORK:
(will be collected on Monday—read the examples above first!) Probability Problems (write the probability as a
quotient of nCr counts as in the above notes and find the probability as a
decimal, either by using the formula for nCr or by using your calculator.
Where AND and OR are capitalized, use multiplication and addition
respectively to put two nCr counts together): 1. Probability of getting a nine in one draw. 2. Probability of getting a spade in one draw. 3. Probability of getting a red King in one draw. 4. Probability of getting a red non-face card in one
draw. 5. Probability of getting 2 red cards in 2 cards
drawn. 6. Probability of getting 2 fives in 2 cards drawn. 7. Probability of getting 2 clubs in 2 cards drawn. 8. Probability of getting five AND a ten in 2 cards
drawn. 9. Probability of getting a spade AND a heart in 2
cards drawn 10. Probability that in 1 card drawn, it will be a
spade OR a heart? |
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|
M 04/20 |
We went over the homework and
1. talked more about
intersections and unions (read p406) 2. about how to write
probabilities using the nCr method with problems not about cards (read 7.4
p383 ex. 8 and 8.1 ex. 8 p402), and 3. briefly looked at the
general addition rule (see p 407 theorem #1). Homework due Weds. 7.4 p385 #38, 40, 42, 62, 64
and 8.2 p416 #54, 56 |
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|
W 04/22 |
We talked about section 8.3
conditional probability today in conjunction with the general multiplication
rule: The general
multiplication rule says that for
events A and B, the probability that both A and B occur at the same time is: P(A and B)=P(A)*P(B
given A), where P(A) is used to mean
“the probability that event A will occur”, P(A and B) means “the
probability that both events A and B will occur”, where P(A and B) = P( A
intersection B) and P( B / A) = P(B given A) is what is called a conditional
probability, and means “the
probability that B occurs, given that we know A has occurred already” or “the
probability that B occurs, given that we know we are making selections only
from A”. Homework due Friday 8.3 p429 # 4, 6, 8, 10, 26,
60, 64 |
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|
F 04/24 |
You have a test scheduled
for Friday 5/1. I will have to wait until Monday to give you the exact format.
It will include some of what we do on Monday, and I have not decided what
that will be. I gave a little quiz like
ex 8 p402 and then took the method further with a look at the California
Lottery-Super Lotto Plus. We can use our “nCr” counting method to form rather
sophisticated counts to verify the probabilities that are listed on the back
of a California Lottery ticket for the general public’s information. THE CALIFORNIA LOTTERY:
SUPER LOTTO PLUS To play the game, you are
asked to pick 5 different numbers choosing from 1 to 47 regular numbers and
one “Mega” number choosing from 1 to 27. The top prize (which is the one
advertised in millions) goes to whoever matches all 5 of 5 winning numbers
and matches the one Mega number. Much smaller prizes are awarded for matching
some of the numbers. Prizes are awarded to the following winning
combinations:
You can find these
percentages by counting with combinations nCr as we did with cards. Here are some
examples (we looked at the first and the sixth ones in class): Getting all 5 of 5 and the Mega (any 5 of the 47 could be chosen, but you
want all 5 of the 5 available winners, and any 1 of the 27 possible Megas
could be chosen, but you want 1 of only 1 successful): ( 5C5 / 47C5 ) * ( 1C1 / 27C1 ) = ( 1 / 1,533,939 )( 1 / 27
) = 1 / 41,416,353 = 0.000000002 as above. Getting all 5 of 5 and not getting the Mega (here, “success”, the top of
the probability fraction for the Megas, is defined as not getting the Mega
but getting one of the 26 losing numbers!): (
5C5 / 47C5 ) *( 26C1 / 27C1 ) = ( 1 / 1,533,939 )( 26 / 27 ) = 26 /
41,416,353 = 0.000000628 as above. Getting any 4 of 5 and the Mega (you must consider that “success” is
getting 4 of 5 winners and 1 of
the other 42 losing numbers, which means multiplying those two in the top of
the first fraction!): ( ( 5C4 * 42C1 ) / 47C5 ) *
( 1C1 / 27C1 ) = ( ( 5 * 42 ) / 1,533,939 ) * ( 1 / 27 )= 210 / 41,416,353
=0.000005070 as above. Getting any 4 of 5 and not getting the Mega: ( ( 5C4 * 42C1 ) / 47C5 ) *
( 26C1 / 27C1 ) = ( ( 5 * 42 ) / 1,533,939 ) * ( 26 / 27 ) = 5460 /
41,416,353 = 0.000131832 which is slightly different from above, because the
odds work out to about 1/7585.41 and the lottery rounds the figure to 1/7585
as above. This difference can be considered negligible! Getting any 3 of 5 and the Mega (notice that you are selecting 3 of 5
winners and 2 of 42 losers): ( ( 5C3 * 42C2 ) / 47C5 ) *
( 1C1 / 27C1 ) = ( ( 10 * 861 ) / 1,533,939 ) * ( 1 / 27 )= 8610 / 41,416,353
= 0.00207889 which is slightly different from above, because the odds work
out to about 1/4810.26 and the lottery rounds the figure to 1/4810 as above. Getting any 3 of 5 and not getting the Mega: ( ( 5C3 * 42C2 ) / 47C5 ) *
( 26C1 / 27C1 ) = ( ( 10 * 861 ) / 1,533,939 ) * ( 26 / 27 ) = 223860 /
41,416,353 = 0.005405111 again because they rounded the odds of 1/185.01 to
1/185. As I have been suggesting
in class, it can be quite helpful to map out the strategy for outcomes by
breaking down each number set into the important subsets and see how many are
to be taken from each. The last probability above would be written as
follows:
Which translates to : Homework Problems for
you to do (due Monday): A. Compute the following
6 probabilities (refer to the 6 examples I did above for guidance): Getting any 2 of 5 and getting
the Mega, Getting
any 2 of 5 and not getting the Mega, Getting any 1 of 5 and getting the Mega,
Getting any 1 of 5 and not getting the Mega, Getting none of the 5 and only
getting the Mega, Getting none of the 5 and not getting the Mega. To make your life a
little easier, here are some answers to the combinations you will use so that
you do not have to find them using factorials and cancellations: nC0=1 for all n nC1=n for all n nCn=1 for all n 5C2=10 42C3=11,480 42C4=111,930 42C5=850,668 47C5=1,533,939 B. Try to explain why 3
of the 6 you computed above are excluded from the prize categories at the top
of this exercise. C. The old California
Lottery did not have the Mega number. You had to pick 6 winning numbers
choosing from numbers 1 to 49. Compute the odds of winning that game. Why do
you think they changed the game? |
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|
M 04/27 |
We looked
at the Lottery, then briefly looked at some set issues from 8.2. Originally,
I was going to put this and independence on the test, but since we did not
get to spend quality time on it, it will not! Homework is to try some of
this, though, and also look at the test format below and let me know if there
are any specific questions on Weds. We will not spend all of Weds. reviewing
for the test. You must go thru your notes and look below to be prepared. The
format is very specific! Please read it and be ready for it! Homework
due Weds. 4/29 8.2
p415 #2, 4, 6, 12, 14 (look at 11 and 13 to help!). Read ex. 2 p421 for next
time. Format of Test
4 scheduled for Fri 05/01: 1. One
problem to use the general multiplication rule and tree diagram to find the
number of possible outcomes. We did some examples of this in lecture and you
can find more examples in the book: 7.3 p369
ex. 3, p370 ex. 4, p373 #35, 36 2. One
problem to compute a given nCr showing the meaning of factorial and how to
cancel part of the numerator with the denominator and finding the value. See 7.3
p385 #16 For
example, 50C3=(50!)/(3!)(47!)=(50*49*48*47****2*1)/(3*2*1)(47*****2*1)
so the (47*****2*1)
in top and bottom cancel to leave (50*49*48)/(3*2*1)=19600 3. about 3
situations where you must decide if order matters or not and indicate the
appropriate nPr or nCr notation without actually computing it (however, you
must include what numbers are used for n and r). See 7.4
p385 #25, 26, 31, 32, 33, 34 For
example: a. In how
many different ways can a radio station select 5 listeners out of the 50 who
entered the contest to receive free passes to a show? Answer: 50C5 b. A lottery
game asks you to pick four numbers from 1 to 40. In how many different ways
can you do this? Answer: 40C4 c. In how
many different ways can Bob pick a best man and a verse reader from his 5
closest friends for his upcoming wedding? Answer: 5P2 4. About 4
or 5 probabilities where you form a quotient of nCr values in the classical
probability sense, e.g., P(one heart in one draw)=13C1/52C1. You are
responsible for knowing the subsets of a deck of cards, such as what the
suits and kinds are, how many of each, what face cards are and how many of
each, etc. See notes
from 4/17 and 8.1 p403 #11-20 5. One
non-card probability using the same nCr counting method as with cards. See 8.1
p402 ex. 8, quiz in class, p405 #91, 92 For
example: A shipment of 120 fasteners that contains 4
defective fasteners was sent to a manufacturing plant. The quality control
manager at the plant randomly selects and inspects five fasteners. In how
many ways could one of the chosen fasteners be defective? Answer:
(4C1*116C4)/120C5 6. One
lottery situation as in the homework from the California Lottery, but with
different pools of numbers: For example: Suppose a lottery that asks you to pick 4 of 35
available regular numbers and 1 of 17 available mega numbers. What is the
probability that you will get 3 of the winning regular numbers and not get
the mega number? (Just write the probability in terms of nCr notation—you
don’t have to compute them). Answer: (4C3*31C1)/35C4 multiplied by 16C1/17C1. 7. A table
from which to write probabilities and demonstrate the multiplication rule, as
in an example in class and 8.3 p429
#1-12. For example:
Let event A be that it is a
Target shopper and event B be that it is a person aged 29-39 a.
Find P(A) Answer: 145/282 b.
Find P(B) Answer: 125/282 c.
Find P(A given B) Answer: 50/125 d.
Find P(B given A) Answer: 50/145 e.
Find P(A and B) using the general multiplication rule and see if it looks
like the appropriate intersection. P(A and B)=P(A)*P(B given A)=
(145/282)(50/145)=50/282 |
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|
W 04/29 |
Test on
Friday will be as above. There will be some homework on independence from 8.3
after the test! |
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|
F 05/01 |
Homework due
Monday 8.3 p431#
14, 16, 18, 20, 59, 64 Hints for
#64 (like #59) (B)
find P(F given C) (C) find
P(C given not F) (D) Is P(C
given not F) equal to P(C) ? (E) Is P(F
given C) equal to P(F)? |
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|
M 05/04 |
We started
section 8.5 as a bridge to getting into the statistical portion of the
course, chapter 11. If we have some time left after ch.11, we may look at
portions of ch9/10, but it is not mandatory! Homework
due Weds. (read section 8.5 first, especially how expected value is
constructed—see definition p442!) 8.5 p446 #
2, 4, 6, 16, 24 (construct a payoff table for each, as they do in the
examples). |
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|
W 05/06 |
We started
ch11 very briefly, talking about frequency and histograms. We will probably
move on to 11.2 and 11.3 on Friday if you want to start reading ahead. Homework
due Friday 8.5 p446
#18, 23 11.1p540
look at the answer to #1 in the back, then try #2. |
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|
F 05/08 |
We talked about 11.2 mean,
median and mode briefly, then we started to look at 11.3 distributions of
data, focusing on continuous data that is “normally” distributed. The normal
distribution has a bell shape with the “mean” as its center (this is where
the data that occurs most frequently lies). The curve falls symmetrically on
either side of the mean. Half of the distribution is to the left of the mean
and half is to the right. Surprisingly, there are only two numbers needed to
describe the normal curve: the mean, and the standard deviation. The mean is the arithmetic mean
that we usually call an “average”. Add the data and divide by the number of
data points, n, to find the mean.
A deviation is the
difference in spaces between a particular data point and the mean. It
measures both position of a data value with respect to the mean and magnitude
of the distance of the value from the mean. The standard deviation is (with a
few difficulties to overcome) essentially an “average” of the deviations of
each data point from the mean.
We looked at an example
briefly in class.
Use the following additional example as
a guide: Example: Given data, find the mean and standard deviation using the formulas above: 9, 7, 11, 10, 13, 7 (data does not have to be in any particular order), Find the mean and then the standard deviation using the formula. Hint: make a
table of x values and deviations from the mean to help organize your data.
Note that the Greek symbol
-------------------------------------------------------------------- 9 9-9.5=-0.5 (-0.5)
2 =0.25 7 7-9.5
=-2.5
(-2.5) 2 = 6.25 11 11-9.5
=1.5
(1.5) 2 = 2.25
10 10-9.5 =0.5 (0.5) 2
= 0.25
13 13-9.5 =3.5
(3.5) 2 =12.25
7
7-9.5=-2.5
(-2.5) 2= 6.25
------------------------------------------------------------------------ so Answer: mean=9.5 and std. deviation=about 2.35 **************************************************** HOMEWORK due Monday: 1. try 11.2 p553 #14 2. find the mean and standard deviation for the following set of data 22,26,35,47,58 ********************************************************** On Monday, we will talk
about how to put these two important numbers to use to “standardize” any
normal distribution (mean of 0 and std. deviation of 1) and look into how to
use a table of std. normal distribution values to find the areas under the
curve that describes the distribution. Note that if we were not able to
standardize and use a table for areas, we would need Calculus to integrate
using the function that describes the normal distribution
Of course, a statistical calculator is programmed
with this function also, so that it can yield areas under this curve in an
instant when it is provided with the proper parameters. We will stick with
the table! |
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|
M 5/11 |
We worked on using those
two important numbers, the mean and the std. deviation from above, to find
areas under the normal curve. You find the z values for each x value in the
original distribution using the formula on p575, then refer to the table on
p655 to find the areas under the normal curve between that value and zero. Homework due Weds. 11.5 p581 # 2, 4, 6 , 8,
12, 14, 18 |
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|
W 5/13 |
We worked on the finding of
areas that are not immediate from the table, and solving of word problems in the
normal distribution. Problems 19-26 are similar to hmk problem 2 below…try
some of them first as a guide! Homework due Friday: 1. Find the area under the
standard normal curve corresponding to the following values: a. What is the area to the
left of z= -1.75 b. What is the area to the
right of z= -1.75? c. What is the area to the
right of z=1.05? d. What is the area between
z= -0.25 and z= -1.97 e. What is the area between
z= -2.09 and z=+3.07? 2.
Draw a picture (as in class) of a normal distribution for each, with both an
x and a z number line below it, and use the table and the standardizing
formula to answer: Suppose a large group of
students took a test (the results are distributed normally), and suppose that
the average score was 73 with std. deviation of 13.5. a. What percent of the
students scored 80 or above? b. What percent of the
students scored more than 70? c. What percent of the
students scored anywhere from 70 to 80? d. Above what score lie the
15% highest scores in the class? |
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|
F 5/15 |
Homework for Monday: Using the last hmk
situation (Suppose a large group of students took a test, the results of
which were normally distributed,
the avg. score was 73 with std. deviation of 13.5), a. What percent of the
students scored anywhere from 80 to 90? b. Above what score lies
the 25% highest scores in the class? c. Below what score lies
the lowest 16% of the scores in the class? and try some more exotic
situations from the book (draw a picture and read the situation carefully; a
couple of tricks in some!): 11.5 p582 #53, 55, 59, 65 Test #5 will be the last
graded activity for the course. It will cover problems like what you did for
homework in sections 8.3 (conditional probabilities and independence), 8.5
(expected values using payoff tables), 11.3 (calculating the mean and
standard deviation for a given set of data), 11.5 (finding standard values
and areas under a normal curve). We will review on Monday 5/18, then you will
take Test 5 on 5/20. Your “final activity” will be to come pick up your tests
and grades during finals week (this is an optional activity on your part!).
If you do not show up for Test 5, you will be offered a chance to take a
longer harder comprehensive final in its place during finals week! |
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|
M 5/18 |
We wrapped up the review
today. Please come on Weds. 5/20 to take your mandatory last test. If you do
not show up and take it, you will have to come take a comprehensive final to
replace it during finals week! Test 5 format: several problems like the
short problems in 8.3 p429 #1-20 one or two problems like
8.5 p446 #3-26 (set up x/p table and find E(x)) a set of data as in 5/8 notes
above to find the mean and std. deviation (this formula will be provided).
Other sets of data to practice on are in 11.3 p560 #13, 14, 16, 17, 18 several problems of the
type 11.5 p581 #1-26 and like the ones above given in hmk. |
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