Chapter 2 Section 3 Question 17
Question
An urn contains 4 black balls and 5 white balls. After a thorough mixing, a ball is drawn from the urn, its color is noted, and the ball is set aside. The remaining balls are then mixed and a second ball is drawn.
  1. What is the probability distribution of the number of black balls observed?
  2. What is the probability distribution of the number of white balls observed?
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 19 2020
(a) What is the probability distribution of the number of black balls observed?
Since all our trials are dependant on each other (except for the first). For example if a white ball is drawn first, then the next draw is less likely to be a white ball.
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 balls $N=9$
M represents the number of desired balls (black) $M=4$
$N-M$ represent the number of undesirable balls(white) $N-M=5$
$n$ represents the length of the subset of elements that we are selecting. In this case we are drawing 2 balls so $n=2$
$x$ represents the number of elements of the type $M$ that we have in this subset. In this case it represents the number of black balls we have drawn.

$$Hypergeometric(N,M,n)$$ $$Hypergeometric(9,4,2)$$ $$P(X=x)=\frac{{M\choose x}{{N-M}\choose{n-x}}}{N \choose n}$$ $$P(X=x)=\frac{{4\choose x}{{5}\choose{2-x}}}{9 \choose 2}$$
(b) What is the probability distribution of the number of white balls observed?
This answer is the same as above with $M$ and $N-M$ swapped

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 balls $N=9$
M represents the number of desired balls (white) $M=5$
$N-M$ represent the number of undesirable balls(white) $N-M=4$
$n$ represents the length of the subset of elements that we are selecting. In this case we are drawing 2 balls so $n=2$
$x$ represents the number of elements of the type $M$ that we have in this subset. In this case it represents the number of black balls we have drawn.

$$Hypergeometric(N,M,n)$$ $$Hypergeometric(9,5,2)$$ $$P(X=x)=\frac{{M\choose x}{{N-M}\choose{n-x}}}{N \choose n}$$ $$P(X=x)=\frac{{5\choose x}{{3}\choose{2-x}}}{9 \choose 2}$$