An article in the Los Angeles Times (Dec. 3, 1993) reports that $1$ in $200$ people carry the defective gene that causes inherited colon cancer. In a sample of $1000$ individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that

• (a) Between $5$ and $8$ (inclusive) carry the gene.
• (b) At least $8$ carry the gene.
   

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter $\lambda=20$ (suggested in the article "Dynamic Ride Sharing: Theory and Practice", J. of Transp. Engr., 1997:308--312). What is the probability that the number of drivers will

• (a) Be at most $10$ ?
• (b) Exceed $20$ ?
• (c) Between $10$ and $20$, inclusive? Be strictly between $10$ and $20$ ?
• (d) Exceed the mean number by more than two standard deviations?
   

Three brothers and their wives decide to have children until each family has two female children. What is the pmf of $X$=the total number of male children born to the brothers? What is $E(X)$, and how does it compare to the expected number of male children born to each brother?
 

Suppose that $p=P(\text{male birth})=.5$. A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled.

• (a) What is the probability that the family has $x$ male children?
• (b) What is the probability that the family has four children?
• (c) What is the probability that the family has at most four children?
• (d) How many male children would you expect this family to have? How many children would you expect this family to have?
   

A geologist has collected 10 specimens(标本/標本) of basaltic rock(玄武岩岩石) and 10 specimens of granite(花岗岩/花崗岩). The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis.

• (a) What is the pmf of the number of granite specimens selected for analysis?
• (b) What is the probability that all specimens of one of the two types of rock are selected for analysis?
• (c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?
   

Each of 12 refrigerators of a certain type has been returned to a distributor because of the presence of a high-pitched oscillating noise(高音调振荡噪声/高音調振盪雜訊/高音振動ノイズ) when the refrigerator is running. Suppose that 5 of these 12 have defective compressors and the other 7 have less serious problems. If they are examined in random order, let $X$=the number among the first 6 examined that have a defective compressor. Compute the following:

• (a) $P(X=1)$
• (b) $P(X\geqslant 4)$
• (c) $P(1\leqslant X\leqslant 3)$
   

The negative binomial rv and distribution are based on an experiment satisfying the following conditions:

1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either a success ($S$) or a failure ($F$).
3. The probability of success is constant from trial to trial, so $P(S\ \text{on trial}\ i)=p$ for $i=1,2,3,\ldots$
4. The experiment continues (trials are performed) until a total of $r$ successes have been observed, where $r$ is a specified positive integer.

The random variable of interest is

$X$=the number of failures that precede the $r$th success; 

$X$ is called a negative binomial random variable because, in contrast to the binomial rv, the number of success is fixed and the number of trials is random.

Possible values of $X$ are $0,1,2,\ldots$. Let $nb(x;r,p)$ denote the pmf of $X$. The event $\{X=x\}$ is equivalent to $\{r-1$ $S$'s in the first $(x+r-1)$ trials and an $S$ on the $(x+r)$th trial $\}$.

\[
\begin{split}
nb(x;r,p)&=P(X=x)\\
&=P(r-1\ S\text{'s}\ \text{on the first}\ x+r-1\ \text{trials})\cdot P(S)\\
&=\binom{x+r-1}{r-1}p^{r-1}(1-p)^x\cdot p\\
&=\binom{x+r-1}{r-1}p^{r}(1-p)^x
\end{split}
\]

Thus we have

Prop. The pmf of the negative binomial rv $X$ with parameters $r=$number of $S$'s and $p=P(S)$ is

\[
nb(x;r,p)=\binom{x+r-1}{r-1}p^{r}(1-p)^x,\quad x=0,1,2,\ldots
\]

In some sources, the negative binomial rv is taken to be the number of trials $X+r$ rather than the number of failures.

In the special case $r=1$, the pmf is

\[
nb(x;1,p)=(1-p)^x p,\quad x=0,1,2,\ldots\tag{*}
\]

Both $X$=number of $F$'s and $Y$=number of trials $(=1+X)$ are referred to in the literatures as geometric random variables(几何随机变量), and the pmf (*) is called the geometric distribution（几何分布）.

Prop. If $X$ is a negative binomial rv with pmf $nb(x;r,p)$, then

\[
E(X)=\frac{r(1-p)}{p},\quad V(X)=\frac{r(1-p)}{p^2}
\]

Finally, by expanding the binomial coefficient in front of $p^r(1-p)^x$ and doing some cancellation, it can be seen that $nb(x;r,p)$ is well defined even when $r$ is not an integer. This generalized negative binomial distribution has been found to fit observed data quite well in a wide variety of applications.

Remark:

负二项分布也称帕斯卡分布（巴斯卡分布）

References:

The above content is copied from the following:

Proposition in section 5 of Chapter 3, BOOK:

《Probability and Statistics For Engineering and The Sciences》(Fifth Edtion) P.131
Author: Jay L. Devore

The assumptions leading to the hypergeometric distribution are as follows:

1. The population or set to be sampled consists of $N$ individuals, objects, or elements (a finite population).
2. Each individual can be characterized as a success ($S$) or a failure ($F$), and there are $M$ successes in the population.
3. A sample of $n$ individuals is selected without replacement in such a way that each subset of size $n$ is equally likely to be chosen.

Prop. If $X$ is the number of $S$'s in a completely random sample of size $n$ drawn from a population consisting of $M$ $S$'s and $(N-M)$ $F$'s, then the probability distribution of $X$, called the hypergeometric distribution, is given by

\[
P(X=x)=h(x;n,M,N)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}
\]

for $x$ an integer satisfying $\max(0,n-N+M)\leqslant x\leqslant\min(n,M)$.

Notation:

$N$: number of individuals

Prop. The mean and variance of the hypergeometric rv $X$ having pmf $h(x;n,M,N)$ are 

\[
E(X)=n\cdot\frac{M}{N},\quad V(X)=\frac{N-n}{N-1}\cdot n\cdot\frac{M}{N}\cdot\biggl(1-\frac{M}{N}\biggr)
\]

The ratio $\frac{M}{N}$ is the proportion of $S$'s in the population. If we replace $M/N$ by $p$ in $E(X)$ and $V(X)$, we get

\[
\begin{aligned}
E(X)&=np,\\
V(X)&=\frac{N-n}{N-1}\cdot np(1-p)=\frac{N-n}{N-1}\cdot npq
\end{aligned}
\]

It shows that the means of the binomial and hypergeometric rv's are equal, whereas the variances of the two rv's differ by the factor $\frac{N-n}{N-1}$, often called the finite population correction factor. 

This factor is less than 1, so the hypergeometric variables has smaller variance than does the binomial rv. The correction factor can be written

\[
\dfrac{1-\frac{n}{N}}{1-\frac{1}{N}},
\]

which is approximately 1 when $n$ is small relative to $N$.

Remark:

An easy way to remember the scope of $x$ is use the formula

\[
\binom{N-M}{n-x}=\binom{N-M}{N-M-(n-x)}=\binom{N-M}{x-(n-N+M)}.
\]

Hence, to make the equation meaningful, $x$ should satisfies $n-N+M\leqslant x$.

References:

Proposition in section 5 of Chapter 3, BOOK:

《Probability and Statistics For Engineering and The Sciences》(Fifth Edtion) P.129
Author: Jay L. Devore

Show that $E(X)=np$ when $X$ is a binomial random variable.

[Hint: First express $E(X)$ as a sum with lower limit $x=1$. Then factor out $np$, let $y=x-1$ so that the sum is from $y=0$ to $n-1$, and show that the sum equals $1$.]
 

The variance is $V(X)=np(1-p)=npq$, and the Standard Deviation (SD) of $X$ is $\sigma_X=\sqrt{npq}$ (where $q=1-p$).

The proof can be seen in Question 32. (Where the variance is denoted by $D(X)$.)

(a) Show that $b(x;n,1-p)=b(n-x;n,p)$.

(b) Show that $B(x;n,1-p)=1-B(n-x-1;n,p)$. [Hint: At most $x$ $S$'s is equivalent to at least $(n-x)$ $F$'s.]