A bag contains 4 printed articles of different kinds: periodicals, novels, news papers and hard covers. The total number of each kind is unknown, and hence the total number also. When 4 articles are drawn from the bag without replacement, the following events are equally likely - selecting:
1. 4 periodicals.
2. 1 novel and 3 periodicals.
3. 1 news paper, 1 novel, and 2 periodicals
4. 1 article each kind.
Q. What is the smallest number of articles in the bag satisfying these conditions? How many of these are of each kind?
First make a shorthand for the different types of articles to make the working clearer:
p - no. of periodicals P
n - no. of novels N
w - no. of newspapers W (because N is already taken)
h - no. of hard covers H
Hence there are (say) x = p + n + w + h articles altogether.
Work out the probability of each statement and express it algebraically. Then equate the probabilities to find (p,n,w,h) and hence also x. All sampling is without replacement.
1) Probability of choosing 4P = (p/x). (p-1)/(x-1) . (p-2)/(x-2) . (p-3)/(x-4) =
Ch(p,4)/Pe(x,4) = p(p-1)(p-2)(p-3)/[x!/(x-4)!]
because we condition on 4 articles of a certain type being chosen from the x articles (they must all be periodicals). Which 4 are chosen is determined by Ch(p,4). Ch - choose, Pe - permutation.
2) Probability of choosing NPPP = (n/x) . p/(x-1) . (p-1)/(x-2) . (p-2)/(x-3)
3) Probability of choosing WNPP = (w/x) . n/(x-1) . p/(x-2) . (p-1)/(x-3)
4) Probability of choosing 1 article of each kind =
Pr(PNWH) + Pr(PNHW) + Pr(PWNH) + Pr(PWHN) + Pr(PHNW) + Pr(PHWN) +
Pr(NPWH) + Pr(NPHW) + Pr(NWPH) + Pr(NWHP) + Pr(NHPW) + Pr(NHWP) + etc
= Ch(p,1).Ch(n,1).Ch(w,1).Ch(h,1)/Pe(x,4) x 4! = 4! (pnwh)/[x(x-1)(x-2)(x-3)]
We condition on 4 articles of a certain type being chosen from the x articles. Which one of each type is determined by Ch(y,1) where 1 is the generic number of any type. The order of the type of articles chosen doesn't matter and there are 4! ways of arranging 4 items.
We can work out what x, n, w and h are by equating all these probabilities:
1) = 2) = 3) = 4) that is (cancelling out the x(x-1)(x-2)(x-3) in all the denominators and p in all the numerators
(p-1)(p-2)(p-3) = n(p-1)(p-2) = wn(p-1) = 4! nwh
n = p-3
w = p-2
h = (p-1)/4!
and hence x = p + (p-3) + (p-2) + (p-1)/4! = (3-1/4!)p - 5 - 1/4!
NB (p-1) must be a multiple of 4! for h to be a whole number.
So the smallest actual number for h is 4! = 24 so that p = 25, n = 22, w = 23.
Smallest actual number, min(x), in the bag is hence given by
min(x) = min(p) + min(n) + min(w) + min(h) = 25 + 22 + 23 + 24 = 94
Other possibilities are multiples of 24, so that x could be 94, 24(94), 24^2(94),...