3
$\begingroup$

There are $n$ urns $u_1,u_2,\cdots,u_n$.Each urn contains $2n+1$ balls.The $ith$ urn contains $2i$ types of balls.$P(u_i)$,i.e. probability of selecting $ith$ urn is proportional to $i^2+2$.If we randomly select one of the urns and draw one ball and probability of ball being white be $P(A)$,then the question is to evaluate $$\lim_{n \to \infty}P(A)$$

I tried to use $$P(A)=\sum_{i=1}^n P(A \cap u_i)P(u_i)$$$P(u_i)=k(i^2+2)$ .I understand that there must be atleast two balls in the urn which has the same colour.But I couldnot proceed with this.Any ideas?Thanks.

  • 1
    For clarification, you say each urn contains $2n+1$ balls, then the $i$-th urn contains $2i$ types of balls. So how many white balls are in the $i$-th urn? – 2017-02-23
  • 1
    I do not understand the wording of the question [what is a "type" of ball?] Yet, I observe that the quantity $P[A]$ does not seem to depend on $n$, while you take a limit as $n\rightarrow \infty$. So as written, clearly $\lim_{n\rightarrow\infty} P[A]=P[A]$ (the limit of a constant is that same constant). – 2017-02-24

1 Answers 1

2

As noted in the comments, the text of this problem lacks some important information. It is stated that each urn contains $2n+1$ balls and that the $i^{th} $ urn contains $2i$ types of balls, but there is no information about the number of white balls contained in the $i^{th} $ urn. This number could range from a minimum of $1$ (assuming that the remaining $2n $ balls in each urn are distributed among the remaining $2i-1 \,$ colors) and a maximum of $2n+1-(2i-1) \,$ (assuming that, within each urn, there is only one ball for each of the $2i-1 \,$ colors different from the white, and that all remaining balls are white). I will provide a solution for these two scenarios to give a bound for the searched probability limit, highligting that this necessarily cannot be considered a complete solution (not achievable without additional data).

Since the probability of selecting the $i^{th} $ urn is proportional to $ i^2+2 \,$, we can write it as $k (i^2+2) \, $. We can then calculate the sum of these probabilities for all urns as

$$\sum_{i=1}^n k(i^2+2)=\frac {1}{6} kn (2n^2+3n+13)$$

Because this sum includes the probabilities of all possible cases, it must be equal to $1$. So we obtain

$$\frac {1}{6} kn (2n^2+3n+13)=1$$

$$ k=\dfrac {6}{ n (2n^2+3n+13)}$$

Therefore, for a given $n $, the probability of selecting the $i^{th} $ urn is

$$ \dfrac {6 (i^2+2)}{ n (2n^2+3n+13)}$$

Now if we consider the scenario where there is only one white ball in each urn, to obtain the probability of choosing a white ball in the $i^{th} $ urn we have to multiply the last quantity by $1/(2n+1) \,$. In this case the searched probability limit is

$$\lim_{n \to \infty} P (A) = \lim_{n \to \infty} \sum_{i=1}^n \dfrac {6 (i^2+2)}{ n (2n^2+3n+13)(2n+1)} \\ = \lim_{n \to \infty} \dfrac {1}{(2n+1)}=0$$

On the other hand, if we consider the scenario where there are $2n+1-(2i-1) \,\,$ white balls in each urn, to obtain the probability of choosing a white ball in the $i^{th} $ urn we have to multiply the last quantity by $ (2n+2-2i) /(2n+1) \,\, $. In this case the searched probability limit is

$$\lim_{n \to \infty} P (A)= \lim_{n \to \infty} \sum_{i=1}^n \dfrac {6 (i^2+2)(2n+2-2i)}{ n (2n^2+3n+13)(2n+1)} \\ = \lim_{n \to \infty} \dfrac {n^3+4n^2 +17n+14}{4 n^3+8 n^2 + 29 n +13} =\dfrac {1}{4}$$