Probability

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

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

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.

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 ???

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):

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

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).