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.

- What is the probability distribution of the number of black balls observed?
- 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.

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