Chapter 2 Section 3 Question 13
Question
Let $X$ ~ $Hypergeometric(20, 7, 8)$. What is the probability that $X = 3$? What is the probability that $X = 8$?
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 16 2020
From Example 2.3.7 in the book we know: $$p\scriptsize{X}\normalsize{(x)=P(X=x)=\frac{{M\choose x}{{N-M}\choose{n-x}}}{N \choose n}}$$ N represents the total number of 'elements'
M represents the number of 'elements' of one type
$N-M$ represent the number of elements of the other type.
$n$ represents the length of the subset of elements that we are selecting
$x$ represents the number of elements of the type $M$ that we have in this subset

We are going to start by plugging in $N=20$, $M=7$, $n=8$, $x=3$ into the equation $$P(X=x)=\frac{{M\choose x}{{N-M}\choose{n-x}}}{N \choose n}$$ $$P(X=3)=\frac{{7\choose 3}{{20-7}\choose{8-3}}}{20 \choose 8}$$ $$P(X=3)=\frac{{7\choose 3}{13\choose 5}}{125970}$$ $$P(X=3)=\frac{35*1287}{125970}$$ $$P(X=3)=\frac{231}{646}$$

P(X=0) is impossible to solve as you cant have 8 of your prefered type of element in a subset of 7 elements.