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Many of you may have seen my other blog "You guessed it, Math Homework." Well, you guessed it, MATH HOMEWORK!
Since this chapter is self-evaluation (Meaning we learn it ourselves), and I don't understand it, I've come to TL for answers.  (Doesn't everyone? XD)
So I'm supposed to do my homework using nCr and nPr (Permutations and Combinations) as needed. I'm like 99% sure that these problems require permutation, except (like last time) I don't know how to set up the equation. T_T
Janice has 8 DVD cases on a shelf, each case is one season of CSI. Her brother accidently knocks over all the cases and puts them back on the shelf randomly.
#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
I got 1/70 for this one. I seriously doubt it's correct though.
#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.
 
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#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
= number of solutions/total number of positions
can you figure out the total number of positions?
can you figure out the total number of solutions?
it's really easy, just think about it.
#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.
same thing.
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yeah this type of stats is common sense. you can find the solutions pretty easily without using choose or permute. just do exactly what ydg said: find # of solutions, # of total positions, and work it out
if you're still having trouble, post back your thought process and we can see where (if anywhere) you go wrong
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For #2 I tried P(8,2)/P(8,6)+P(8,2)
P(8,2) because 8 total DVD's and 2 are in the correct positions and then P(8,6) equals the number that are in the wrong order (failure rate)
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I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D=
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On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D=
well ok
what constitues a solution in part 1? 1234XXXX
how many ways can you arrange ur dvd's such that this is the case
so you have
1234 5678
1234 5687
etc.
keep thinking you almost got it
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On February 23 2009 09:25 JeeJee wrote:Show nested quote +On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D= well ok what constitues a solution in part 1? 1234XXXX how many ways can you arrange ur dvd's such that this is the case so you have 1234 5678 1234 5687 etc. keep thinking you almost got it
Hmmm would it be p(8,4) = 1680? and then divide that by 8! so 1/24?
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That's not quite right. You want to count the number of arrangements 1234xxxx. How many ways can you put 5,6,7,8 into xxxx?
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um 4! = 4*3*2*1 = 24 ways?
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On February 23 2009 09:32 OmgIRok wrote:Show nested quote +On February 23 2009 09:25 JeeJee wrote:On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D= well ok what constitues a solution in part 1? 1234XXXX how many ways can you arrange ur dvd's such that this is the case so you have 1234 5678 1234 5687 etc. keep thinking you almost got it Hmmm would it be p(8,4) = 1680? and then divide that by 8! so 1/24?
yes.... since they first 4 are already in position.... u just need to worry about the last 4.... soo 4P4 or 4!
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so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something...
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huh?? you question is the number of ways u can arrange the last 4 cases right?
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On February 23 2009 09:41 Monoxide wrote:
huh?? you question is the number of ways u can arrange the last 4 cases right?
the question asks whats the probability that the 8 cases are put back so that 1234 are in the correct positions, cases 5678 dont matter
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On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something...
You've pretty much got it with "successful possibilities/total possibilities"
The ways 5678 can be arranged is your successful possibilities.
Total possibilities is the number of ways you can arrange 12345678.
The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. (But you should understand what nCr nPr actually do, you'll realise you might not want/need it)
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On February 23 2009 09:51 Nytefish wrote:Show nested quote +On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects.
But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720
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second question? 6! / 8! ??
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On February 23 2009 09:58 OmgIRok wrote:Show nested quote +On February 23 2009 09:51 Nytefish wrote:On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720
P(8,2) means the number of ways of picking 2 objects from 8, where order does matter, that doesn't really help.
Okay I was being vague, but what you're really doing is fixing 1 and 8. And permuting the 6 in between. So you're really doing P(6,6) = 6!.
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wait yeah order does matter because 1 and 8 have to be in the correct positions
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On February 23 2009 10:07 Nytefish wrote:Show nested quote +On February 23 2009 09:58 OmgIRok wrote:On February 23 2009 09:51 Nytefish wrote:On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720 P(8,2) means the number of ways of picking 2 objects from 8, where order does matter, that doesn't really help. Okay I was being vague, but what you're really doing is fixing 1 and 8. And permuting the 6 in between. So you're really doing P(6,6) = 6!.
yes... soo 6! / 8!
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You really need to check some definitions.
nCr means the number of ways of choosing r objects from n objects, where order doesn't matter. How does the number of ways of choosing 2 cases from 8 help? (it doesn't)
I gave you the answer in the last line of my previous post O_o. And in the one before that told you that nCr and nPr aren't needed.
@monoxide I know you got the answer but he clearly doesn't understand why
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On February 23 2009 10:16 Monoxide wrote:Show nested quote +On February 23 2009 10:07 Nytefish wrote:On February 23 2009 09:58 OmgIRok wrote:On February 23 2009 09:51 Nytefish wrote:On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720 P(8,2) means the number of ways of picking 2 objects from 8, where order does matter, that doesn't really help. Okay I was being vague, but what you're really doing is fixing 1 and 8. And permuting the 6 in between. So you're really doing P(6,6) = 6!. yes... soo 6! / 8!
OH wAIT YEAAA Okay so the success rates are all the positions where its 1XXXXXXX8 and since 1 & 8 are fixed you just find how many possible ways you can arrange 23456 which is 6! YES I SEE WHAT YOU DID THERE
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Okay the next question is P(all evens in the correct position) and I got 4!/8! <-- correct answer amirite?
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Yeah, these are the "easy" questions right?
The later ones will need you to use nCr and nPr?
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On February 23 2009 10:26 Nytefish wrote:
Yeah, these are the "easy" questions right?
The later ones will need you to use nCr and nPr?
Um... I think we are supposed to use ncr and npr on all of them as a shortcut to the counting principle but yeah whatever cuz i know npr and ncr formulas [not exactly sure what they are used for yet though, pretty sure used for what i mentioned earlier]
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well on to the next ones!!
I already did some of them, please check my answers? XD
3 people are chosen randomly out of 3 sophomores and 3 juniors
#1 P(0 sophmores)
3C0 / 6C3 = 1/20
2 P(1 sophmore)
3C1 / 6C3 = 3/20
#3 P(2 sophomore)
3 C 2 / 6C3 = 3/20
#4 P(3 sophomores)
3C3 / 6C3 = 1/20
#5 P(2 juniors)
Currently on #5
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The P(0 sophomores) + P(1) + P(2) + P(3) should add up to 1, so your answers to 1-4 don't make sense.
They don't add up because there are more than 3 ways to make a group with 1 or 2 sophomores. You also have to account for the number of different ways to pick the juniors.
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=O
okay I found out that P(1) and p(2) sophomores is actually supposed to be 9/20 chance to be picked, but where is that logic found?
6C1 gives me 6
6C2 gives me 15 15-6 = 9 
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P(1)=3C1*3C2/6C3=9 because there are 3C1 ways to choose 1 sophomore and 3C2 ways to choose the remaining juniors. The same applies to P(2) and then P(0) + P(1) + P(2) + P(3)=1
#5 is easy if you use the same its P(2)=3C1*3C2/6C3=9
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Oh I see. thankyou for explaining! 
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I note a lack of understanding in the multiplication principle when counting. The basic is that if an event A can occur in a different ways and an event B can occur in b different ways THEN the numbers of ways that the event A and B can occur is a*b.
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On February 23 2009 11:15 malongo wrote:
I note a lack of understanding in the multiplication principle when counting. The basic is that if an event A can occur in a different ways and an event B can occur in b different ways THEN the numbers of ways that the event A and B can occur is a*b.
haha yeah totally forgot about counting principle X_X eff me
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On February 23 2009 09:00 OmgIRok wrote:
#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
I got 1/70 for this one. I seriously doubt it's correct though.
#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.
Does it specifically state those are the only ones in the correct positions? Because P(1 & 8 correct) can have other seasons in correct spots the way you worded it.
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the first one is
8 nPr 5 or 8x7x6x5
the second one is
8 nPr 1 or 8! or 8x7x6x5x4x3x2x1
Each problem follows the same principles; you have a 1/8 chance of randomly placing the first dvd correctly, then a 1/7 chance of the next dvd and so on...
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I dont know if you know this problem but its a nice problem about "simple" probabilities:
The birthday problem:
In a room there are n people, what is the probability that at least 2 of them have a common birthday?
hint
+ Show Spoiler +use the complement probability, that is find the P(no pair of them have the same birthday), then P(at least 2 of them have a common birthday)= 1-P(no pair of them have the same birthday)
solution
+ Show Spoiler +since there are 365 posible birthdays then the total number of cases is 365!. The number of ways that no posible repetition of birthdays occur is 365*364*...*(365-n+1) because the first one have 365 posible birthdays, the second has 364 posible birthdays (because we want that he doesnt have the same birthday as the first), the third 363 and so on.
then P(no common birthday)= 365*364*...*(365-n+1)/365!=(365Cn)*n!/365! and the desired probablity is 1-(365Cn)*n!/365!.
why is this problem interesting? because its highly counterintuitive, for n=40 P>0.5 so for n=90 its almost sure that there are 2 with the same birthday.
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@malongo
check your equations i think you typo'd something
and 23 is the magic number not 40 last i checked
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ya the number is 23... and 23 gives a result of just over 50% actually
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On February 23 2009 13:33 JeeJee wrote:
@malongo
check your equations i think you typo'd something
and 23 is the magic number not 40 last i checked
?? please what is wrong? i rechecked but cant find it. And for the actual numebers i just put numbers that i was sure to fit, i dont remember them from memory i took probabilities about3 years ago dont ask that much ahaha
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EPT
