0

An urn has $n-3$ green balls and 3 red balls. Draw $l$ balls with replacement. Let $B$ denote the event that a red ball is seen at least once. Find $P(B)$ using inclusion exclusion with events $B_i= $ {ith draw is red}

I understand most of the problem except for the beginning. So to apply inclusion-exclusion we need to find the probabilities $P(A_{i_1}\cap\cdots \cap A_{i_k})$ for each choice of indices $1\leq i_1<\cdots<i_k\leq l$ (to be honest, I don't really understand what the inequality means when used with union or intersection symbols).

Let's derive the example $$P(A_2\cap A_5)=P(\text{the 2nd and 5th draw are red})$$

by counting favorable outcomes and total outcomes. Each of the $l$ draws comes from a set of $n$ balls, so $|\Omega|=n^l$. Now here is where I'm confused. The number of favorable outcomes is according to my book is $n\cdot 3\cdot n\cdot n\cdot 3\cdot n\cdots n = n^{l-2}3^2$.

Why are the other position not restricted to be $n-3$? If we want the probability that the 2nd and 5th draw are red, would that not imply that the other positions are not red? Or is that implication wrong? Are the events where other spots are red also counted when we find $P(A_2\cap A_5)$?

The rest of the problem makes sense, it's just this that I'm struggling to understand.

rudytheduck
  • 387
  • See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Jun 03 '24 at 05:00
  • 1
    The expression $~1 \leq i_1 < i_2 < \cdots < i_k \color{red}{\leq} l ~$ is used to represent that there are $~\displaystyle \binom{l}{k}~$ choices for the ordered $~k$-tuple $~(i_1, i_2, \cdots, i_k),~$ where the $~k~$ components of the $~k$-tuple are all distinct, and are in ascending order. The $~\displaystyle \binom{l}{k}~$ expression represents how many ways that there are of selecting $~k~$ elements from the set $~{1,2,\cdots,l},~$ sampling without replacement, where order of selection is deemed irrelevant. – user2661923 Jun 03 '24 at 05:04
  • Re previous comment, note that in your posted question, you specify $~1 \leq i_1 < \cdots < i_k \color{red}{<} l,~$ which I strongly suspect is a typo. – user2661923 Jun 03 '24 at 05:05
  • Not relevant, but I really hate problems that forces you to use a methodology. Here, PIE is totally unnecessary and the problem can be solved directly with very basic probability knowledge. The authors should make an effort in setting problems that make the wanted technique a game changer for the solution. Here is the opposite, using PIE is so convoluted that I doubt a student can understand when it's useful. – nicola Jun 03 '24 at 05:44
  • That makes a lot of sense. Thank you! I've also seen this notation with sums, what do they mean there? @user2661923 – rudytheduck Jun 03 '24 at 20:24
  • @nicola Yes, it would be nice if the wanted technique was a game changer; agreed. However, in rebuttal [1] PIE not so convoluted, because of considerations of symmetry. [2] The very fact that the problem is easily solved by alternate means is an advantage to teaching PIE. The student is forced to (a) verify that the two answers are equivalent and (b) presume and resolve analytical mistakes if the two answers are not equivalent. Good teaching tool for PIE. – user2661923 Jun 03 '24 at 21:31
  • To the original poster: $~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq l} ~{\text{compute something}}~~~$ represents that the computation is performed repeatedly, once for each of the $~\displaystyle \binom{l}{k} ~k$-tuples that I mentioned in my second comment. So, you end up with a summation that has exactly $~\displaystyle \binom{l}{k}~$ terms. – user2661923 Jun 03 '24 at 21:36
  • @user2661923 I didn't mean that PIE is convoluted by itself of course. However, for this problem, it would be very unnatural to use it. I doubt that, if taught just like this, PIE would become the powerful tool that is for the student. There are tons of problems that are very complicated and PIE can simplify tremendously. Here it just overcomplicates things, so my fear is that the student might think about it as just a math theorem that they are forced to learn, but without any real utility. – nicola Jun 04 '24 at 05:18

1 Answers1

1

The event $A_2 \cap A_5$ only "says something" about the positions 2 and 5. The others can be whatever. For example in the draw $G{\color{red}R}RG{\color{red}R}GRR$ 2nd and 5th are green so $A_2 \cap A_5$ is satisfied.

That is the thing that "makes inclusion-exclusion work"; we focus on particular things and the whole is then calculated as this intersection sum of them.

ploosu2
  • 12,367