Imagine you have a large bag full of skittles of X different colors. There are n skittles in total. In the bag there are n_{1} skittles of one specific color, n_{2} of another color, n_{3} of yet another and so on until n_{x} skittles of the last color.

Imagine you stick your hand into the bag without looking, and draw k skittles at random. You now want to know the probability of getting k_{1} skittles of the first color, k_{2} skittles of colur number two, and so forth until k_{x }skittles of color number X. I think the answer should be:

P(A) = [ ∏(from i = 1 to X) (n_{i }C k_{i}) ]/ (n C k)

Or, (if shown correctly on the computer screen):

X

P(A) = ∏ (n_{i }C k_{i}) /

i = 1 / (n C k)

Is this written correctly, or have I made a mistake somewhere?

Guest Nov 23, 2014

#5**+5 **

Thanks for all the answers!

I used the capital pi notation, which workes the same way as sigma notation, except that one uses products, not sums:

∏(i=1 to 5) 2i = 2*4*6*8*10 = 3840

So, after some trying and failing:

$$P(A) =\displaystyle \;\;\frac{Pi \limits_{n=1}^X\ ^{n{_i}}C_{k_i}}{^nC_k}$$

I'm really not able to write ∏ in the formula :(

Anyway, I really hope it's written correctly :)

Guest Nov 23, 2014

#1**+5 **

This is all too much for me at the moment but I would try using known quatities of things.

If I could work out what to do for the known quatities then I would extrapolate to get a formula for the unknown quanities.

I am not saying it will necessarily work but that is how I would try to attack the problem.

good luck

Melody
Nov 23, 2014

#2**+5 **

As far as notation goes I think that this is what you are after?

$$P(A)=\displaystyle \lim_{i=1}^X}\;\;\frac{^{n{_i}}C_{k_i}}{^nC_k}$$

this is written in LaTex. I opened the LaTex formula button and typed in

P(A)=\displaystyle \lim_{i=1}^X}\;\;\frac{^{n{_i}}C_{k_i}}{^nC_k}

Melody
Nov 23, 2014

#3**+5 **

Shouldn't each of the individual factors in your product have this form?

(n-sub-i C k-sub-i) · (n minus n-sub-i C k minus k-sub-i) / n C k

geno3141
Nov 23, 2014

#4**0 **

Beats me - i havent even thought about it and if i did it may not help much. :)

I just reproduced what was there.

Melody
Nov 23, 2014

#5**+5 **

Best Answer

Thanks for all the answers!

I used the capital pi notation, which workes the same way as sigma notation, except that one uses products, not sums:

∏(i=1 to 5) 2i = 2*4*6*8*10 = 3840

So, after some trying and failing:

$$P(A) =\displaystyle \;\;\frac{Pi \limits_{n=1}^X\ ^{n{_i}}C_{k_i}}{^nC_k}$$

I'm really not able to write ∏ in the formula :(

Anyway, I really hope it's written correctly :)

Guest Nov 23, 2014