# 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...

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?

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### 2 Answers

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

implying

**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.**

**Sources:**

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),...