From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: high schooler needs help w/ probability
Date: 6 Jul 1998 16:17:24 GMT

In article <359c8f1e.43387331@news.pcmagic.net>,
Chen <kchen@pcmagic.com> wrote:

>N unique stickers are printed on a sticker sheet. X of the stickers
>are printed once, Y of the stickers are printed twice. The stickers
>are then cut individually, mixed up, and sold in groups of Z.
>
>i'm hoping for a general formula that can be used to plug in actual
>numbers. it should be able to answer the following questions:
>
>Q1. What is the probability of receiving a SPECIFIC once-printed
>sticker?
>
>Q2. What is the probability of receiving a SPECIFIC twice-printed
>sticker? (i'm guessing its twice that of q1)


No, no, you can't state it this way. It's required by law that you
discuss the removal of marbles from urns -- consult any combinatorics text.

So you have  X+2Y marbles in a big urn, of a grand total of N=X+Y different
colors. You're randomly reaching in and selecting  Z  of them. Is that
the picture?

If so, there are many different sets of marbles which may be selected,
namely  C( X+2Y, Z ), where  C(a,b) = a!/b!/(a-b)! is the number of subsets of
cardinality  b  in a set of cardinality of  a  (assuming  0<=b<=a, of course).

Of those, how many sets include the lone chartreuse marble? Well, the
sets in question are formed by selecting  Z-1  marbles from among the
X+2Y-1  non-chartreuse marbles, along with the one chartreuse one. There
are  C( X+2Y-1, Z-1 ) ways of doing so. Thus your probability of
selecting a chartreuse-containing set when you select a subset randomly is
C( X+2Y-1, Z-1 )/C( X+2Y, Z ) = Z/(X+2Y).

Similarly, you may ask how many sets include one of the two
heliotrope marbles. Your question is a tad ambiguous: do you care
_which_ of the heliotrope ones you get? do you mind getting _both_?
The sets containing both heliotrope marbles number  C(X+2Y-2, Z-2).
The sets containing the chipped heliotrope marble but not the smooth one
number  C(X+2Y-2, Z-1), as do the sets containing the smooth one but not
the chipped one.

Thus the probabilities involving the heliotrope marbles are
	Z(Z-1)/(X+2Y)/(X+2Y-1)		to get both
	Z(X+2Y-Z)/(X+2Y)/(X+2Y-1)	to get chipped not smooth
	Z(X+2Y-Z)/(X+2Y)/(X+2Y-1)	to get smooth not chipped
	2Z(X+2Y-Z)/(X+2Y)/(X+2Y-1)	to get just one
	Z(2X+4Y-Z-1)/(X+2Y)/(X+2Y-1)	to get at least one

None of these is equal to twice the probability of getting the chartreuse
marble if  X>0, Y>0  and  Z>1, but if  Z  is small relative to  X+2Y  then you 
have _roughly_ twice the chance of drawing a heliotrope marble as drawing the 
chartreuse one, so your intuition is good.

dave