From d9bb34c054dc7dd5da6b5e6f6f5df73af09b1c6d Mon Sep 17 00:00:00 2001 From: Simon Veitner Date: Tue, 9 Jul 2024 01:41:36 +0200 Subject: [PATCH] Add --- _posts/2024-07-08-combination-permutation.md | 2 +- _posts/2024-07-08-introduction-probability.md | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) diff --git a/_posts/2024-07-08-combination-permutation.md b/_posts/2024-07-08-combination-permutation.md index daa6c97..b3d5eff 100644 --- a/_posts/2024-07-08-combination-permutation.md +++ b/_posts/2024-07-08-combination-permutation.md @@ -134,7 +134,7 @@ event = {h for h in hands if h[0].rank == h[1].rank} How many elements to be have in the event? We choose one length and than can choose two suits so we have -$\#E = {13 \choose 1}{4 \choose 2}$ +#E $ = {13 \choose 1}{4 \choose 2}$ ```python len(event) == get_number_k_out_of_n(13,1)*get_number_k_out_of_n(4,2) diff --git a/_posts/2024-07-08-introduction-probability.md b/_posts/2024-07-08-introduction-probability.md index 92a69b8..3e30e5e 100644 --- a/_posts/2024-07-08-introduction-probability.md +++ b/_posts/2024-07-08-introduction-probability.md @@ -49,7 +49,7 @@ event {('H', 'H', 'H'), ('H', 'H', 'T'), ('H', 'T', 'H'), ('T', 'H', 'H')} ``` The *frequentist* approach states the following: -If all possible outcomes are equally likely an event E occurs with frequency $\#E/\#S$ where $\#E$ is the number of elements in the Event and #S is the number of elements in the State Space. +If all possible outcomes are equally likely an event E occurs with frequency #E/#S where #E is the number of elements in the Event and #S is the number of elements in the State Space. For our particular example we can do it very simply in Python. ```python