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[Partial Gamification]

thanks for the advice guys. I've got the truth tables and euclids symbols kind of down already... so yea I guess it's mostly just practice (and yes I took discrete because I thought intro to proofs would be a pain, never should have minored in math!).

Anyway, seems like I'm getting confused on the last steps... like I can follow the logic, step for step, but then we get to the end and everybody says "Yes, this proves that, we're done." and I just don't see it.

For example, (here's the problem I'm working on now, I hope it comes out here in a way that is readable))

Theorm:
For every n in the integers such that n is greater than or equal to 1, 2+4+6+...+2n=n^2+n

Proof:
P = 2+4+6+...+2n = (n^2)+n
P(1) = 2(1) = ((1)^2)+1, so P(1) is true.
P(k) = 2+4+6+...+2k = (k^2)+k, assume P(k) is true.
P(k+1) = 2+4+6+...+2(k+1) = ((k+1)^2)+(k+1) the next step is where I go wrong, on the left side of the equation I want to multiply out 2(k+1) to 2k+2, and then fill in the previous integers in the sequence to get:
= 2+4+6+...+2k+(2k+1)+(2k+2) = ((k+1)^2)+(k+1) and then I can substitute P(k) and simplify both sides, but it doesn't work out. The book says the sequence for P(k+1) simplifies to 2+4+6+...+2k+2(k+1) "by making the next to last term explicit", but I don't see how 2k comes before 2(k+1)..... oh son of bitch, I get it now... fucking MATH! Damn you. I'm not going to delete all this, I'm just going to post it, because fucking math, that's why.

fake edit: In case anyone read this and can't figure it out: I screwed up because I kept thinking (2k+1) is in the sequence, but it's not because the sequence is only the even numbers and (2k+1) is odd.

I'm not sure where you're going wrong, its like you have it but its just not organized right or you just have done enough induction yet. This is what I got out of your post:

 

[MisterLuffy]

I need help on this problem, I just want to get an idea how to answer this question because the other problems are similar. And thanks cpp_is_king for answering my previous question.

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries.

a) If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this?
b) If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this?
c) If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety?
d) If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?
e) If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

 

[XenodudeX]

So I'm trying to factor this : 30w^3 +25w^2-30w

I pulled out the gcf of 5 to get 5w(6w^2 +5w -6)

I don't think I can factor this any further, right?

 

[DEATH™]

So I'm trying to factor this : 30w^3 +25w^2-30w

I pulled out the gcf of 5 to get 5w(6w^2 +5w -6)

I don't think I can factor this any further, right?

You still can. Use quadratic equation if you are trapped... computed it and it's still clean...

 

[Memento_Mori15]

I'm not sure where you're getting the 2's from, but I get an equivalent answer that doesn't use any 2's.

Fix a starting seat, and then place the women in a circle starting from that seat. There are 4! ways to do that. Once its done, there are an additional 4! ways to place the men. This is a total of 4! * 4! = 576 ways to place the people at the round table. However, there are 8 ways of symmetry (for each configuration, the configuration obtained by rotating the entire table 1,2,3,...,8 spaces to the right is equivalent. So we divide by 8. This gives (4! *4!) / 8 = 72.

This is equivalent to your idea about (2*4!*4!) / (2*8), but I'm not sure where the 2 is coming from, so perhaps it's easier to understand using my explanation.

Forgot I posted this. But the way I got the 2's is that there are 2 ways you can start. You can start with a man or woman. Thus 2 choices. But the part I'm getting confused in was why are we dividing by 8. But as you said, it's equivalent(right?). I do like the way you thought this out though. It makes it much simpler. Thank you.

So I'm trying to factor this : 30w^3 +25w^2-30w

I pulled out the gcf of 5 to get 5w(6w^2 +5w -6)

I don't think I can factor this any further, right?

You definitely can:
Possible pairs: 1 and 6. 2 and 3.
It can't be 6 or one, no matter if you negate 6 or 1, the middle value will never be 5. So then play around with 2 and 3. There's probably a better way to do this but you get better at it the more you do it and figuring out which pair is good.
Answer:

Spoiler

(2w+3) (3w-2)so:
5w(2w+3) (3w-2)

@Ketch
Good luck with Discrete Math. I'm in the second portion of it( Discrete Math I is proofs, Discrete Math II is probability, I think, where I am). I took the first portion over the summer but I had to attend office hours and ask questions here. I couldn't find any resource for discrete math online. It does get easier as you go along though but you'll find yourself doubting and wanting to double check. Or maybe not lol. I miss Calc compared to discrete math lol.

 

[cpp_is_king]

So I'm trying to factor this : 30w^3 +25w^2-30w

I pulled out the gcf of 5 to get 5w(6w^2 +5w -6)

I don't think I can factor this any further, right?

Sadly, not enough people know about the rational root theorem. It makes factoring so easy. Every time someone posts a factoring question, I always take the opportunity to explain the rational root theorem. It's overkill for quadratics because you can alwys apply the quadratic formula, but for anything higher than quadratics it's indispensable.

We've got this polynomial 6w^2 + 5w - 6. The first coefficient is 6. the factors of 6 are +-1, +-2, +-3, +-6. The last coefficient is -6, and has the same factors.

Any rational root will be of the form (factor of last coefficient) / (factor of first coefficient).

So we need to find all fractions that you can generate as a result of taking the numbers from one set divided by number from another set. Doing this, we get the following:

+-1/6, +-1/3, +-1/2, +-2/3, +-1, +-3/2 +-2, +-6

So that seems like a shitload of possibilites to check, but we can prune alot of the wrong answers easily by plugging in all the integer values.

P(0) = -6
P(1) = 5

We see that the sign of the result changes (from -6 to 5). That's what we're looking for. At this point, you can delete every single possibility from that candidate list that doesn't fall in that range. This means our root has to be between 0 and 1. So it's either 1/6, 1/3, 1/2, 2/3.

Use a binary search approach by always choosing a value from the middle of the result set to reduce the posibilities by the most. So let's try 1/2. If you do this you get -2. Now the results are:

P(0) = -6
P(1/2) = -2
P(1) = 5

So the answer is between 1/2 and 1. The only choice remaining is 2/3, which is the root.


This probably sounds complicated, and indeed it is overkill for this particular problem (the quadratic formula is easier). But this is absolutely the best way to factor higher degree polynomials when it's not obvious from inspection what the roots are. For example, here's a polynomial:

x^3 - 4x^2 - 11x + 30

Turns out this is really easy to factor with the RRT. The possibilities are +-1, +-2, +-3, +-5, +-10, +-15, +-30

P(0) > 30
P(1) > 0
P(2) = 0

So 2 is a root. factoring 2 out you get:

(x-2)(x^2-2x-15)

and you can use RRT again or get (x+3)(x-5) by inspection.

 

[Ketch]

@Ketch
Good luck with Discrete Math. I'm in the second portion of it( Discrete Math I is proofs, Discrete Math II is probability, I think, where I am). I took the first portion over the summer but I had to attend office hours and ask questions here. I couldn't find any resource for discrete math online. It does get easier as you go along though but you'll find yourself doubting and wanting to double check. Or maybe not lol. I miss Calc compared to discrete math lol.

I hated calc.

Had my first exam in discrete this morning... think I did alright, but that's probably a bad sign.

 

[cyborg009]

I need help solving the congruence and I'm not sure Iif this is the correct answer

solve 5x + 3 ≡ 19 mod(6)
I'm getting x ≡ 8 mod (6) while my friend is getting x ≡ 2 mod (6)

 

[big ander]

I need help solving the congruence and I'm not sure Iif this is the correct answer

solve 5x + 3 ≡ 19 mod(6)
I'm getting x ≡ 8 mod (6) while my friend is getting x ≡ 2 mod (6)

those are the same, really. Honestly I've never set up an equation the way that one is so maybe your prof considers one more right, but 8 and 2 are congruent in mod 6. 8–6 = 2.

I believe the way you'd figure this out is to subtract 19 from each side and then find an x that makes your new LHS a multiple of 6. So you'd now have 5x – 16. You could use 2 because 5*2 – 16 = -6 = -1 * 6, you could use 8 because 5*8 – 16 = 24 = 4 * 6, you could even use 20 because 5*20 – 16 = 84 = 14 * 6. Shot in the dark, but I'm guessing the answer your prof likes "most" is 2 because it's "on the clock." Remember that the best way for many to think of modular arithmetic is as a clock where your modulus (here 6) is at the 0 or 12 o'clock position. So if you were to go 8 ticks forward/clockwise, you'd be in the same spot as if you had gone 2 ticks.

 

[cyborg009]

those are the same, really. Honestly I've never set up an equation the way that one is so maybe your prof considers one more right, but 8 and 2 are congruent in mod 6. 8–6 = 2.

I believe the way you'd figure this out is to subtract 19 from each side and then find an x that makes your new LHS a multiple of 6. So you'd now have 5x – 16. You could use 2 because 5*2 – 16 = -6 = -1 * 6, you could use 8 because 5*8 – 16 = 24 = 4 * 6, you could even use 20 because 5*20 – 16 = 84 = 14 * 6. Shot in the dark, but I'm guessing the answer your prof likes "most" is 2 because it's "on the clock." Remember that the best way for many to think of modular arithmetic is as a clock where your modulus (here 6) is at the 0 or 12 o'clock position. So if you were to go 8 ticks forward/clockwise, you'd be in the same spot as if you had gone 2 ticks.

The way he show us was really long and the way the book did it way better. And the sad thing is when I ask my professor about this he kind of ignore me.

edit: Also thank you. I also have another one if you don't mind answering.

 

[Banglish]

edit: nvm

 

[Memento_Mori15]

So while doing my probability hw, this problem came up

...I don't what's it really asking. It just seems really vague to me. Any hints?

 

[DarkLinkage]

So while doing my probability hw, this problem came up

...I don't what's it really asking. It just seems really vague to me. Any hints?

It wants the value of p.

If there are two outcomes then you can say outcome one + outcome two must be equal to 1, since there is a 100% chance of either of those two outcomes occurring.

So:
(1-p)+(2p^2)=1

Solve for p.

I haven't done this in a while but I'm quite sure I'm correct.

 

[TCRS]

I can't bloody find the eigenvectors for this (it's already in A-lambdaE form):

0 1 1
0 0 1
0 0 -1

gaussian elimination gives me (basically L3+L2 and L1-L2):

0 1 0
0 0 1
0 0 0

does that mean t2=0 (t being the eigenvector), t3=0 and t1=alpha? In which case the eigenvectors would be

1
0
0

I'm confused.

 

[TUSR]

If you write me out the whole question I'll solve it for you.

 

[TCRS]

well I'm asked to solve

2 1 1 * y = y'
0 2 1
0 0 1

I've already calculated the eigenvalues which are 2, 2, 1. the calculation above is for the first eigenvectors of 2 (I already have them for lambda=1). The first column being 0 is confusing me.

 

[TUSR]

well I'm asked to solve

2 1 1 * y = y'
0 2 1
0 0 1

I've already calculated the eigenvalues which are 2, 2, 1. the calculation above is for the first eigenvectors of 2 (I already have them for lambda=1). The first column being 0 is confusing me.

Yeah if you do cofactor expansion along column 1. By inspection it looks like 2,2,1. I'll write you out a good solution after I'm done classes today if you still haven't got it down

 

[Memento_Mori15]

It wants the value of p.

If there are two outcomes then you can say outcome one + outcome two must be equal to 1, since there is a 100% chance of either of those two outcomes occurring.

So:
(1-p)+(2p^2)=1

Solve for p.

I haven't done this in a while but I'm quite sure I'm correct.

So it's really that simple? I don't know, it just seems really easy among the list of problem we have lol. But my TA did say not to over think it and that it was supposed to be easy...

 

[DarkLinkage]

So it's really that simple? I don't know, it just seems really easy among the list of problem we have lol. But my TA did say not to over think it and that it was supposed to be easy...

Not much information is given, so naturally the question shouldn't be complicated.

 

[Petrichor]

So this is probably going to be super rudimentary for you guys but I'm racking my brains over this - maybe I'm tired.

It's just simplifying the expression 10x^5 / 2x^-3

So I get as far as simplifying it to 5x^5 / x^-3 just by dividing through, and x^a / x^b = x^a-b would lead me to believe that the answer is 5x^8 only it clearly isn't. WHAT AM I DOING WRONG WHY IS MATHS LIKE THIS.

The only other way I can think of to tackle it is rejigging the divisor so that its 5x^5 / (1/x^3) but it all seems so arbitrary and hnnnng.

 

[Okamid3n]

So this is probably going to be super rudimentary for you guys but I'm racking my brains over this - maybe I'm tired.

It's just simplifying the expression 10x^5 / 2x^-3

So I get as far as simplifying it to 5x^5 / x^-3 just by dividing through, and x^a / x^b = x^a-b would lead me to believe that the answer is 5x^8 only it clearly isn't. WHAT AM I DOING WRONG WHY IS MATHS LIKE THIS.

The only other way I can think of to tackle it is rejigging the divisor so that its 5x^5 / (1/x^3) but it all seems so arbitrary and hnnnng.

Assuming you mean 10x^5 / (2x^-3) and not (10x^5 / 2) * x^-3, this is indeed 5x^8.

x^a / x^b = x^a-b is right, so you did it correctly.

A trick to confirm that you did it correctly would just be inputting a number or two in place of x in your calculator, but don't forget the parenthesis or else you'll end up calculating (10x^5 / 2) * x^-3 which is most likely not what you want.

 

[thequickandthedead]

Anyone have any experience with the poisson distribution? Stuck on this:

"Customers arrive at post office at avg. rate of 0.6 per hour.

Using tables, estimate the greatest period of time such that the probability of more than 5 customers arriving is less than 0.1"

 

[Okamid3n]

Anyone have any experience with the poisson distribution? Stuck on this:

"Customers arrive at post office at avg. rate of 0.6 per hour.

Using tables, estimate the greatest period of time such that the probability of more than 5 customers arriving is less than 0.1"

I don't remember how these things work at all, so don't listen to me making a fool of myself, I'm just having fun with wild guesswork and googling everything I don't remember.

But here's what I'd do:

You want P(X > 5) < 0.1 for a given period of time. Using a Poisson table with &#955; = 3.1, I got P(X>5) = 0.0942, which is just a little less than 0.1.

So, you want &#955; = 3.1. As we recall, &#955; is your mean rate. Your mean, here, is 0.6/h, or 0.6 / 60mins.

0.6/60mins is equivalent to 3.1/310mins. If you want &#955; = 3.1, your period of time must be 310 minutes.

 

[MisterLuffy]

I have a question. Thanks in advance.

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A) = .4 and P(B) = .7.

a) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.
b) What is the probability that at least one of the two projects will be successful?
c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

 

[cpp_is_king]

c)Ignore this. Will fix tomorrow. I would've said 1-0.7=0.3, but who knows?

Consider all 4 outcomes:

A & E : P = (.4)(.7) = .28
A & ~E : P = (.4)(.3) = .12
~A & E : P = (.6)(.7) = .42
~A & ~E : P = (.6)(.3) = .18


Since we're given that at least one is successful, we know it wasn't the last outcome. So there are only 3 possibilities:

A & E : P = (.4)(.7) = .28
A & ~E : P = (.4)(.3) = .12
~A & E : P = (.6)(.7) = .42

The sum of all these probabilities is .28 + .12 + .42 = .82. It's asking for the probability out of these 3 that only the Asian project was successful, which corresponds to the second event in the above list. So the probability is .12 / .82 = .146

 

[MisterLuffy]

It's possible the answers (especially the last) are wrong as it's pretty late over here and I'm pretty tired, but for some reason I felt like doing some math before going to bed. Sorry. Actually I'm pretty sure c) is wrong, but I really need some sleep. Curse you conditional probabilites!

It's cool, I blame myself for not getting a tutor for help. Also, I don't have the courage to ask my instructor for help because he usually replies with another question which I'm unable to answer.

Consider all 4 outcomes:

A & E : P = (.4)(.7) = .28
A & ~E : P = (.4)(.3) = .12
~A & E : P = (.6)(.7) = .42
~A & ~E : P = (.6)(.3) = .18


Since we're given that at least one is successful, we know it wasn't the last outcome. So there are only 3 possibilities:

A & E : P = (.4)(.7) = .28
A & ~E : P = (.4)(.3) = .12
~A & E : P = (.6)(.7) = .42

The sum of all these probabilities is .28 + .12 + .42 = .82. It's asking for the probability out of these 3 that only the Asian project was successful, which corresponds to the second event in the above list. So the probability is .12 / .82 = .146

Thank you.

 

[RyanDG]

Okay. I have a series of 64 numbers and I am told that it is a hill cipher using a matrix of unknown size to decipher it. I do know that the resulting numbers will range from 0-27 with 0 representing a space and then the following 1-26 representing a - z in the English alphabet. What steps could you take to help crack the cipher?

I know that one approach would be a brute force approach (looking at number repetition and corresponding it to the frequency in letters appearing in common words and trying to work backwards from there), but are there any other techniques that I can use in this case?

 

[luoapp]

Nvm.

 

[Memento_Mori15]

Can someone tell me if I did this correctly? This is one of the few I didn't really get.

My attempt:
I can describe the probability space by splitting it like so:
C: space of colors{red, green, yellow}
S: Size{small, big}
Thus CxS:
(Red,small): 15/20
(Red,big): 5/20
(Green, small): 20/30
(Green, big): 10/30
(Yellow, small): 25/50
(Yellow, big): 25/50.

A)Probability space:
It would be similiar to the above but the denominator would be 100?

B)61/100
C)15/100?

And one other question:

Now this seems like a simple problem, it would be choose(5 2). Yet, I feel like there's more to it. Since the order doesn't matter, I feel like I have to do something with the choose...I feel like I have to divide choose(5 2) with 5....but not entirely sure...

 

[youngwerther]

We just went over the quotient rule and I'm not really comfortable with it yet.

But see I looked at that kinda like you said with the dividing to simplify but I wasn't sure how exactly to do that with the -8/x. I get it now, but for whatever reason that part was blowing my mind :lol

Thanks for the help, I appreciate it.

I pretty much never use quotient rule.
If you don't like it just rewrite the expression as a product instead of a quotient and then use product rule.

So the one you gave rewritten would be

(3x^3 + 3x^2 + 6x -8) *x^-1

 

[Rctdaemon]

*snip*
A)Probability space:
It would be similiar to the above but the denominator would be 100?

B)61/100
C)15/100?

And one other question:
Now this seems like a simple problem, it would be choose(5 2). Yet, I feel like there's more to it. Since the order doesn't matter, I feel like I have to do something with the choose...I feel like I have to divide choose(5 2) with 5....but not entirely sure...

I'm going to guess that A is asking you need to find the probability of colors and the probability of sizes without being dependent on each other. Your probabilities are correct, but I don't know how you have to present them. B should be 60/100 and C is correct.

As for your second question, it's not going to be choose(5 2). That would be selecting two people out of a group of five. You're on the right path with the choose, but you need different numbers.

 

[Memento_Mori15]

I'm going to guess that A is asking you need to find the probability of colors and the probability of sizes without being dependent on each other. Your probabilities are correct, but I don't know how you have to present them. B should be 60/100 and C is correct.

As for your second question, it's not going to be choose(5 2). That would be selecting two people out of a group of five. You're on the right path with the choose, but you need different numbers.

I guess I'll have one as separate probabilities and then a combination of them and go from there.

After you wrote it out, I figured it had to be choose (10 5). Thanks for the help.

But now I think I got a previous problem wrong...or rather over-counting.

For this one, say we have [1 2]. We can make that into [2 1]. So that's a pair on it's own. The pair [2 1] is not the same as [1 2], so we must count that as well. I thought it would be choose (9 2). It just being choose (9 2) just doesn't really seems right. I think I may be over-counting.

I hate these counting problems...

 

[coldfoot]

But now I think I got a previous problem wrong...or rather over-counting.

For this one, say we have [1 2]. We can make that into [2 1]. So that's a pair on it's own. The pair [2 1] is not the same as [1 2], so we must count that as well. I thought it would be choose (9 2). It just being choose (9 2) just doesn't really seems right. I think I may be over-counting.

I hate these counting problems...

The wording of this question is unclear. First it says 1:2 and 2:1 are used interchangably, then it asks how many different kinds of pieces are there, but it doesn't explicitly state whether 1:2 and 2:1 should be counted as 1 or 2, I know they're used interchangably.

My interpretation is that 1:2 is the same piece as 2:1 and should only be counted as 1, which results in
9+8+7+6+5+4+3+2+1= 45 different dominos.

 

[Memento_Mori15]

The wording of this question is unclear. First it says 1:2 and 2:1 are used interchangably, then it asks how many different kinds of pieces are there, but it doesn't explicitly state whether 1:2 and 2:1 should be counted as 1 or 2, I know they're used interchangably.

My interpretation is that 1:2 is the same piece as 2:1 and should only be counted as 1, which results in
9+8+7+6+5+4+3+2+1= 45 different dominos.

Now I'm confused with this question...there are 81 possible pairs where order matters, isn't there? Hmm...not quite sure.

 

[Okamid3n]

Can someone tell me if I did this correctly? This is one of the few I didn't really get.

My attempt:
I can describe the probability space by splitting it like so:
C: space of colors{red, green, yellow}
S: Size{small, big}
Thus CxS:
(Red,small): 15/20
(Red,big): 5/20
(Green, small): 20/30
(Green, big): 10/30
(Yellow, small): 25/50
(Yellow, big): 25/50.

A)Probability space:
It would be similiar to the above but the denominator would be 100?

B)61/100
C)15/100?

And one other question:

Now this seems like a simple problem, it would be choose(5 2). Yet, I feel like there's more to it. Since the order doesn't matter, I feel like I have to do something with the choose...I feel like I have to divide choose(5 2) with 5....but not entirely sure...

B is 60/100, but I really don't think C is 15/100 unless I'm misunderstanding something big. You are told that A happens, so the only possibilities are the 60 small balls. Of the 60 small balls, 15 are small red balls. The probability is 15/60, not 15/100. You're in the subset of situations where you know the ball is small, so all the big balls are out of the equation!

Now I'm confused with this question...there are 81 possible pairs where order matters, isn't there? Hmm...not quite sure.

Yes, there's 81 possiblities if order matters.

If order DOESN'T matter, though, you have

1 and 123456789 (9 possibilities starting with 1 on top)

2 and 23456789 (8 possibilities starting with 2 on top; we took out 2/1 because it was counted above already and order doesn't matter)

3 and 3456789 (7 possibilities; we took out 3/1 and 3/2 because they got counted already)

...

9 and 9

So it's 9+8+7+6+5+4+3+2+1 = 45

There's no way they would have worded the question that way if you needed to count 2/1 and 1/2 as two different kinds of pieces.

 

[alazz]

B is 60/100, but I really don't think C is 15/100 unless I'm misunderstanding something big. You are told that A happens, so the only possibilities are the 60 small balls. Of the 60 small balls, 15 are small red balls. The probability is 15/60, not 15/100. You're in the subset of situations where you know the ball is small, so all the big balls are out of the equation!

Yeah, I agree with you. My interpretation is that it's asking for the probability of Red Small given only Small, which would be 15/60. It's like drawing a face card from a deck with only one suit remaining.

As for those basketball players, it's definitely not 5C2. I'm not sure (I haven't taken stats in like two years), but shouldn't it be 10C5 * 2? My reasoning is that you choose your first five, and the others go to the other team. So you need to know the amount of combinations for the first team, and then double it to account for the second team? Still, that doesn't totally make sense because there are only 5 unique players for the second team and not 10 (as per the * 2)...

Talk to your professor, if possible, about that last one.

 

[Okamid3n]

Yeah, I agree with you. My interpretation is that it's asking for the probability of Red Small given only Small, which would be 15/60. It's like drawing a face card from a deck with only one suit remaining.

As for those basketball players, it's definitely not 5C2. I'm not sure (I haven't taken stats in like two years), but shouldn't it be 10C5 * 2? My reasoning is that you choose your first five, and the others go to the other team. So you need to know the amount of combinations for the first team, and then double it to account for the second team? Still, that doesn't totally make sense because there are only 5 unique players for the second team and not 10 (as per the * 2)...

Talk to your professor, if possible, about that last one.

Naw, it's 10C5 x 5C5, which is just 10C5 x 1. Imagine, you choose 5 out of 10 for the first team, then you choose 5 out of the remaining 5 for the last team. Basically, after you form the first team, the second team is automatically formed, there's nothing to choose. For every first team of 5 you form, you automatically form the other team as well.

For example, if it was 12 players and you had to make two teams of 5, it'd be 12C5 x 7C5. But here, it's just 10C5 x 5C5 = 10C5

 

[Memento_Mori15]

B is 60/100, but I really don't think C is 15/100 unless I'm misunderstanding something big. You are told that A happens, so the only possibilities are the 60 small balls. Of the 60 small balls, 15 are small red balls. The probability is 15/60, not 15/100. You're in the subset of situations where you know the ball is small, so all the big balls are out of the equation!




Yes, there's 81 possiblities if order matters.

If order DOESN'T matter, though, you have

1 and 123456789 (9 possibilities starting with 1 on top)

2 and 23456789 (8 possibilities starting with 2 on top; we took out 2/1 because it was counted above already and order doesn't matter)

3 and 3456789 (7 possibilities; we took out 3/1 and 3/2 because they got counted already)

...

9 and 9

So it's 9+8+7+6+5+4+3+2+1 = 45

There's no way they would have worded the question that way if you needed to count 2/1 and 1/2 as two different kinds of pieces.

For C, that's what I originally had but then I thought about applying the conditional probability formula and got 3/20...which is 15/100 unless I used it incorrectly for this case...or did I use baye's rule...

I see, that would make sense then.

Are there any resource I could use to improve my way of thinking for counting? We're not using a textbook for this class

 

[cpp_is_king]

Can someone tell me if I did this correctly? This is one of the few I didn't really get.

My attempt:
I can describe the probability space by splitting it like so:
C: space of colors{red, green, yellow}
S: Size{small, big}
Thus CxS:
(Red,small): 15/20
(Red,big): 5/20
(Green, small): 20/30
(Green, big): 10/30
(Yellow, small): 25/50
(Yellow, big): 25/50.

A)Probability space:
It would be similiar to the above but the denominator would be 100?

B)61/100
C)15/100?

B should be 60/100 = 3/5. For part C, let's change the denominators to 100 and then erase all the events that don't have a small ball since we're told a small ball is chosen.

(Red,small): 15/100 = .15
(Green, small): 20/100 = .2
(Yellow, small): 25/100 == .25

These 3 lines now consist of the entire sample space due to being given that we drew a small ball. The probabiliby that it was red is thus .15 / (.15 + .2 + .25) = .25

And one other question:

Now this seems like a simple problem, it would be choose(5 2). Yet, I feel like there's more to it. Since the order doesn't matter, I feel like I have to do something with the choose...I feel like I have to divide choose(5 2) with 5....but not entirely sure...

Let's think about it a different way. How many different ways are there to choose the members of the first team? This would be (10; 5). Once you choose the first 5 members, the second team is decided. So the answer is (10; 5) = 252

But now I think I got a previous problem wrong...or rather over-counting.

For this one, say we have [1 2]. We can make that into [2 1]. So that's a pair on it's own. The pair [2 1] is not the same as [1 2], so we must count that as well. I thought it would be choose (9 2). It just being choose (9 2) just doesn't really seems right. I think I may be over-counting.

I hate these counting problems...

A good way to do this one is to assume that the squares are ordered. This solves the order problem by only considering dominos where the first square is less than or equal to the second square.

Suppose the first square is 1, there are 9 choices for the second square. If the first square is 2, there are 8 choices for the second square (we're ignoring 2 / 1 because 1/2 was already counted by the previous step). etc. So we've got 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 9*10 / 2 = 45

Are there any resource I could use to improve my way of thinking for counting? We're not using a textbook for this class

http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting

I admit that this series of books is geared towards competition mathematics, so it contains some very difficult problems, but at the same time this is the easier of the two probability books, and it does in fact go through all the basics. More importantly is that the explanations are excellent and very clear, and it teaches very intuitive and easy to understand strategies for solving counting problems. Plus, when you understand the explanations for the difficult problems, then the easy problems become trivial. FWIW, the problems don't just immediately start difficult, it just offers the more difficult ones for those who want to push their limits a bit further.

 

[Okamid3n]

For C, that's what I originally had but then I thought about applying the conditional probability formula and got 3/20...which is 15/100 unless I used it incorrectly for this case...or did I use baye's rule...

Conditional probability formula works here.

We have A = {a small ball is chosen} and let's name B = {a red ball is chosen}.

P(B|A) = P( B and A both happens) / P(A happens)

P(B and A happens) is just the probability of having a small ball, with that ball being red. There's 15/100=0.15 chance of picking a small red ball.

P(A) = 60/100 = 0.6 as calculated before.

As such, P(B|A)= 0.15/0.6 = 15/60 = 1/4.

 

[alazz]

Can someone briefly tell me why Trapezoidal, Midpoint, and Simpson rules are taught if series are so much more accurate? Are there times when they're preferred? (My class just finished the three rules, but we don't start series until next Thursday.)

Naw, it's 10C5 x 5C5, which is just 10C5 x 1. Imagine, you choose 5 out of 10 for the first team, then you choose 5 out of the remaining 5 for the last team. Basically, after you form the first team, the second team is automatically formed, there's nothing to choose. For every first team of 5 you form, you automatically form the other team as well.

For example, if it was 12 players and you had to make two teams of 5, it'd be 12C5 x 7C5. But here, it's just 10C5 x 5C5 = 10C5

Oooh, thanks for clearing that up. I originally though it was 10C5 * 5C5 but because 5C5 is 1 I didn't think that that made much sense at the time the question was posted. I was thinking about the second team in terms of permutations, which is why I got confused at the order of the second team.

I should really leaf through my old notes before I try to help with combinatorics and permutations again, heh.

 

[DEATH™]

Can someone briefly tell me why Trapezoidal, Midpoint, and Simpson rules are taught if series are so much more accurate? Are there times when they're preferred? (My class just finished the three rules, but we don't start series until next Thursday.)

There are some functions that is waaay complicated to be able to fully integrate or even be transformed into series. So you will still gonna go back to these numerical methods if things don't go right.

(And nowadays computers do this for us though...)

 

[cuevas. PhD.]

Is there a maths help thread

 

[alazz]

There are some functions that is waaay complicated to be able to fully integrate or even be transformed into series. So you will still gonna go back to these numerical methods if things don't go right.

(And nowadays computers do this for us though...)

I see, thanks! I don't like doing the homework when n = 32, but it makes me feel better knowing that it's not an obsolete method.

 

[youngwerther]

We are learning U-substitution and I'm finding it a bit tricky.

What would I substitute in this example?

f(x) = sinxcos(cosx)dx

Should I make u be cos?
If I did that I would get du = -sin(x)dx.

Then I could write -u(u)du

??
But what does it mean to write u(u)? That's why I'm confused.

 

[Leezard]

We are learning U-substitution and I'm finding it a bit tricky.

What would I substitute in this example?

f(x) = sinxcos(cosx)dx

Should I make u be cos?
If I did that I would get du = -sin(x)dx.

Then I could write -u(u)du

??
But what does it mean to write u(u)? That's why I'm confused.

Let u be cos(x) instead. that way you get -cos(u)du

 

[youngwerther]

Let u be cos(x) instead. that way you get -cos(u)du

Ok.
Now the only thing I'm not understanding is I seem to be left over with an extra negative.

When I integrate -cos(u)du, I can pull the -1 out in front of the integral.

Then I integrate and get -sin(u), then plug back in for u and get

-1(-sin(cos(x))

leaving me with

sin(cos(x))

I must be doing something wrong though because the answer is supposed to be

-sin(cos(x))

Where did I mess up?

 

[hemtae]

Ok.
Now the only thing I'm not understanding is I seem to be left over with an extra negative.

When I integrate -cos(u)du, I can pull the -1 out in front of the integral.

Then I integrate and get -sin(u), then plug back in for u and get

-1(-sin(cos(x))

leaving me with

sin(cos(x))

I must be doing something wrong though because the answer is supposed to be

-sin(cos(x))

Where did I mess up?

If you pull the negative sign out of the integral, you should just get sin(u) once you integrate.

 

[youngwerther]

*sigh*
duh.
thanks.

 

[youngwerther]

If I need to find the new limits of integration for the function that I talked about above, do you just plug in the old limits of integration into the "u" term?

If so, how do I know whether to use degrees or radians?
The old limits of integration was [-2,3].