Probability Theory 6 | Hypergeometric Distribution

Поделиться
HTML-код
  • Опубликовано: 18 дек 2024

Комментарии • 23

  • @jordanfernandes581
    @jordanfernandes581 3 года назад +8

    This series has a really nice balance between abstraction and application. I'm enjoying it and looking forward to learning and developing an intuition for probability theory. Thank you 🙂

    • @brightsideofmaths
      @brightsideofmaths  3 года назад +2

      Thank you very much for your kind words. I really want to combine different aspects of probability theory. I would be a pity if one knows a lot about measure theory and stochastic processes but never heard of the hypergeometric distribution :)

    • @jordanfernandes581
      @jordanfernandes581 3 года назад +2

      @@brightsideofmaths This is the first time I'm hearing of the hypergeometric distribution. As someone learning just out of curiosity I often miss out on some essential ideas because of a lack of understanding or awareness. Having a more complete overview like this is definitely much appreciated. I'll be going through your course on measure theory as well along with this. 🙂

  • @surendrabarsode8959
    @surendrabarsode8959 3 года назад +2

    Very well explained!! You are doing a great favor to those who are learning mathematics and statistics. One can learn and appreciate the concepts easily and clearly from your videos. Thanks !!

  • @IgorVladK
    @IgorVladK 3 года назад +3

    great series, great explanations! R is a nice touch.

  • @jaimelima2420
    @jaimelima2420 3 года назад +3

    Very nice, as always. As a suggestion, this series would be a good match for a another one about Stochastic Processes, which in turn would connect the series Distribution Theory already available. It would be good to be able to learn Stochastic Processes from a more mathematical theory point of view.

    • @jaimelima2420
      @jaimelima2420 3 года назад

      Sorry, forgot to mention …. That suggestion above also should have included also spices of
      Ergodic theory…

  • @cowgomoo444
    @cowgomoo444 Год назад

    took me a second to realize that the hypergeometric distribution with 2 colors is different from the binomial distribution because there is no replacement. in the binomial distribution the probabilities didnt change over the process because we kept replacing balls in the urn. here once we take them out we keep them out, so it's not the same.
    great video as always. just leaving this comment in case anyone is a little slow like me haha

  • @rodilrun9195
    @rodilrun9195 2 года назад

    Thank you sir

  • @minglee5164
    @minglee5164 3 года назад

    What is the relationship between this one and Dirchlet distribution?

  • @starryzhang8194
    @starryzhang8194 Год назад +1

    I am not a native speaker of English. I noticed that you canceled the production of English subtitles in the following video, which greatly hindered my smooth learning from the video. I hope you can produce English subtitles in each video, which will make the subtitles automatically translated by youtube more accurate, which will help promote your videos to non-native English speakers.
    I like your videos very much and recommend you to my classmates. I will support you when I have the ability!

    • @brightsideofmaths
      @brightsideofmaths  Год назад +1

      You are right! A lot of subtitles are missing and usually some nice people in the community do them. I try to write more!

  • @EmanueleBonardi80
    @EmanueleBonardi80 6 месяцев назад

    I still don't understand the sample space.. I mean i am focusing on # (success ) / #(possible outcomes )... I don't understand how #(possible outcomes) can just be Cn(N, n) when this number contain no information about colours .. but it is just counting the way you can pick n balls out of N balls.. no matter the colour.
    I tried to figure out a formula by my self.. and in case | C | = 2, it is Cn(n, k) * Cn(N-n, K-k) / sum_i^(min(n,K) Cn(n,i)*Cn(N-n,K-i)

    • @brightsideofmaths
      @brightsideofmaths  6 месяцев назад

      But the sample has the colors in the tupel given.

    • @EmanueleBonardi80
      @EmanueleBonardi80 6 месяцев назад

      @@brightsideofmaths thank you for your kind answer. i don't understand how you figure out Cn(N, n) , it is just counting the way you pick up n balls from N balls, no matter the colour.
      I don't understand the relationship between the sample space and the Cn(N, n)

    • @EmanueleBonardi80
      @EmanueleBonardi80 6 месяцев назад

      I don't understand why Cn(N, n) is representative of "all possible outcomes" since there is no information about colours in the CN(N, n) expression.

    • @brightsideofmaths
      @brightsideofmaths  6 месяцев назад

      @@EmanueleBonardi80 The sample space consists of tupels (k_1, k_2, k_3, k_4). Each position represents one colour.

    • @EmanueleBonardi80
      @EmanueleBonardi80 6 месяцев назад

      Ok. So i don't understand how to count tuples