If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
Precalculus II
Asst # 18
04-28-2002
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16:20:12
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**** Query 11.7.6 find the value of P(9,0).
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16:21:38
Your solution:
Confidence Assessment:
Given Solution:
P ( 9, 0 ) = ( 9! ) / ( 9 - 0 ) ! = 9! / 9! = 1
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16:21:39
Self-critique (if necessary):
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**** Query 11.7.16 C(18,9). **** What is the value of the expression and how did you obtain it? Your solution:
Confidence Assessment:
Given Solution:
PROBLEM NOW READS **** Query 12.2.12 (5th ed 11.7.12) C(6,2).
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16:25:37
Your solution:
Confidence Assessment:
Given Solution:
** Be sure you howed the process, which is illustrated here for C(18,9)
C (18,9) = 18! / [ (18-9)! (9!) ]
= ( 18*17*16*15*14*13*10 * (9!)) / (9! *9! )
= ( 18*17*16*15*14*13*10) / 9!
Note that ( 18*17*16*15*14*13*10) / 9! = ( 18*17*16*15*14*13*10) / (9*8*7*6*5*4*3*2*1) and that every term in the entire denominator divides evenly into the numerator (9 into 18, 8 into 16, 7 into 14, 6 into 12, 5 into 10, 4 into any two of the 2's you get from the last e divisions, 2 into the other one and 3 into the 3 you get when you divide 15 by 5.
To avoid loss of information when the number are big you should always simplify as much as possible before dividing.
Of course the C(n, r) button on your calculator takes all that into account and gives you an accurate result whenever the result is within the greatest number of digits displayed. However when doing a problem for the test you are being tested on your ability to evaluate the expression yourself, not on your ability to use the C(n, r) button, so you need to show the details.
Self-critique (if necessary):
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**** Query 12.2.24 (5th ed 11.7.24) combinations of 3 elements of a,b,c,d,e,f
**** How many combinations are there and how did you obtain your result?
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15:19:19
Your solution:
Confidence Assessment:
Given Solution:
When listing you need to be very systematic. I would advise listing the numbers in numerical order, something like this:
123
124
125
126
134
135
136
145
146
234
235
236
245
246
256
345
346
356
456
The above listing left off one combination. See if you can find it before I tell you below. *&*&
*&*& There are C(6, 3) = 6 ! / [ (6-3)! * 3! ] = 6*5*4*3*2*1 / (3*2*1 * 3*2*1) = 6*5*4 / (3*2*1) = 2*5*2 = 20 possibilities. *&*&
*&*& The missing combination is 156 *&*&
Self-critique (if necessary):
Self-critique Rating:
**** Query 12.2.40 (5th ed 11.7.40) mult choice 5 quest. 4 possible answers each
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23:11:15
**** How many possible combinations of answers are possible?
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23:11:35
Your solution:
Confidence Assessment:
Given Solution:
** For each question there are 4 possible answers. So to answer the questions we have to 'fill in' 5 blanks, each with 4 possible answers.
By the fundamental counting principle there are 4 * 4 * 4 * 4 * 4 = 1024 ways to do this. **
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16:29:50
16:30:05
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**** Query 11.7.38 committee of 4 from department of 8. **** How many committees can be chosen from the given number, and how did you obtain you result?
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16:34:44
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION
I used combinations to find the number of possibilities because order is not important.
nCr ( 8,4) = 8! / ( (8-4) ! (4!) )
= (8*7*6*5) / 4!
= 1680 / 24
= 70
Self-critique (if necessary):
Self-critique Rating:
**** Query 12.2.40 (5th ed 11.7.40) mult choice 5 quest. 4 possible answers each
**** How many possible combinations of answers are possible?
Your solution:
Confidence Assessment:
Given Solution:
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15:23:39
The answer chosen on a given question is unrelated to the answer chosen on any other question, so the choices here are unrestricted.
There are thus four possible answers on each of the 5 questions, so by the Fundamental Counting Principle there are
4 * 4 * 4 * 4 * 4 or 4^5 = 1024 possible combinations.
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16:34:45
16:34:57
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**** Query 11.8.8 spinner 3 (Forward, Backward) then 2 (red, yellow, green). **** What is the probability of getting Forward then Green?
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16:42:37
Your solution:
Confidence Assessment:
Given Solution:
** There are 3 positions. The probability of getting Forward on the first spin is 1/3. The probabiity of getting Green on the second spin is 1/3.
So the probability of getting Forward then Green is 1/3 * 1/3 = 1/9. **
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16:42:57
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**** Query 11.8.16 assignment of probabilities if tails is twice is likely as heads. **** List the probabilities of HH, HT, TH and TT under the given conditions.
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16:43:47
Your solution:
Confidence Assessment:
Given Solution:
** If Tails is twice as likely as Heads then the probability of Tails is 2/3 and the probability of Heads is 1/3 on any spin.
HH has probability 1/3 * 1/3 = 1/9.
HT has probability 1/3 * 2/3 = 2/9.
TH has probability 2/3 * 1/3 = 2/9.
TT has probability 2/3 * 2/3 = 4/9. **
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16:43:48
16:43:52
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Self-critique (if necessary):
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**** Query 11.8.28 9 white, 8 green, 3 orange. **** What is the probability that a randomly selected ball will be white?
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16:49:32
Your solution:
Confidence Assessment:
Given Solution:
GOOD STUDENT SOLUTION
First I figured the probability of a ball not being white, because the book worded it that way.
P (Green or Orange) = p ( g U o)
P (g U o) = (8 green balls) / (20 balls) + (3 orange balls) / (20 balls) = 11/20
Then for the probability of a white ball I subtracted P (g U o) from 1.
P (w) = 1 - P (g U o)
= 1 - 11/20
= 9/20
** Very good. This can also be reasoned out directly, and should be understood both ways. It’s obvious that you’ll understand the following:
There are 9 + 8 + 3 = 20 balls, 9 of which are white.
So if the selection is random the probability of white will be
prob of whie = # white / total # = 9 / 20. **
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16:49:33
16:50:43
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**** Query 11.8.36 prob of 1,2,3,>4 are .1, .15, .2, .24, .31. **** What is the probability that at most 2 people will be in the line?
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16:52:31
Your solution:
Confidence Assessment:
Given Solution:
P (<=2) = P(0) + P(1) + P(2) = P(0 U 1 U 2)
= .10 + .15 + .20
= .45
** If P(n) is the probability of n then we want P( <= 2).
P (<=2) = P(0) + P(1) + P(2)
= .10 + .15 + .20
= .45 **
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16:52:32
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**** What is the probability that at least 2 people will be in the line?
**** What is the probability that at least 1 person will be in the line?
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16:54:10
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16:54:11
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Your solution:
Confidence Assessment:
Given Solution:
P(>=2 ) = P( 2 U 3 U >4)
= .20 + .24 + .31
= .75
P(>=1 ) = P( 1 U 2 U 3 U >4)
= .15 + .20 + +.24 + .31
= .90
** The probability of at least 1 person in line is the sum of the probabilities of having 1, 2, 3 and > 4 people in line:
P( >=1) = P(1) + P(2) + P(3) + P( >= 4) = .1 + .15 + .2 + .24 + .31 = .90. **
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