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Math Problems - Manu - 2010-11-17



Math Problems - KajitiSouls - 2010-11-17

I'm getting 20 people without tokens as well =/

 Work

EDIT: Stereo seems to have nailed it. You really have to hate when the problem writers, though they are technically correct, lead you into false assumptions, and I know this will get you fired in some professions.

Math Problems - Stereo - 2010-11-17

5: every color (these guys have red + white, and blue + white, and red + blue)
10: red + white
20: blue + white
15: red + blue
15: red
10: white
20: blue
= 5 without any tokens.

Math Problems - Kaasoljoyyx - 2010-11-17

Stereo Wrote:5: every color (these guys have red + white, and blue + white, and red + blue)
10: red + white
20: blue + white
15: red + blue
15: red
10: white
20: blue
= 5 without any tokens.

This. There's a formula for these types of problems, it goes like

no one getting anything = + (A + B + C +...) - (AB + AC + ... + BC + BD + ...) + (ABC + ABD +... + BCD + BCE + ...) - (ABCD + ...) etc...

Drawing a "venn diagram" helps too. Start from the middle and work outwards

Math Problems - Hazzy - 2010-11-17

Kaasoljoyyx Wrote:Drawing a "venn diagram" helps too. Start from the middle and work outwards

Seconding this.

Flawed method... that worked?
 Spoiler

How did that work when I pushed subtracting 5 to the end and how did you guys get 20?

Math Problems - KajitiSouls - 2010-11-17

Since there's a thread like this already...

I'm trying to do maffs on this Plinko game I found on newgrounds.com, which is an imitation of The Price is Right. It's basically a giant binomial distribution machine (durrr) and where the chips fall are based on where you drop it.

Pic. The one I'm modeling after is shorter than that.

I already calculated the odds of a single chip starting in the middle falling into these categories:
$100 - 2/256
$500 - 16/256
$1000 - 56/256
0 - 112/256
$10,000 - 70/256

This is only one of many probabilities I'm trying to calculate: What are the odds of you winning $30,100 or more using 5 plinko chips?

Math Problems - Noah - 2010-11-17

KajitiSouls Wrote:Since there's a thread like this already...

I'm trying to do maffs on this Plinko game I found on newgrounds.com, which is an imitation of The Price is Right. It's basically a giant binomial distribution machine (durrr) and where the chips fall are based on where you drop it.

Pic. The one I'm modeling after is shorter than that.

I already calculated the odds of a single chip starting in the middle falling into these categories:
$100 - 2/256
$500 - 16/256
$1000 - 56/256
0 - 112/256
$10,000 - 70/256

This is only one of many probabilities I'm trying to calculate: What are the odds of you winning $30,100 or more using 5 plinko chips?

I'll answer in probability, not odds. I don't like odds, they are weird.

First of all, you have to get at least 3 $10,000 in order to get more than $30,100. Then you know that you need at least one of the other chips to not be 0. The last one is don't care.

so basically, this problem is
[Image: 2wr59wz.png]

Which in this case equal to
[Image: 2eoa434.png]

Which turns into 1929375/8388608 or ~0.23

Noah

Math Problems - KajitiSouls - 2010-11-17

How about in non-probability equations plz? I can't read that...

Math Problems - Russt - 2010-11-17

http://en.wikipedia.org/wiki/Binomial_coefficient ?

Math Problems - Hazzy - 2010-11-17

How did you get your answers without being able to read probability equations? o.o

Math Problems - Stereo - 2010-11-17

[Image: 29qj3q8.png]

Is equivalent to "a choose b", or [Image: 3ydlm4t.png] - the number of ways you can choose b items from a set of a elements if you don't care about order.


They're related to Pascal's triangle directly via the bth element in the ath row, for the binomial probability bit. Noting of course that a = 0, 1, 2, ..., while b = 0, 1, ..., a, and taking the identity 0c0 = 1

Math Problems - Noah - 2010-11-18

KajitiSouls Wrote:How about in non-probability equations plz? I can't read that...

You could eventually do this through dynamic programming (Which would be easier to generalize), but I don't know if that's easier.

Noah

Math Problems - Noah - 2010-11-18

Sorry for being spammy: Here's probably (pun not intended) what you want.

 Spoiler

Turns out my answer's wrong too, so there's probably some case I've forgot, or some miscalculation on my side.

Noah

Math Problems - Manu - 2010-11-26



Math Problems - Hazzy - 2010-11-26

I found the triangle DCP:
The bottom, DC, is 3QC. (DQ is 2QC, and QC is QC)
The right, PC, is 3PB. (Given)

The right side is 4 PB (3PB + PB)

Both sides are equal to k, so 3QC = 4PB

QC = 4/3 * PB

tangent (3PB/3QC) = tan(PB/QC) = tan(4/3*QC/QC) = tan(3/4) ~= 36.87 degrees.

You know an angel and a side, I think that's enough to break down the rest of the triangle's sides / area. I haven't done this in two years, so iunno.

Edit: Nope, you need another side. Rawr. I'll get back to this.
Leaving this here for reference. http://gcse.wikia.com/wiki/Area_of_a_triangle

Math Problems - Manu - 2010-11-26

Yeah, you're given that the base (DQ) is 2k/3. So you need to figure out the height, using a x/y axis system, I --- just found out by writing this that I screwed up xD, I think I know how to do it now.
Wheee got it Big Grin

I used a system on x/y axis.
basically:
AQ segment is y=-3/2x + k
and DP is y=3/4x

you put it as y=y and get M (x,y)
3/4x = -3/2x + k
therefor x = 4k/9, replace in either equation and y=k/3

Y in M is the height of the triangle

base * height /2

2k/3 * k/3 = 2k^2/9 divide by 2 and you get the result k^2/9

Math Problems - Corn - 2010-11-26

Another math problem: Calculus this time.

"A rectangular post has an area of 200 cm^2 is inscribed in a circle. (So imagine a rectangle in a circle) Find the dimensions of the post and the radius r of the circle to have a minimum circular area."

Math Problems - Hazzy - 2010-11-26

If a and b are sides of the rectangle and r is the radius of the circle, then the point from the corner of the rectangle to the center of the rectangle is equal to r.

Or:
(sqrt ((a/2)^2+b^2)) = r

Using that relation between the rectangle's dimensions and the circle's radius, you could write a function for the circumference of the circle given rectangle side lengths.
and 200/a=b, so...

2*pi*(sqrt( (a/2)^2 + (200/a)^2 ) = circumference. Differentiate that with respect to a to find the minimum value. Not entire sure what the limits would be though. That could get messy.

Math Problems - Corn - 2010-11-26

I have to find the area, but the derivative of the area is the circumference anyways, so w/e. I'm attempting to find it, and I got 0=2pi (sqrt w^4+40000) / 4x^2. It's...impossible to solve for w O_o.
.
Also here: 2*pi*(sqrt( (a/2)^2 + (200/b/2)^2 ) and here: (sqrt ((a/2)^2+b/2^2)) = r. Isn't there supposed to be a b/2? Why is that left out to find the radius?

Math Problems - Salguod - 2010-11-26

I hope you have the answers, because I'm not sure if it's right. Seems to work out though.

First, I wrote Min: Area of circle

Then I wrote the relations
(2r)^2 = x^2 + y^2
xy = 200

Next, I did A = pi (r ^2)

I rewrote r in terms of x using the two relations and I got r = (1/2) sqrt [(x^2 + (200/x)^2]
I plugged that into the equation for the area and differentiated.

Hard to type out the steps, but I got x^4 = 40000, so x = 10 root 2

I think you should use a closed interval, but I used the first derivative test to determine that's the minimum because I'm confused by domains.

Plugged x in to get y, which also came out to 10 root 2, so I guess it's a square.

Plugged in x and y to get r, which came out to 20.