Math Problems

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

 Work

Code:

Given:
--100 People
--150 Tokens
----45 Red Tokens
----45 White Tokens
----60 Blue Tokens

Possible grabs:
--R
--W
--B
--RW
--RB
--WB
--RWB
--0

    R    W    B    T Peop    T Tok
15 RW    15    15        15    30
25 BW        25    25    25    50
20 RB    20        20    20    40
5 RWB    5    5    5    5    15
    ________________________    ___________
Count    40    45    50        135
    
R    5            5    5
W        0        
B            10    10    10
                ___________________
                80    150

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.

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.

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

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

Seconding this.

Flawed method... that worked?

 Spoiler

I made 3 circles in a box. The box represents everyone. Those in the box and not a circle get nothing. Each circle represents a color. Each circle over-laps the other two circles.

Where the Red and White circles overlap, label "15". Red and Blue 20, Blue and White 25. These numbers are given.

Since 45 people got red, one circle, including the bits shared with other circles, must be 45. You have 15+20=35 already accounted for in the Red circle, so you can conclude that 45-35=10 people got exclusively red. Repeat for the two remaining circles.

My end result was 10 with no tokens. When I reread the problem and post, I saw "5 with all three". 10-5 = 5.

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

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?

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


Which in this case equal to


Which turns into 1929375/8388608 or ~0.23

Noah

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

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

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

Is equivalent to "a choose b", or - 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

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

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

 Spoiler

Code:

def CDF (P, V, minGoal, N):
  lookup = {}
  PV = zip(P, V)
  def recCDF(val, n):
    if (val, n) in lookup:
      pass
    
    elif n == 0:
      if minGoal <= val:
        lookup[(val, n)] = 1
      else:
        lookup[(val, n)] = 0
    
    else:
      lookup[(val, n)] = sum([p * recCDF(val + v, n - 1)
                              for (p, v) in PV])
    
    return lookup[(val, n)]
  
  return recCDF(0, N)

VV = [100, 500, 1000, 0, 10000]
PP = [2./256, 16./256, 56./256, 112./256, 70./256]

print CDF(PP, VV, 30100, 5)

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

Noah

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

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

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

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."

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.

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?

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.