All Answers

+893

There are usually two methods of solution for his type of problem, a method based on counting, as used by Melody and CPhill, and a basic probability method.

The counting method has already been discussed, so here is the probability method.

Suppose that the six smarties fall in the sequence BBBBOO, where B represents a blue smartie and O one of the other colours.

The probability of that sequence happening will be

(6/19)*(5/18)*(4/17)*(3/16)*(13/15)*(12/14).

If you consider any other sequence consisting of four blues and two others, the individual fractions will be different but the combined top and bottom lines will be the same.

So, for example, if the sequence happened to be OBBOBB, the probability would be (13/19)*(6/18)*(5/17)*(12/16)*(4/15)*(3/14).

Different fractions, but the combined top line is still 13*12*6*5*4*3 and the bottom line 19*18*17*16*15*14.

So, the overall probability will be this probability multiplied by the number of possible sequences consisting of four blues and two others, that is,

(6/19)*(5/18)*(4/17)*(3/16)*(13/15)*(12/14)*6C4 = 0.0028748*15 = 0.0431(4dp).

BertieApr 12, 2014

fukin b***h can't even do that s**t?

1+1=2!!

GuestApr 12, 2014

+3146

enter: e^(3/4)

$${{\mathtt{e}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} = {\mathtt{2.117\: \!000\: \!016\: \!612\: \!675}}$$

adminApr 12, 2014

+3146

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{2}}$$

adminApr 12, 2014

5.0*10^13 = .029/x(.18)^4 Flip both sides to get x on top

x(.18)^4/0.29= 1/5.0*10^13 multiply by 0.29

x(.18)^4= .29/5.0*10^13 Divide by (.18)^4

and you get x

GuestApr 12, 2014

+118725

This is the LaTex code that i used. It didn't display properly if I didn't use display mode (with $$)
I didn't need to use all of those packages, they are just a part of quick start with Texmaker, which is the version of LaTex that I use.

Latex - sum of display.JPG

MelodyApr 12, 2014

+118725

Thank you Alan,
I will have to try and remember that.

MelodyApr 12, 2014

+33666

CPhill:
I'm not a probability "guru," but I believe that we are just counting "sets" of things. (BTW...the marbles could have hit the ground one by one, but maybe not....maybe 6 of them (or any subset of such) struck the ground almost simultaneously !!!........ )

So, the total number of "sets" that we could have (the sample space) is just C(19, 6).
And we're interested in choosing 4 of the 6 blue ones and the other 2 out of the remaining 13.

So we have

C(6 , 4) * C( 13, 2) / C(19, 6)

Which I believe is the same thing that Melody found, so I won't bother with any computations !!!

Anyone else have any thoughts on this one ???

Yes, Melody is right. The probability of selecting a blue one changes each time a smartie is removed from the bag. Sometimes both the number of blues and the total number reduces; sometimes the number of blues stays the same and just the total reduces. Rom's method assumes the probability, p, is constant.

By the way, to answer Melody's query about using combination notation here (though it still appears in factorial form in the black bar):

nCr(6,2)*nCr(13,2)/nCr(19,6)

AlanApr 12, 2014

+130561

I'm not a probability "guru," but I believe that we are just counting "sets" of things. (BTW...the marbles could have hit the ground one by one, but maybe not....maybe 6 of them (or any subset of such) struck the ground almost simultaneously !!!........ )

So, the total number of "sets" that we could have (the sample space) is just C(19, 6).
And we're interested in choosing 4 of the 6 blue ones and the other 2 out of the remaining 13.

So we have

C(6 , 4) * C( 13, 2) / C(19, 6)

Which I believe is the same thing that Melody found, so I won't bother with any computations !!!

Anyone else have any thoughts on this one ???

CPhillApr 12, 2014

+118725

Hi Rom,
I want you to display the code that you have used please. Can you take a snip of it please?

I keep meaning to write a proper LaTex post. i just haven't gotten around to it yet.
I often refer to the Latex cookbook but it is not very comprehensive and finding code for obscure things can be quite difficult and time comsuming.
I thought eventually we could build up our own LaTex cookbook.
We would have the code on one side and the display on the other. I thought we could give it a go anyway. It is just time consuming to set up and to keep managing.
Anyway, it is a thought.

hi Chris,
It doesn't look like you used Latex. Did you use MathType or something differerent again.

MelodyApr 12, 2014

+118725

Hi Rom,
I question your solution because I don't think that you can use a binomial distibution. By definition with a binomial each selection must be independant of the last one.
If the marbles roll on the floor one at a time then the probability of the next marble being blue is different each time.

MelodyApr 12, 2014

+6252

Clipboard01.jpg

is this what you mean?

RomApr 12, 2014

+118725

SassuWunnu:
We have 19 smarties in a bag left. 6 spills on the floor. What are the chances of 4 out of that 6 being blue, if there were 2 red 6 blue 5 brown 3 yellow and 3 green in the bag?

Hi Sassu,
I don't have a lot of confidence with probability questions but I am pretty sure this is right.
There will be other replies. Probably quite soon so you will be able to see if other people agree with me.

There are 19C6 ways of choosing 6 marbles from 19
You want 4 blue and 2 a different colour.
the number of ways of choosing 4 blue from 6 is 6C4
the number of ways of choosing 2 from 13 other colours is 13C2
so the probability of getting exactly 6 blue and 2 others is

(6C4 * 13C2) / 19C6 = (15 * 78) / 27132 = 0.0431225 or 4.31225% correct to 6 significant figures

(6!/(4!*2!)*(13!/(2!*11!))/(19!/(6!*13!))

Can anyone tell me if I can enter this using combination notation and have it display straight onto the forum? Thank you.

MelodyApr 12, 2014

+6252

Quote:
We have 19 smarties in a bag left. 6 spills on the floor. What are the chances of 4 out of that 6 being blue, if there were 2 red 6 blue 5 brown 3 yellow and 3 green in the bag?

let p = Pr[smartie in the bag is blue] = 6/19

then the number of blue smarties out of the 6 spilled is a binomial random variable

Pr[k blue smarties out of 6 spilled] = Choose(6,k) p ^k(1-p) ^(6-k)

Choose(n,k) is "n choose k", or n!((n-k)!k!)

Pr[0, 1, ... 6] = (0.102598, 0.284117, 0.327828, 0.20174, 0.0698331, 0.0128923, 0.000991713)

Pr[4 of 6 spilled were blue] = 0.0698331

RomApr 12, 2014

+118725

CPhill:
15
4 * Σ 3^{(n -1)} * (-1)^(n-1)
n = 4

I can't get this to display correctly...The "15" is supposed to be directly over the summa sign and the "n = 4" is supposed to be directly under it....I think you get the idea......

Hi Chris,
What have you used to try and display this. Maybe someone can help you to display it properly and then we may all know for next time.
Melody.

MelodyApr 12, 2014

+118725

bump

MelodyApr 12, 2014

#97

+118725

Hi everyone,
It has been a quiet 3 days. That is to be expected at this time of year. Many young people would be on holidays.
There sere some good questions and they were answered by Alan, CPhill, Reinout-g, and DavidQD. Thank you.
The question titled Surface Area and Volumes of Spheres attracted some good debate.
I don't think anything else stood out.
Enjoy your weekend.
Melody.

MelodyApr 12, 2014

#13

+118725

Thanks reinout-g,

I also refer people to the mathisfun site all the time. It is a great site.

MelodyApr 12, 2014

+118725

bump

MelodyApr 12, 2014

+118725

reinout-g:
mariahmorgan143:
-2-5

-2-5 = -7

http://www.mathsisfun.com/positive-negative-integers.html

Also take a look at this ( Just highlight-> left click -> choose goto )
http://web2.0calc.com/questions/positive-and-negative-numbers-directed-numbers

If you post a response, I will be glad to check your answers.

MelodyApr 12, 2014

+118725

Thanks Chris.

MelodyApr 12, 2014

#14

+330

Alan:
DavidQD:

...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).

Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! ...