0

We are given a stack of presents received in a bit hday. Each of them contains one of $5$ different figures $F_1,...,F_5.$ We assume that the probability of finding any of those figures is the same, $\mathbb{P} = \frac{1}{5}.$ I want to use the inclusion-exclusion principle in calculating the probability of finding only the figures $F_1,F_2,F_3$ in 6 sequential presents. In case one would write down the inclusion-exclusion principle given 6 events, it becomes too complex. I do not understand how to define appropriately the events we need in finding the required probability. I will appreciate your support in providing some hints or a solution proposal. Thanks.

  • 1
    To clarify... do you want only those outcomes where all three of these occurred? Or at most these three having occurred? If the latter, then this is just a binomial distribution. If it is the former... then I recommend using Stirling Numbers of the Second Kind if you wish to avoid a (more obviously) inclusion-exclusion style argument. – JMoravitz Apr 20 '23 at 20:02
  • Thanks. I want only the outcomes such that out of the 6 presents one gets the outcomes containing $F_1, F_2$ and $F_3.$ That means that each of these had occurred at least once. But I want to use the inclusion-exclusion principle. I do not see if I have to express the union as the union of $3$ or 6 events. In the latter case the expansion of that formula gets quite complicated. – user996159 Apr 20 '23 at 21:22

2 Answers2

1

Less elegant Inclusion-Exclusion approach:

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

The desired probability may be expressed as

$$\frac{N}{D} ~: ~D = 5^6.$$

That is, the denominator above represents that you have $~5~$ choices for each of the $~6~$ gifts.

So, the problem reduces to computing $~N.$

Following the model in the second link:

  • Let $~S~$ denote the set of all the ways that the $~6~$ sequential presents contain no gifts other than $~F_1, ~F_2, ~$ or $~F_3.$
    Then
    $|S| = 3^6.~$

  • For $~k \in ~\{1,2,3\}~$ let $~S_k~$ denote the subset of $~S~$ such that the constraint that the $~6~$ sequential gifts include at least one gift $~F_k~$ is violated.

Then,

$$N = |S| - |S_1 \cup S_2 \cup S_3|.$$

Let :

  • $T_0~$ denote $~|S|.$

  • $T_1~$ denote $~|S_1| + |S_2| + |S_3|.~$

  • $T_2~$ denote $~|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|.$

  • $T_3~$ denote $~|S_1 \cap S_2 \cap S_3|.$

Then, per Inclusion-Exclusion theory,

$$N = T_0 - T_1 + T_2 - T_3.$$

By symmetrical considerations:

  • $T_1 = 3 \times (2)^6.$
    That is, to compute $~|S_1|~$ note that you have exactly $~2~$ choices for each of the $~6~$ gifts.
    Further, by symmetrical considerations, $~|S_1| = |S_2| = |S_3|.$

  • By similar considerations of symmetry,
    $T_2 = 3 \times (1^6).$

  • $T_3 = (0)^6 = 0.$
    That is, since the set $~S~$ represents all of the ways that there can be $~6~$ sequential gifts, each of which is an element in $~\{F_1, F_2, F_3\}~$ it is not possible to have any such group of $~6~$ sequential gifts, taken from the set $~S,~$ where each of the elements in the set $~\{F_1, F_2, F_3\},~$ is excluded.

Therefore,

$$N = 3^6 - [~3 \times 2^6~] + [~3 \times 1^6].$$

Therefore, the probability is

$$\frac{3^6 - [~3 \times 2^6~] + [~3 \times 1^6]}{5^6}.$$

user2661923
  • 35,619
  • 3
  • 17
  • 39
  • Thanks for your post. What is the meaning of the sets $S_k, k\in { =1,2,3}? $ If the constraint is violated, the sets $S_k$ can not be subsets of S, right? – user996159 Apr 21 '23 at 08:27
  • @user996159 $S_1~$ is the subset of $~S~$ where none of the $~6~$ sequential gifts were the gift $~F_1,~$ so $~S_1~$ represents that all of the $~6~$ sequential gifts were either $~F_2~$ or $~F_3.~$ $~S_2~$ and $~S_3~$ are similarly defined. – user2661923 Apr 21 '23 at 08:31
  • Why is $N$ not simply equal to $|S|$? – user996159 Apr 21 '23 at 09:12
  • @user996159 Because of one of your comments, following the posting, where you specified that the $~6~$ gifts must include at least one occurrence of each of $~F_1, F_2, F_3.~$ The set $~S,~$ which represents all distributions of the $~6~$ gifts that do not include any occurrence of either $~F_4~$ or $~F_5,~$ includes distributions where (for example) $~F_1~$ is also excluded. Hence the application of Inclusion-Exclusion against the subsets $~S_1, S_2, S_3.$ – user2661923 Apr 21 '23 at 09:17
  • Thanks. I do not get the correct result, though. Should one also subtract from $|S|$ the elements of subsets of $S$ which are missing two and not only one of $F_1, F_2, F3$ ? – user996159 Apr 21 '23 at 13:12
  • @user996159 : I advise against trying to take shortcuts, as you study the material. First, review the two Inclusion-Exclusion links fairly carefully. Then review my answer fairly carefully. The question that you just asked seems to indicate that you didn't focus on the fact that $~N = T_0 - T_1 + T_2 - T_3.$ – user2661923 Apr 21 '23 at 13:19
  • Thanks. Anyway the answer is deviating by almost a factor of $10.$ Subtracting didn't help much. And maybe the answer that I have is wrong and your result is true. – user996159 Apr 21 '23 at 13:22
  • @user996159 What is the official answer to the problem? Alternatively, what do you think that the answer should be? – user2661923 Apr 21 '23 at 13:34
  • $0.34947$ vs $0.03436.$ – user996159 Apr 21 '23 at 13:37
  • @user996159 Please click on the following link, so that we can discuss the problem in a chat room: Chat Room Link. – user2661923 Apr 21 '23 at 13:53
0

Building off of this CV answer (which builds off this post that uses inclusion-exclusion):

The probability of observing $k$ unique letters in $m$ random uniform samples from an alphabet of size $n$ is:

$$\bigg\{\!{m\!\atop{k}}\bigg\}\binom{n}{k}\frac{k!}{n^m}$$

Where $\big\{\!{m\!\atop{k}}\big\}$ is the Stirling number of the second kind.

But you don't want just any set of $k$ letters, you want one specific set, so divide by the number of ways to get $k$ out of $n$ letters:

$$\bigg\{\!{m\!\atop{k}}\bigg\}\frac{k!}{n^m}$$

jblood94
  • 316