# Solutions To Mathematics Textbooks/Probability and Statistics for Engineering and the Sciences (7th ed) (ISBN-10: 0-495-38217-5)/Chapter 3

Let X repress the difference between the number of head and the number of tail optained .when a coin is tossed n time what are the possible values of X? For n=3 if the coin is assumed fair what are the probabilities associated with the values that X can take on.

## Section 3.1 - Random Variables
### Exercise 1

Let $X$ be the number of beams that failed due to shear (S). Three beams are sampled, and they will fail either from shear (S) or from flexure (F). The sample space is: $S={SSS,SSF,SFS,SFF,FSS,FSF,FFS,FFF}.$
The random variable maps to the sample space as:
* $X=0:\ FFF$
* $X=1:\ SFF,\ FSF,\ FFS$
* $X=2:\ FSS,\ SFS,\ SSF$
* $X=3:\ SSS$

### Exercise 2

Three examples of Bernoulli random variables:
* Toss a coin once, let a success be having a the coin lands heads up. $X={\begin{cases}1&{\text{head}}\\0&{\text{tails}}\end{cases}}$
* Ask credit card users if their primary card is Visa or Mastercard. $X={\begin{cases}1&{\text{Visa}}\\0&{\text{Mastercard}}\end{cases}}$
* Inspecting pennies in a coin collection for steel pennies. $X={\begin{cases}1&{\text{steel penny}}\\0&{\text{copper penny}}\end{cases}}$

### Exercise 3

Consider an Experiment in which the number of pumps in use at each of two gas stations was determined. Define the random variables:
* $W=$ the number of pumps with a waiting line of vehicles waiting to fuel up.
* $Z=$ the number of pumps providing both Diesel and Gasoline.

### Exercise 4

Let $X$ be the number of nonzero digits in a randomly selected zip code.
$X={\begin{cases}0&{\text{zip digits are: all zeros, e.g., 00000}}\\1&{\text{zip digits are: 4 zero, 1 nonzero, e.g, 00100}}\\2&{\text{zip digits are: 3 zeros, 2 nonzero, e.g., 10050}}\\3&{\text{zip digits are: 2 zeros, 3 nonzero, e.g., 90210}}\\4&{\text{zip digits are: 1 zero, 4 nonzero, e.g., 80223}}\\5&{\text{zip digits are: 0 zeros, 5 nonzero, e.g., 76491}}\end{cases}}$

### Exercise 5

Consider a sample space for the number of rolls of a set of dice until the sum of 2 is rolled. Let $X={\begin{cases}1&{\text{a sum of 2 is rolled before the 8th roll}}\\0&{\text{a sum of 2 is not rolled until the 8th or later roll}}\end{cases}}$. In this case the random variable for mapping the sample space to the real numbers is no infinite, it is discrete and dichotomous.

### Exercise 6

$X$ is the number of cars observed at an intersection until a car turns Left ($L$). (Cars can turn right ($R$), left ($L$) or go straight ($A$)).
$X$ can take on the values 1, 2, 3, 4, ..., for example:
$X={\begin{cases}1&\{L\}\\2&\{RL,\ AL\}\\3&\{AAL,\ ARL,\ RAL,\ RRL\}\\4&\{AAAL,\ AARL,\ ...\}\\...\end{cases}}$

### Exercise 7

* Part a. Let $X$ be the number of unbroken eggs in a standard egg carton. The random $X$ is a discrete random variable.
$X\in \{0,1,2,3,4,5,6,7,8,9,10,11,12\}$.
* Part b. Let $Y$ be the number of students on a class list who are absent on the first day of class. Say there are $n$ students on the class list. $Y$ is a discrete random variable.
$X\in \{0,1,2,3,...,n\}$
* Part c. Let $U$ be the number of swings before hitting a golf ball. $U$ is a discrete random variable.
$U\in \{1,2,3,4,5,6,\ldots \}$
* Part d. Let $X$ be the length of a rattlesnake. $X$ is a continuous random variable.
$X\in (0,\infty )$
* Part e. Let $Z$ Royalties earned from the sale of the first 10,000 copies of a textbook. For this question we will assume there is a standard royalty, $x, earned per book sold. So $Z$ is a discrete random variable and the values of $Z$ are
$Z\in \{0,x,2x,3x,4x,\ldots ,10000x\}$
* Part f. Let $Y$ be the pH of a soil sample. $Y$ is a continuous random variable.
$Y\in (0,14)$
* Part g. Let $X$ be the tension measured in psi of a string on a tennis racket. $X$ is a continuous random variable.
$X\in (0,\infty )$
* Part h. Let $X$ be the total number of tosses of coins for three people to get the same outcome, i.e., $HHH$ or $TTT$. $X$ is a discrete random variable and can take on the values
$X\in \{3,6,9,12,15,18,\ldots \}$

### Exercise 8

Let $Y$ be the total number of trails until you have three successes in a row. $Y$ will take on the values $3,4,5,6,\ldots$
$Y={\begin{cases}3&\{SSS\}\\4&\{FSSS\}\\5&\{SFSSS,FFSSS\}\\6&\{SSFSSS,FFFSSS,SFFSSS,FSFSSS\}\\7&\{SSFFSSS,SFSFSSS,SFFFSSS,FSSFSSS,FSFFSSS,FFSFSSS,FFFFSSS\}\\\ldots \end{cases}}$

### Exercise 9

* Part a. $X$ is the number of moves until Claudius returns back to 0. $X$ is discrete and takes on the values $X\in \{2,4,6,8,\ldots \}$
* Part b. $X\in \{2,3,4,5,6,\ldots \}$

### Exercise 10

The number of pumps in use at both a six-pump (Station 1) and a four-pump (Station 2) gas station will be examined.
* Part a. $T$ is the total number of pumps in use between the two stations.
$T\in \{0,1,2,3,4,5,6,7,8,9,10\}$
* Part b. $X$ is the difference between the number of pumps in use between station 1 and station 2.
$X\in \{-4,-3,-2,-1,0,1,2,3,4,5,6\}$
* Part c. $U$ is the maximum number of pumps in use at either station.
$U\in \{0,1,2,3,4,5,6\}$
* Part d. $Z$ is the number of stations with two pumps in operation.
$Z\in \{0,1,2\}$

## Section 3.2 - Probability Distributions for Discrete Random Variables
### Exercise 11

Let $X$ be the number of cylinders on the next car to be serviced.
* Part a. The probability mass function of $X$ is:
$P(X=x)={\begin{cases}0.45&x=4\\0.40&x=6\\0.15&x=8\\0&{\text{otherwise}}\end{cases}}$
* Part b.
* Part c. $P(X\geq 6)=P(X=6)+P(X=8)=0.40+0.15=0.55$

### Exercise 12

Let $Y$ be the number of ticketed passengers who show up for a flight.
* Part a. The probability the flight will accommodate all ticketed passengers who show is:
$P(Y\leq 50)=\sum _{y=45}^{50}P(Y=y)=0.83$
* Part b. The probability that not all ticketed passengers who show up for the flight will get a seat on the plane:
$P(Y>50)=1-P(Y\leq 50)=1-0.83=0.17$
* Part c. The probability the first person on standby will get on the flight is:
$P(Y\leq 49)=\sum _{y=45}^{49}P(Y=y)=0.66$
The probability the third person on standby gets on the flight is:
$P(Y\leq 47)=\sum _{y=45}^{47}P(Y=y)=0.27$

### Exercise 13

Let $X$ be the number of phone lines (out of six total lines) in use at any one time in the office. Using the given probability mass function:
* Part a. The probability of at most three lines in use is:
$P(X\leq 3)=\sum _{x=0}^{3}P(X=x)=0.10+0.15+0.20+0.25=0.70$
* Part b. The probability of fewer than three lines in use is:
$P(X<3)=\sum _{x=0}^{2}P(X=x)=0.10+0.15+0.20=0.45$
* Part c. The probability at least three lines are in use is:
$P(X\geq 3)=1-P(X<3)=1-0.45=0.55$
* Part d. The probability between two and five lines, inclusive, are in use is:
$P(2\leq X\leq 5)=\sum _{x=2}^{5}P(X=x)=0.20+0.25+0.20+0.06=0.71$
* Part e. The probability between two and four lines, inclusive, are not in use is:
$P(6-X\geq 2{\text{ or }}6-X\leq 4)=P(X\leq 4{\text{ or }}X\geq 2)=P(X=2)+P(X=3)+P(X=4)=0.2+0.25+0.2=0.65$
* Part f. The probability of at least four lines are not in use is:
$P(6-X\geq 4)=P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)=0.10+0.15+0.2=0.45$

### Exercise 14

Let $Y$ be the number of forms the contractor needs to submit. Given the probability $y$ forms are required is proportional to $y$ via the function $p(y)=ky$ for $y=1,2,3,4,5$.
* Part a. Find the value of $k$ such that $p(y)$ is a probability mass function.
${\begin{array}{rccl}\sum _{y}P(Y=y)&=&1&{\text{by definition}}\\\sum _{y=1}^{5}ky&=&1&{\text{input values for this exercise}}\\k\left(1+2+3+4+5\right)&=&1&\\15k&=&1&\\k&=&{\frac {1}{15}}&\\\end{array}}$
* Part b. The probability at most three forms are required is:
${\begin{array}{rl}P(Y\leq 3)&=\sum _{y=1}^{3}P(Y=y)\\&={\frac {1}{15}}+{\frac {2}{15}}+{\frac {3}{15}}\\&=0.40\end{array}}$
* Part c. The probability of between 2 and 4 forms (inclusive) will be submitted is:
${\begin{array}{rl}P(2\leq Y\leq 4)&=P(Y=2)+P(Y=3)+P(Y=4)\\&={\frac {2}{15}}+{\frac {3}{15}}+{\frac {4}{15}}\\&=0.60\end{array}}$
* Part d. The function $p(y)=y^{2}/50$ for $y=1,2,3,4,5$ cannot be a probability mass function because the sum over all the possible values of $y$ does not equal 1 as seen here:
$\sum _{y=1}^{5}p(y)={\frac {1}{50}}\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\right)={\frac {55}{50}}\neq 1$

### Exercise 15

Receive a shipment of 5 boards, select 2 boards for inspection at random.
* Part a. The ten different possible outcomes are:
$\left\{(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)\right\}$
* Part b. If boards 1 and 2 are the only two defective boards and $X$ is the number of defective boards selected, then the probability mass function for $X$ is:
$P(X=x)={\begin{cases}0.10&x=2\\0.6&x=1\\0.3&x=0\\0&{\text{otherwise}}\end{cases}}$
* Part c.
The cumulative distribution function, cdf, is:
$F(x)=P(X\leq x)={\begin{cases}0&x<0\\0.3&0\leq x<1\\0.9&1\leq x<2\\1.0&2\leq x\end{cases}}$

### Exercise 16

Let $X$ be the number of homeowners who have insurance. The probability any one homeowner has insurance is 0.30.
* Part a. The distribution of $X$ is:
$P(X=x)={\begin{cases}{\binom {4}{0}}(0.30)^{0}(1-0.30)^{4-0}=0.2401&x=0,i.e.,FFFF\\{\binom {4}{1}}(0.30)^{1}(1-0.30)^{4-1}=0.4116&x=1,i.e.,FFFS,FFSF,FSFF,SFFF\\{\binom {4}{2}}(0.30)^{2}(1-0.30)^{4-2}=0.2646&x=2,i.e.,FFSS,FSFS,SFFS,FSSF,SFSF,SSFF\\{\binom {4}{3}}(0.30)^{3}(1-0.30)^{4-3}=0.0756&x=3,i.e.,SSSF,SSFS,SFSS,FSSS\\{\binom {4}{4}}(0.30)^{4}(1-0.30)^{4-4}=0.0081&x=4,i.e.,SSSS\end{cases}}$
* Part b.
* Part c. The most likely observed value is 1 with probability 0.4116. (The expected observation is 1.20).
* Part d.
${\begin{array}{rl}P\left(X\geq 2\right)&=P(X=2)+P(X=3)+P(X=4)\\&=0.2646+0.0756+0.0081\\&=0.3483\end{array}}$

### Exercise 17

90% of batteries have acceptable voltages. Let $Y$ be the number of batteries tested to find two acceptable batteries.
* Part a. The probability that you need to sample exactly two batteries is:
$P\left(Y=2\right)=P(AA)=0.90^{2}=0.81$
* Part b. The probability that you need to sample exactly three batteries is:
$P\left(Y=3\right)=P(AUA)+P(UAA)=0.90\times 0.10\times 0.90+0.10\times 0.90\times 0.90=2\times 0.10\times 0.90^{2}=0.1620$
* Part c. The probability that you need to sample exactly five batteries is:
${\begin{array}{rl}P(Y=5)&=P(AUUUA)+P(UAUUA)+P(UUAUA)+P(UUUAA)\\&=4\times 0.1^{3}\times 0.90^{2}\\&=0.0032\end{array}}$
* Part d. The probability mass function for $Y$ is:
$P(Y=y)={\begin{cases}(y-1)\times 0.1^{y-2}\times 0.90^{2}&y=2,3,4,\ldots \\0&{\text{otherwise}}\end{cases}}$

### Exercise 18

Let $M$ be the maximum of two tosses of a fair six-sided die.
* Part a. The probability mass function of $M$ is:
$P(M=m)={\begin{cases}1/36&m=1\\3/36&m=2\\5/36&m=3\\7/36&m=4\\9/36&m=5\\11/36&m=6\\0&{\text{otherwise}}\end{cases}}$
* Part b. The cumulative distribution function for $M$ is:
$F_{M}(m)=P\left(M\leq m\right)={\begin{cases}0&m<1\\1/36&1\leq m<2\\4/36&2\leq m<3\\9/36&3\leq m<4\\16/36&4\leq m<5\\25/36&5\leq m<6\\36/36&6\leq m\end{cases}}$

### Exercise 19

Let $Y$ be the number of days after Wednesday it takes for both magazines to arrive. $Y$ can take on the values 0, 1, 2, or 3. The probability mass function is:
$P(Y=y)={\begin{cases}0.09&y=0,i.e.,(W,W)\\0.40&y=1,i.e.,(W,Th),(Th,W),(Th,Th)\\0.32&y=2,i.e.,(W,F),(Th,F),(F,W),(F,Th),(F,F)\\0.19&y=3\end{cases}}$

### Exercise 20

Three couples and two individuals are going to a seminar. Let $X$ be the number of people who arrive late for the seminar.
* Part a. The probability mass function of $X$ is:
$P(X=x)={\begin{cases}0.0778&x=0,{\text{ no one is late}}\\0.1037&x=1,{\text{ one individual is late}}\\0.1901&x=2,{\text{ one couple is late, or both individuals are late}}\\0.2074&x=3,{\text{ one couple and one individual are late}}\\0.1728&x=4,{\text{ two couples or one couple and both individuals are late}}\\0.1382&x=5,{\text{ two couples and one individual are late}}\\0.0691&x=6,{\text{ all three couples, or two couples and both individuals are late}}\\0.0307&x=7,{\text{ all three couples and one individual are late}}\\0.0102&x=8,{\text{ everyone is late}}\end{cases}}$
* Part b.
The CDF is:
$F(x)=P(X\leq x)={\begin{cases}0&x<0\\0.0778&0\leq x<1\\0.1815&1\leq x<2\\0.3716&2\leq x<3\\0.5790&3\leq x<4\\0.7518&4\leq x<5\\0.8900&5\leq x<6\\0.9591&6\leq x<7\\0.9898&7\leq x<8\\1.0000&8\leq x\end{cases}}$
$P(2\leq X\leq 6)=P(X\leq 6)-P(X<2)=0.9591-0.1815=0.7776$

### Exercise 21

Benford's Law, $p(x)=P(1^{st}{\text{ digit is }}x)=log_{10}(1+1/x),x=1,2,\ldots ,9$
* Part a.
The individual probabilities from the probability mass function are:
$P(X=x)={\begin{cases}0.3010&x=1\\0.1761&x=2\\0.1249&x=3\\0.0969&x=4\\0.0792&x=5\\0.0669&x=6\\0.0580&x=7\\0.0512&x=8\\0.0458&x=9\\0.0000&{\text{otherwise}}\end{cases}}$
Part b. The CDF is:
$F(x)=P(X\leq x)={\begin{cases}0.0000&x<1\\0.3010&1\leq x<2\\0.4771&2\leq x<3\\0.6020&3\leq x<4\\0.6989&4\leq x<5\\0.7781&5\leq x<6\\0.8450&6\leq x<7\\0.9030&7\leq x<8\\0.9542&8\leq x<9\\1.0000&9\leq x\end{cases}}$
* Part c.
** $P(X\leq 3)=0.6020$
** $P(X\geq 5)=1-P(X<5)=1-0.6989=0.3011$

### Exercise 22

The CDF from Exercise 13 is:
$F(x)=P(X\leq x)={\begin{cases}0.00&x<0\\0.10&0\leq x<1\\0.25&1\leq x<2\\0.45&2\leq x<3\\0.70&3\leq x<4\\0.90&4\leq x<5\\0.96&5\leq x<6\\1.00&6\leq x\end{cases}}$
* at most three lines are in use: $P(X\leq 3)=0.70$
* fewer than three lines are in use: $P(X<3)=0.45$
* at least three lines are in use: $P(X\geq 3)=1-P(X<3)=1-0.45=0.55$
* between two and five lines, inclusive: $P(2\leq X\leq 5)=P(X\leq 5)-P(X<2)=0.96-0.25=0.71$

### Exercise 23

Using the CDF provided
* Part a.
$P(X=2)=P(X\leq 2)-P(X<2)=0.39-0.19=0.20$
* Part b.
$P(X>3)=1-P(X\leq 3)=1-0.67=0.33$
* Part c.
$P(2\leq X\leq 5)=P(X\leq 5)-P(X<2)=0.97-0.19=0.78$
* Part d.
$P(2<X<5)=P(X<5)-P(X\leq 2)=0.92-0.39=0.53$

### Exercise 24

* Part a. The pmf is:
$P(X=x)={\begin{cases}0.30&x=1\\0.10&x=3\\0.05&x=4\\0.15&x=6\\0.40&x=12\end{cases}}$
* Part b.
** $P(3\leq X\leq 6)=P(X\leq 6)-P(X<3)=0.60-0.30=0.30$
** $P(4\leq X)=1-P(X<4)=1-0.40=0.60$

### Exercise 25

If a the boy is born first: $P(Y=y)=P(Y=0)=P(B)=p$
If a the boy is born second then we have: $P(GB)=P(Y=1)=(1-p)p$
If the boy is born third we have: $P(GGB)=P(Y=2)=(1-p)^{2}p$
If the boy is born fourth we have: $P(GGGB)=P(Y=3)=(1-p)^{3}p$
The probability mass function for $Y$ is:
$P(Y=y)={\begin{cases}(1-p)^{y}p&y=0,1,2,3,4,\ldots \\0&{\text{otherwise}}\end{cases}}$
This is a preview of a distribution call the Geometric distribution.

### Exercise 26

* Part a. Let $X$ be the number of times Alvie visits a friend.
** Alvie visits one friend: $1/3$
** Alvie visits two friends: $(2/3)(1/3)$
** Alvie visits three friends: $(2/3)^{2}(1/3)$
** Alvie visits four friends: $(2/3)^{3}(1/3)$
** Alvie visits $x$ friends:
$P(X=x)={\begin{cases}(2/3)^{x-1}(1/3)&x=1,2,3,4,\ldots \\0&{\text{otherwise}}\end{cases}}$
* Part b.
Alvie will travel on $x+1$ straight line segments. He must travel $x$ segments to visit $x$ friends and then travel one more segment to get back home. Thus $Y=X+1$. The probability mass function for $Y$ is:
$P(Y=y)={\begin{cases}(2/3)^{y-2}(1/3)&y=2,3,4,5,6,\ldots \\0&{\text{otherwise}}\end{cases}}$
* Part c. If two of the locations are female and two locations are male. Let $Z$ is the number of female friends Alvie will visit.
If he visits a male first and then goes home, $Z=0$: probability of $1/4\times 1/3+1/4\times 1/3=1/6$
For visiting one female:

### Exercise 27

Let the string (2, 4, 3, 1) denote the order that four textbooks are handed back to four students who claim to be missing their books. Let $Y$ be the number of students who get their book back when the books are handed back in a random order.
* Part a. The 24 possible outcomes and corresponding $Y$ values are:
${\begin{array}{cc||cc}{\text{Outcome}}&y&{\text{Outcome}}&y\\\hline 1234&4&3124&1\\1243&2&3142&0\\1324&2&3214&2\\1342&1&3241&1\\1423&1&3412&0\\1432&2&3421&0\\2134&2&4123&0\\2143&0&4132&1\\2314&1&4213&1\\2341&0&4231&2\\2413&0&4312&0\\2431&1&4321&0\\\end{array}}$
* Part b. The probability mass function for $Y$ is:
$P(Y=y)={\begin{cases}9/24&y=0\\8/24&y=1\\6/24&y=2\\1/24&y=4\\0&{\text{otherwise}}\end{cases}}$

### Exercise 28

Show that the cdf, $F(X)$, is a non-decreasing function. Let $x_{1}<x_{2}$ then:
${\begin{array}{rl}F(x_{2})&=P\left(X\leq x_{2}\right)\\&=P\left[\left(X\leq x_{1}\right)\cup \left(x_{1}<X\leq x_{2}\right)\right]\\&=P\left[X\leq x_{1}\right]+P\left[x_{1}<X\leq x_{2}\right]\\&=F\left(x_{1}\right)+P\left[x_{1}<X\leq x_{2}\right]\\&\geq F(x_{1})\end{array}}$
The equality holds if $P\left(x_{1}<X\leq x_{2}\right)=0$.

## Section 3.3 - Expected Values
### Exercise 29

For the given probability mass function find the expectation and variance.
* Part a.
${\begin{array}{rl}E(X)=\mu _{X}&=\sum _{x}xP(X=x)\\&=(0)(0.08)+(1)(0.15)+(2)(0.45)+(3)(0.27)+(4)(0.05)\\&=0.00+0.15+0.90+0.81+0.20\\&=2.06\end{array}}$
* Part b.
${\begin{array}{rl}Var(X)=\sigma _{X}^{2}&=\sum _{x}\left(x-\mu _{X}\right)^{2}P(X=x)\\&=(0-2.06)^{2}(0.08)+(1-2.06)^{2}(0.15)\\&\ +(2-2.06)^{2}(0.45)+(3-2.06)^{2}(0.27)+(4-2.06)^{2}(0.05)\\&=0.3395+0.1685+0.0016+0.2386+0.1882\\&=0.9364\end{array}}$
* Part c.
${\begin{array}{rl}\sigma _{X}={\sqrt {Var(X)}}&={\sqrt {0.9364}}=0.9677\end{array}}$
* Part d.
${\begin{array}{rl}Var(X)=\sigma _{X}^{2}&=\left(\sum _{x}x^{2}P(X=x)\right)-\mu _{X}^{2}\\&=\left((0^{2})(0.08)+(1^{2})(0.15)+(2^{2})(0.45)+(3^{2})(0.27)+(4^{2})(0.05)\right)-2.06^{2}\\&=5.18-4.2436\\&=0.9364\end{array}}$

### Exercise 30

* Part a.
    ${\begin{array}{rl}E(Y)&=\sum _{y}yP(Y=y)\\&=(0)(0.60)+(1)(0.25)+(2)(0.10)+(3)(0.05)\\&=0.60\end{array}}$
* Part b.
    ${\begin{array}{rl}E\left(100Y^{2}\right)&=\sum _{y}100y^{2}P(Y=y)\\&=(100\times 0^{2})(0.60)+(100\times 1^{2})(0.25)+(100\times 2^{2})(0.10)+(100\times 3^{2})(0.05)\\&=110\end{array}}$

### Exercise 31.

The expectation for the random variable in exercise 12 is:
    ${\begin{array}{rl}E(Y)&=\mu _{y}=\sum _{y}yP(Y=y)\\&=(45)(0.05)+(46)(0.10)+(47)(0.12)+(48)(0.14)+(49)(0.25)\\&\quad +(50)(0.17)+(51)(0.06)+(52)(0.05)+(53)(0.03)+(54)(0.02)+(55)(0.01)\\&=48.84\\E(Y^{2})&=\sum _{y}y^{2}P(Y=y)\\&=(45^{2})(0.05)+(46^{2})(0.10)+(47^{2})(0.12)+(48^{2})(0.14)+(49^{2})(0.25)\\&\quad +(50^{2})(0.17)+(51^{2})(0.06)+(52^{2})(0.05)+(53^{2})(0.03)+(54^{2})(0.02)+(55^{2})(0.01)\\&=2389.84\\Var(Y)&=\sigma _{y}^{2}=E(Y^{2})-E(Y)^{2}\\&=2389.84-48.84^{2}\\&=4.4944\\\sigma _{y}&={\sqrt {4.4944}}=2.12\end{array}}$
Find the probability $Y$ is within one standard deviation of the mean.
    ${\begin{array}{rl}P\left(\mu _{y}-\sigma _{y}<Y<\mu _{y}+\sigma _{Y}\right)&=P\left(48.84-2.12<Y<48.84+2.12\right)\\&=P(46.72<Y<50.96)\\&=P(Y=47)+P(Y=48)+P(Y=49)+P(Y=50)\\&=0.12+0.14+0.25+0.17\\&=0.68\end{array}}$

### Exercise 32.

For the pmf:
    ${\begin{array}{l|ccc}x&13.5&15.9&19.1\\\hline P(X=x)&0.20&0.50&0.30\\\end{array}}$
* Part a.
    ${\begin{array}{rl}E(X)&=\sum _{x}xP(X=x)\\&=(13.5)(0.20)+(15.9)(0.50)+(19.1)(0.30)\\&=16.38\\E(X^{2})&=\sum _{x}x^{2}P(X=x)\\&=(13.5^{2})(0.20)+(15.9^{2})(0.50)+(19.1^{2})(0.30)\\&=272.298\\Var(X)&=E(X^{2})-E(X)^{2}\\&=272.298-16.38^{2}\\&=3.9936\end{array}}$
* Part b.
    $E\left(25X-8.5\right)=25E(X)-8.5=25\times 16.38-8.5=401$
* Part c.
    $Var\left(25X-8.5\right)=25^{2}Var(X)=2496$
* Part d.
    $E\left(X-0.01X^{2}\right)=E(X)-0.01E\left(X^{2}\right)=16.38-0.01\times 272.298=13.6570$

### Exercise 33.

Let $X$ be a Bernoulli random variable with success probability $p$.
* Part a.
    $E\left(X^{2}\right)=0^{2}(1-p)+1^{2}p=p$
* Part b.
    $Var\left(X\right)=E\left(X^{2}\right)-E(X)^{2}=p-p^{2}=p(1-p)$
* Part c.
    $E\left(X^{79}\right)=0^{79}(1-p)+1^{79}p=p$

### Exercise 34.

For $P(X=x)={\begin{cases}c/x^{3}&x=1,2,3,\ldots \\0&{\text{otherwise}}\end{cases}}$
The expectation is:
    $E(X)=\sum _{x=1}^{\infty }xP(X=x)=c\sum _{x=1}^{\infty }{\frac {1}{x^{2}}}={\frac {c\pi ^{2}}{6}}$
The expectation is finite.

### Exercise 35.

For the pmf for the quantity of magazines is: ${\begin{array}{l|cccccc}x&1&2&3&4&5&6\\\hline P(X=x)&{\frac {1}{15}}&{\frac {2}{15}}&{\frac {3}{15}}&{\frac {4}{15}}&{\frac {3}{15}}&{\frac {2}{15}}\\\end{array}}$
If you make one dollar profit for each mag sold and you have to determine the number to order you have: ${\begin{array}{l|cccccc| c }x&1&2&3&4&5&6&{\text{Expectation}}\\\hline \hline {\text{Profit if you stock 1 mag}}&1&1&1&1&1&1&1\\\hline {\text{Profit if you stock 2 mags}}&0&2&2&2&2&2&1.86667\\\hline {\text{Profit if you stock 3 mags}}&-1&1&3&3&3&3&2.46667\\\hline {\text{Profit if you stock 4 mags}}&-2&0&2&4&4&4&2.66667\\\hline {\text{Profit if you stock 5 mags}}&-3&-1&1&3&5&5&2.33333\\\hline {\text{Profit if you stock 6 mags}}&-4&-2&0&2&4&6&1.6\\\hline P(X=x)&{\frac {1}{15}}&{\frac {2}{15}}&{\frac {3}{15}}&{\frac {4}{15}}&{\frac {3}{15}}&{\frac {2}{15}}\\\end{array}}$
The largest expected profit comes from purchasing 4 magazines to sell.

### Exercise 36.

Expected damages incurred is:
    ${\begin{array}{rl}E(X)&=\$0\times 0.80+(\$1000-\$500)\times 0.10+(\$5000-\$500)\times 0.08+(\$10000-\$500)\times 0.02\\&=0+50+360+190\\&=600\\\end{array}}$
For an expected profit of $100 the premium amount charged should be $700.

### Exercise 37.

Find the expectation and variance for the discrete uniform distribution.
    ${\begin{array}{rl}E(X)&=\sum _{x=1}^{n}x{\frac {1}{n}}\\\\&={\frac {1}{n}}\sum _{x=1}^{n}x\\\\&={\frac {1}{n}}{\frac {n(n+1)}{2}}\\\\&={\frac {n+1}{2}}\\\\\\E(X^{2})&=\sum _{x=1}^{n}x^{2}{\frac {1}{n}}\\\\&={\frac {1}{n}}\sum _{x=1}^{n}x^{2}\\\\&={\frac {1}{n}}{\frac {n(n+1)(2n+1)}{6}}\\\\&={\frac {(n+1)(2n+1)}{6}}\\\\\\Var(X)&=E(X^{2})-E(X)^{2}\\\\&={\frac {(n+1)(2n+1)}{6}}-{\frac {(n+1)^{2}}{4}}\\\\&={\frac {2n^{2}+n+2n+1}{6}}-{\frac {n^{2}+2n+1}{4}}\\\\&={\frac {4n^{2}+6n+2}{12}}-{\frac {3n^{2}+6n+3}{12}}\\\\&={\frac {n^{2}-1}{12}}\end{array}}$

### Exercise 38.

The expected value of $1/X$ where $X$ is the roll of a fair die is:
    $E(1/X)=1/6\left(1/1+1/2+1/3+1/4+1/5+1/6\right)=0.4083$
The guaranteed amount is $1/3.5=0.2857$ So you are better off taking the gamble, the expected winnings are higher.

### Exercise 39.

Let $X$ be the number of lots ordered.
    ${\begin{array}{rl}E(X)&=1\times 0.2+2\times 0.4+3\times 0.3+4\times .1\\&=2.3\\E(X^{2})&=1^{2}\times 0.2+2^{2}\times 0.4+3^{2}\times 0.3+4^{2}\times .1\\&=6.1\\Var(X)&=E(X^{2})-E(X)^{2}\\&=6.1-2.3^{2}\\&=0.81\\\end{array}}$
Let $h(x)=100-5x$ be the number of pounds left after $X$ lots are purchased.
    ${\begin{array}{rl}E(h(X))&=100-5E(X)=88.5\\Var(h(X))&=Var(100-5X)=5^{2}Var(X)=25\times 0.81=20.25\end{array}}$

### Exercise 40.

The variance, a measure of spread, is the same for both $X$ and $-X$.
Note that
    ${\begin{array}{rl}Var(aX+b)&=a^{2}Var(X)\\Var(-X)&=(-1)^{2}Var(X)\\Var(-X)&=Var(X)\\\end{array}}$

### Exercise 41.

Prove $Var(aX+b)=a^{2}\sigma ^{2}$. Let $Y=aX+b$.
    ${\begin{array}{rl}Var(Y)&=\sum _{y}(y-\mu _{y})^{2}P(Y=y)\\&=\sum _{x}\left((ax+b)-(a\mu _{x}+b)\right)^{2}P(X=x)\\&=\sum _{x}\left(ax-a\mu _{x}\right)^{2}P(X=x)\\&=\sum _{x}a^{2}\left(x-\mu _{x}\right)^{2}P(X=x)\\&=a^{2}\sum _{x}\left(x-\mu _{x}\right)^{2}P(X=x)\\&=a^{2}Var(X)\\&=a^{2}\sigma _{x}^{2}\end{array}}$

### Exercise 42.

Let $E(X)=5$ and $E(X(X-1))=E(X^{2}-X)=E(X^{2})-E(X)=27.5$
* Part a.
    ${\begin{array}{rl}E\left(X^{2}\right)&=E(X(X-1))+E(X)\\&=E(X^{2})-E(X)+E(X)\\&=27.5+5\\&=32.5\end{array}}$
* Part b.
    ${\begin{array}{rl}Var(X)&=E(X^{2})-E(X)^{2}\\&=32.5-5^{2}\\&=7.5\\\end{array}}$
* Part c.
    $Var\left(X\right)=E(X(X-1))+E(X)-E(X)^{2}$

### Exercise 43.

$E(X-c)=E(X)-E(c)=E(X)-c$
When $c=\mu$ then $E(X-c)=0$

### Exercise 44.

Chebyshev's Inequality: $P\left(\left|X-\mu \right)\geq k\sigma \right)\leq {\frac {1}{k^{2}}}$
* Part a.
    ${\begin{array}{l|ccccc}k&2&3&4&5&10\\\hline {\text{Upper Bound}}&{\frac {1}{4}}&{\frac {1}{9}}&{\frac {1}{16}}&{\frac {1}{25}}&{\frac {1}{100}}\\\end{array}}$
* Part b.
    $\mu =2.64\quad \sigma ^{2}=2.37\quad \sigma =1.54$
    Thus
    ${\begin{array}{rl}P(|X-\mu |\geq 2\sigma )&=P(X\leq \mu -2\sigma {\text{ or }}X\geq \mu +2\sigma )\\&=P(X\leq -0.44)+P(X\geq 5.72)\\&=P(X=6)\\&=0.04\\\end{array}}$
    The Chebyshev bound is overly conservative. Notices of the other values of $k$ the probability $X$ is within $k$ standard deviations of the mean is 0.
* Part c.
    For $\mu =0$ and $\sigma =1/3$ we have:
    $P(|X-0|\geq 3\sigma )=P(|X|\geq 1)=P(X=-1)+P(X=1)=1/9$ same as the Chebyshev bound.
* Part d.
    For $P(|X-\mu |\geq 5\sigma )=0.04$
    Let $P(X=-1)=P(X=1)=1/50\quad P(X=0)=48/50$
    $E(X)=\mu =0\quad \sigma ={\sqrt {2/50}}=1/5$
    $P(|X|\geq 1)=2/50=0.04$

### Exercise 45.

If $a\leq X\leq b$ then show $a\leq E(X)\leq b$.
Intuitively, we can consider the expectation to be the center of mass of the random variable. So if the mass of the random variable is between $a$ and $b$, then the center of mass, the expectation, will also be between $a$ and $b$.
The proof is easy. You'll need to remember that $\sum _{x}P(X=x)=1$
    ${\begin{array}{rcccl}a&\leq &X&\leq &b\\\sum _{x}aP(X=x)&\leq &\sum _{x}xP(X=x)&\leq &\sum _{x}bP(X=x)\\a\sum _{x}P(X=x)&\leq &E(X)&\leq &b\sum _{x}P(X=x)\\a&\leq &E(X)&\leq &b\end{array}}$

## Section 3.4 - The Binomial Probability Distribution
### Exercise 46.

Find the following binomial probabilities.
* Part a. Let $X\sim {\text{Binomial}}\left(n=8,p=0.35\right)$
    $b(3;8,0.35)=P(X=3)={\binom {8}{3}}\left(0.35\right)^{3}\left(1-0.35\right)^{8-3}=0.2786$
* Part b. Let $X\sim {\text{Binomial}}\left(n=8,p=0.6\right)$
    $b(5;8,0.6)=P(X=5)={\binom {8}{5}}\left(0.6\right)^{5}\left(1-0.6\right)^{8-5}=0.2787$
* Part c. For $X\sim {\text{Binomial}}\left(n=7,p=0.60\right)$
    ${\begin{array}{rl}P\left(3\leq X\leq 5\right)&={\binom {7}{3}}\left(0.6\right)^{3}\left(1-0.60\right)^{7-3}+{\binom {7}{4}}\left(0.6\right)^{4}\left(1-0.60\right)^{7-4}+{\binom {7}{5}}\left(0.6\right)^{5}\left(1-0.60\right)^{7-5}\\&=0.1935+0.2903+0.2613\\&=0.7451\end{array}}$
* Part d. For $X\sim {\text{Binomial}}\left(n=9,p=0.1\right)$
    ${\begin{array}{rl}P\left(X\geq 1\right)&=1-P\left(X<1\right)\\&=1-P(X=0)\\&=1-{\binom {9}{0}}(0.1)^{0}\left(1-0.1\right)^{9-0}\\&=1-(0.9)^{9}\\&=0.6126\end{array}}$

### Exercise 47.

Find the following probabilities:
* Part a. $B(4;15,0.3)=0.5155$
* Part b. $b(4;15,0.3)=0.2186$
* Part c. $b(6;15,0.6)=0.0116$
* Part d. For $X\sim {\text{Binomial}}\left(n=15,p=0.3\right)$
    ${\begin{array}{rl}P\left(2\leq X\leq 4\right)&=P(X\leq 4)-P(X<2)\\&=P(X\leq 4)-P(X\leq 1)\\&=0.5155-0.0353\\&=0.4802\end{array}}$
* Part e. For $X\sim {\text{Binomial}}\left(n=15,p=0.3\right)$
    ${\begin{array}{rl}P(2\leq X)&=1-P(X<2)\\&=1-P(X\leq 1)\\&=1-0.0353\\&=0.9647\end{array}}$
* Part f. For $X\sim {\text{Binomial}}\left(n=15,p=0.7\right)$
    ${\begin{array}{rl}P(X\leq 1)&=0.0000005165607\approx 0.0000\end{array}}$
* Part f. For $X\sim {\text{Binomial}}\left(n=15,p=0.3\right)$
    ${\begin{array}{rl}P(2<X<6)&=P(X\leq 5)-P(X\leq 2)\\&=0.7216-0.1268\\&=0.5948\end{array}}$

### Exercise 48.

Let $X$ be the number of defective compact disc players circuit boards in a random sample of size $n=25$. The distribution of $X$ is:
    $X\sim {\text{Binomial}}\left(n=25,p=0.05\right)$
* Part a.
    ${\begin{array}{rl}P\left(X\leq 2\right)&=P(X=0)+P(X=1)+P(X=2)\\&={\binom {25}{0}}(0.05)^{0}(1-0.05)^{25-0}+{\binom {25}{1}}(0.05)^{1}(1-0.05)^{25-1}+{\binom {25}{2}}(0.05)^{2}(1-0.05)^{25-2}\\&=0.2774+0.3650+0.2305\\&=0.87277\end{array}}$
* Part b.
    ${\begin{array}{rl}P\left(X\geq 5\right)&=1-P\left(X\leq 4\right)\\&=1-0.9928\\&=0.0072\\\end{array}}$
* Part c.
    ${\begin{array}{rl}P(1\leq X\leq 4)&=P(X=1)+P(X=2)+P(X=3)+P(X=4)\\&=0.3650+0.2305+0.0930+0.0269\\&=0.7154\end{array}}$
* Part d. The probability that none of the boards are defective is:
    ${\begin{array}{rl}P(X=0)&=(1-0.05)^{25}=0.2774\end{array}}$
* Part e. The expectation, variance and the standard deviation of $X$ are:
    $E(X)=\mu _{X}=np=25\times 0.05=1.25$
    $Var(X)=\sigma _{X}^{2}=np(1-p)=25\times 0.05\times (1-0.05)=1.1875$
    $\sigma _{X}={\sqrt {np(1-p)}}={\sqrt {1.1875}}=1.0897$

### Exercise 49.

Let $X$ be the number of goblets which have flaws. $X\sim {\text{Binomial}}\left(n=6,p=0.10\right).$
* Part a. What is the probability one of 6 has a flaw?
    $P(X=1)={\binom {6}{1}}\left(0.10\right)^{1}\left(1-0.10\right)^{6-1}=0.354294$
* Part b. What is the probability at least two have flaws?
    ${\begin{array}{rl}P(X\geq 2)&=1-P(X<2)=1-P(X=0)+P(X=1)\\&=1-{\binom {6}{0}}\left(0.10\right)^{0}\left(1-0.10\right)^{6-0}+{\binom {6}{0}}\left(0.10\right)^{1}\left(1-0.10\right)^{6-1}\\&=1-0.531441+0.354294\\&=0.114265\end{array}}$
* Part c. If we examine the goblets one by one, find the probability at most five are examined to fine four that are not flawed. To do this think about the situation, if we need to find four without flaws then we need to sample at least four.
    If we sample four then all four need to be flawless. Call this event $A.$
    If we sample five, then the first four sampled need to have a three flawless and one flawed and the fifth goblet sampled needs to be flawless. Call this event $B.$ (If the first four all were flawless we wouldn't have to sample a fifth time.) So we have the following:
    Let $Y\sim {\text{Binomial}}\left(n=4,p=0.90\right)$ be the number of flawless goblets in a sample of size 4.
    $P(A)=P(Y=4)={\binom {4}{4}}\left(0.90\right)^{4}\left(1-0.10\right)^{4-4}=0.6561$
    $P(B)=P(Y=3)\times 0.10=\left({\binom {4}{3}}\left(0.90\right)^{3}\left(1-0.10\right)^{4-3}\right)\times 0.10=2.6244\times 0.10=0.26244$
    and the total probability for sampling at most five goblets to get four flawless ones is:
    $P\left(A\right)+P\left(B\right)=0.6561+0.26244=0.91854$

### Exercise 50.

Let $X$ be the number of calls involving a fax message. $X\sim {\text{Binomial}}\left(n=25,p=0.25\right).$
* Part a. What is the probability at most 6 of the calls involve a fax message?
    ${\begin{array}{rl}P\left(X\leq 6\right)&=\sum _{x=0}^{6}{\binom {25}{x}}\left(0.25\right)^{x}\left(1-0.25\right)^{25-x}\\&=0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828\\&=0.5611\end{array}}$
* Part b. What is the probability exactly 6 of the calls involve a fax message?
    ${\begin{array}{rl}P\left(X=6\right)&={\binom {25}{6}}\left(0.25\right)^{6}\left(1-0.25\right)^{25-6}\\&=0.1828\end{array}}$
* Part c. What is the probability at least 6 of the calls involve a fax message?
    ${\begin{array}{rl}P\left(X\geq 6\right)&=1-P(X<6)=1-P(X\leq 5)\\&=1-\sum _{x=0}^{5}{\binom {25}{x}}\left(0.25\right)^{x}\left(1-0.25\right)^{25-x}\\&=1-\left(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645\right)\\&=1-0.3783\\&=0.6217\end{array}}$
* Part d. What is the probability more than 6 of the calls involve a fax message?
    ${\begin{array}{rl}P\left(X>6\right)&=1-P(X\leq 6)\\&=1-0.5611\\&=0.4389\end{array}}$

### Exercise 51.

Let $X$ be the same as in Exercise 50.
* Part a. The expectation of $X$ is:
    $E\left(X\right)=\mu _{x}=n\times p=25\times 0.25=6.25$
* Part b. The standard deviation of $X$ is:
    $\sigma _{x}={\sqrt {np(1-p)}}={\sqrt {25\times 0.25\times (1-0.25)}}=2.1651$
* Part c. Find the probability $X$ exceeds the mean by more than two standard deviations.
    ${\begin{array}{rl}P\left(X-\mu _{x}>2\sigma _{x}\right)&=1-P\left(X-\mu _{x}\leq 2\sigma _{x}\right)\\&=1-P\left(X\leq \mu _{x}+2\sigma _{x}\right)\\&=1-P\left(X\leq 6.25+2(2.1651)\right)\\&=1-P\left(X\leq 10.5802\right)\\&=1-P\left(X\leq 10\right)\\&=1-\sum _{x=0}^{10}{\binom {25}{x}}\left(0.25\right)^{x}\left(1-0.25\right)^{25-x}\\&=1-0.97033\\&=0.02967\end{array}}$

### Exercise 52.

Let $X$ be the number of students who want a new textbook. $X\sim {\text{Binomial}}\left(n=25,p=0.30\right)$
* Part a. Find the mean and standard deviation of $X$
    ${\begin{array}{rcl}E\left(X\right)=\mu _{x}&=&np=25\times 0.30=7.5\\\\Var\left(X\right)=\sigma _{x}^{2}&=&np(1-p)=25\times 0.30\times (1-0.30)=5.25\\\\StdDev\left(X\right)=\sigma _{x}&=&{\sqrt {np(1-p)}}={\sqrt {5.25}}=2.2913\end{array}}$
* Part b. What is the probability the number who want a new copy is more than two standard deviations from the mean?
    ${\begin{array}{rcl}P\left(\left|X-\mu _{x}\right|>2\sigma _{x}\right)&=&1-P\left(\left|X-\mu _{x}\right|\leq 2\sigma _{x}\right)\\&=&1-P\left(\mu _{x}-2\sigma _{x}\leq X\leq \mu _{x}+2\sigma \right)\\&=&1-P\left(7.5-2(2.2913)\leq X\leq 7.5+2(2.2913)\right)\\&=&1-P\left(2.9174\leq X\leq 12.0826\right)\\&=&1-P\left(3\leq X\leq 12\right)\\&=&1-\sum _{x=3}^{12}{\binom {25}{x}}\left(0.30\right)^{x}\left(1-0.30\right)^{25-x}\\&=&1-0.9735697\\&=&0.0264303\end{array}}$
* Part c. If there are 15 new copies and 15 used copies in stock, what is the probability everyone of the 25 people coming into the store will be happy? To solve this we need to think about the number of people coming in for a new book and how many that leaves wanting a used book. If, for example, 9 people come in wanting a new book then we can accommodate them, but this leaves 25 - 9 = 16 wanting a used book and there is not enough used books in stock to meet this need. It is easy to see if fewer than 9 people come in wanting a new book the same problem exists; there will not be enough used books. However, if there are 10 people who come in looking for a new book then we will have enough new books and 25 - 10 = 15 will want a used book and we have enough stock to meet both needs. Consider $X=\{10,11,12,13,14,15\}$ will result in having enough used and new books to meet the needs of the customers.
    $P\left(10\leq X\leq 15\right)=\sum _{10}^{15}{\binom {25}{x}}\left(0.30\right)^{x}\left(1-0.30\right)^{25-x}=0.1889825$
* Part d. New books cost $100 and used cost $70. What is the expected revenue for selling 25 books? (Assume there is enough stock that everyone of the 25 people will get the book they want). Let $h(x)$ be the revenue.
    $h\left(X\right)=100X+70(25-X)=1750+30X$
    $E\left(h(X)\right)=1750+30E(X)=1750+30\times 7.5=1975$
    The expected revenue is $1975.

### Exercise 53.

Using the information in Exercise 30, the probability for any one person to be cited for $y$ moving violations while driving in the last three years is defined as:
${\begin{array}{l|cccc}y&0&1&2&3\\\hline P(Y=y)&0.60&0.25&0.10&0.05\\\end{array}}$
Find the following probabilities for a group of 15 such persons.
* Part a. At least 10 have no citations. - Let $X$ be the number of person with no citations.
    $X\sim {\text{Binomial}}\left(n=15,p=0.60\right)$
    ${\begin{array}{rcl}P\left(X\geq 10\right)&=&\sum _{x=10}^{15}{\binom {15}{x}}\left(0.60\right)^{x}\left(1-0.60\right)^{15-x}\\&=&0.1859+0.1268+0.0634+0.0219+0.0047+0.0005\\&=&0.4032\end{array}}$
* Part b. Fewer than half have at least one citation. This is a little different than part a. Now the success probability is 0.40 for having at least one citation. Let $Y$ be the number of persons with at least one citation.
    $Y\sim {\text{Binomial}}\left(n=15,p=0.40\right)$
    ${\begin{array}{rcl}P\left(Y<15/2\right)&=&P\left(Y<7.5\right)\\&=&\sum _{y=0}^{7}{\binom {15}{y}}\left(0.40\right)^{y}\left(1-0.40\right)^{15-y}\\&=&0.0005+0.0047+0.0219+0.0634+0.1268+0.1859+0.2066+0.1771\\&=&0.7869\end{array}}$
* Part c. The number of persons with at least one citation is between 5 and 10 inclusive?
    ${\begin{array}{rcl}P\left(5\leq Y\leq 10\right)&=&\sum _{y=5}^{10}{\binom {15}{y}}\left(0.40\right)^{y}\left(1-0.40\right)^{15-y}\\&=&0.1859+0.2066+0.1771+0.1181+0.0612+0.0245\\&=&0.7734\end{array}}$

### Exercise 54.

Let $X$ be the number of people who want an oversize version of a tennis racket.
$X\sim {\text{Binomial}}\left(n=10,p=0.60\right)$
* Part a. What is the probability that of ten people at least 6 will want the oversize version?
    ${\begin{array}{rcl}P(X\geq 6)&=&\sum _{x=6}^{10}{\binom {10}{x}}\left(0.60\right)^{x}\left(1-0.60\right)^{10-x}\\&=&0.2508+0.2150+0.1209+0.0403+0.0060\\&=&0.6330\end{array}}$
* Part b. What is the number who want the oversize version of the racket will be within one standard deviation of the mean?
    First find the mean and the standard deviation.
    ${\begin{array}{rl}E(X)=\mu &=np=10\times 0.60=6\\Var(X)=\sigma ^{2}&=np(1-p)=2.4\\StdDev(X)=\sigma &={\sqrt {np(1-p)}}={\sqrt {2.4}}=1.5492\end{array}}$
    Now we can answer the question asked.
    ${\begin{array}{rcl}P\left(\left|X-\mu \right|<\sigma \right)&=&P\left(\mu -\sigma <X<\mu +\sigma \right)\\&=&P\left(6-1.5492<X<6+1.5492\right)\\&=&P\left(4.4508<X<7.5492\right)\\&=&P(5\leq X\leq 7)\\&=&P(X=5)+P(X=6)+P(X=7)\\&=&0.2007+0.2508+0.2150\\&=&0.6665\\\end{array}}$
* Part c. If there are 7 normal sized and 7 oversize rackets in the store, what is the probability all 10 customers will get what they want? To solve this think about what needs to happen for the stock in the store to meet the needs of the customers. If $X=3$ then there are three people getting the oversize version and this leaves 7 to get the normal size version. If only two people wanted the oversize version then there would not be enough of the normal size rackets in stock to satisfy the 8 people who would want the normal size racket. It is easy to see the max for $X$ is 7, any more and there would not be enough oversize versions in stock.
    ${\begin{array}{rcl}P\left(3\leq X\leq 7\right)&=&\sum _{x=3}^{7}{\binom {10}{x}}\left(0.60\right)^{x}\left(1-0.60\right)^{10-x}\\&=&0.0425+0.1115+0.2007+0.2508+0.2150\\&=&0.8205\end{array}}$

### Exercise 55.

Note: There are two solution methods presented here. The first is a little more complex than needed, but is a good way to illustrate several concepts in probability.
We are told that 20% of all telephones of a certain type will be submitted for service under warranty. Of the telephones submitted for service 60% are repaired and 40% are replaced. If we by 10 telephones of this type, what is the probability that exactly 2 will be replaced under warranty?
Start with defining some random variables. Let $X$ be the number of phones submitted for service. Let $R$ be the number of telephones which are replaced after being submitted for service. We have some conditional probabilities here and will need to use the law of total probability to find the final answer.
We have the following distributions:
$X\sim {\text{Binomial}}\left(n_{x}=10,p_{x}=0.20\right)$
$R|X\sim {\text{Binomial}}\left(n_{r}=X,p_{r}=0.40\right)$
Use the law of total probability to find the answer. Note that the first two probabilities in the sum are both zero, i.e., it is not possible to have $R=2$ if $X=0$ or $X=1$.
${\begin{array}{rcl}P(R=2)&=&P(R=2|X=0)P(X=0)+P(R=2|X=1)P(X=1)\\&&+P(R=2|X=2)P(X=2)+P(R=2|X=3)P(X=3)+\cdots \\&&+P(R=2|X=10)P(X=10)\\&=&\sum _{x=2}^{10}P\left(R=2|X=x\right)P\left(X=x\right)\\&=&\sum _{x=2}^{10}\underbrace {{\binom {x}{2}}\left(p_{r}\right)^{2}\left(1-p_{r}\right)^{x-2}} _{P(R=2|X=x)}\underbrace {{\binom {10}{x}}\left(p_{x}\right)^{x}\left(1-p_{x}\right)^{10-x}} _{P(X=x)}\\&=&0.1478070\end{array}}$
Now that you've seen this solution, there is another way to do this that is a lot easier. Look at the probability that a telephone is replaced. Again, use the law of total probability.
${\begin{array}{rcl}P({\text{replaced}})&=&P({\text{replaced}}|{\text{submitted}})\times P({\text{submitted}})\\&&+P({\text{replaced}}|{\text{not submitted}})\times P({\text{not submitted}})\\&=&0.40\times 0.20+0\times 0.80\\&=&0.08\end{array}}$
Now, let $W$ be the number of telephones replaced under warranty.
$W\sim {\text{Binomial}}\left(n=10,p=0.08\right)$
and the probability that 2 telephones are replaced is:
$P(W=2)={\binom {10}{2}}\left(0.08\right)^{2}\left(1-0.08\right)^{10-2}=0.1478070$

### Exercise 56.

Let $X$ be the number of students who receive special accommodations.
$X\sim {\text{Binomial}}\left(n=25,p=0.02\right)$
* Part a. The probability exactly one received special accommodations?
    $P(X=1)={\binom {25}{1}}0.02^{1}(1-0.02)^{25-1}=0.3078902$

### Exercise 57.

We know that any one battery will have acceptable voltage with probability 0.90. A flashlight requires two batteries of acceptable voltage to work. If we assume that the batteries are independent the probability the batteries in a flashlight are acceptable is $0.90\times 0.90=0.81$. Let $X$ be the number of working flashlights in a sample of size $n=10$.
$X\sim {\text{Binomial}}\left(n=10,p=0.81\right)$
The probability at least nine of the ten flashlights will work is:
${\begin{array}{rcl}P\left(X\geq 9\right)&=&P(X=9)+P(X=10)\\&=&{\binom {10}{9}}\left(0.81\right)^{9}\left(1-0.81\right)^{10-9}+{\binom {10}{10}}\left(0.81\right)^{10}\left(1-0.81\right)^{10-10}\\&=&0.2852+0.1216\\&=&0.4068\end{array}}$

### Exercise 58.

Let $X$ be the number of defective components.
$X\sim {\text{Binomial}}\left(n=10,p\right)$
Find the probability the batch is accepted given the different success probabilities $p$.
$P(X\leq 2)={\begin{cases}0.9998862\approx 0.9999&p=0.01\\0.9884964\approx 0.9885&p=0.05\\0.9298092\approx 0.9298&p=0.10\\0.6777995\approx 0.6778&p=0.20\\0.5255928\approx 0.5256&p=0.25\end{cases}}$
The sampling plan which maximizes the acceptance for $p<0.10$ and minimizes the acceptance for $p>0.10$ is the best option. Of the three methods suggested, it appears that method d, sampling 15 and accepting if less than 2 are defective is the best call. The ideal case would be having a step function with probability 1 of acceptance for $p<0.10$ and probability 0 of acceptance for $p\geq 0.10$.

### Exercise 59.

This exercise is a prelude to hypothesis testing in the later chapters. In this case we have
$X\sim {\text{Binomial}}\left(n=25,p\right)$
where $p$ is an unknown value. We will reject the claim of $p\geq 0.80$ if we observe $X\leq 15$.
* Part a. What is the probability that the claim is rejected if the true value is $p=0.80:$. This is what is called a Type I Error, rejecting the claim when it is true. In this case we have a probability of rejecting the claim if $X\leq 15$ is:
    $P(X\leq 15)=\sum _{x=0}^{15}{\binom {25}{x}}\left(p\right)^{x}\left(1-p\right)^{25-x}=0.01733187$.
* Part b. What is the probability of not rejecting the claim when
    $p=0.7$
    $1-P(X\leq 15)=\sum _{x=0}^{15}{\binom {25}{x}}\left(p\right)^{x}\left(1-p\right)^{25-x}=1-0.1894360=0.810564$.
    or
    $p=0.6$.
       $P(X\leq 15)=\sum _{x=0}^{15}{\binom {25}{x}}\left(p\right)^{x}\left(1-p\right)^{25-x}=1-0.575383=0.424617$.
    Note: the probabilities here are also error, failure to reject the claim when it is false. These are known as Type II errors. You'll learn more about this in later chapters of this book on Hypothesis testing.
* Part c. How do the error probabilities change in the above if the rejection of the claim is done when $X\leq 14$?
** for part a: the probability is 0.00555492
** for part b: with $p=0.7$ the probability is 0.9022
** for part b: with $p=0.6$ the probability is 0.585775

### Exercise 60.

Let $h(X)$ be the function for the total revenue from a total of $X$ passenger cars.
$X\sim {\text{Binomial}}\left(n=25,p=0.60\right)$
${\begin{array}{rcl}h(X)&=&1.00X+2.5\times (25-X)\\&=&62.5-1.5X\\\\E(h(X))&=&62.5-1.5E(X)\\&=&62.5-1.5np\\&=&62.5-1.5\times 25\times 0.60\\&=&40.00\end{array}}$
The expected revenue of for 25 vehicles over the toll bridge is $40.00.

### Exercise 61.

Let $A$ be the number of books received if the student writes on topic A. Let $B$ be the number of books received if the student writes on topic B.
$A\sim {\text{Binomial}}\left(n=2,p\right)\quad B\sim {\text{Binomial}}\left(n=4,p\right)$
if $p=0.9$ then the probability of having at least half of the books ordered received is
${\begin{array}{rcl}P(A\geq 1)&=&0.99\\P(B\geq 2)&=&0.9963\end{array}}$
The probability of getting at least half of the ordered books for Topic B is slightly higher than for Topic A and thus Topic B should be selected.
If the success probability for receiving a book is reduced from 0.90 to 0.50 the probabilities are:
${\begin{array}{rcl}P(A\geq 1)&=&0.75\\P(B\geq 2)&=&0.6875\end{array}}$
In this case the probability of getting at least half of the ordered books is higher for Topic A than B and Topic A should be selected.

### Exercise 62.

Let $X$ be a binomial random variable. For a fixed value of $n$ is there a value of $p$ that will allow the variance to be zero? It turns out that there are two possible values for $p$ that will allow the variance to be zero.
$Var(X)=np(1-p)=0\quad {\text{if }}p=0{\text{ or }}1.$
This should make sense. If the success probability is 0 then the out come of the experiment will be all failures, all the values are the same and the variance, a measure of spread, is zero since the spread of the data is zero. Same if the success probability is 1, all the outcomes are success, there is no spread in the data.
To maximize the variance we need a value of $p=1/2$. Here is the proof:
${\begin{array}{rcl}{\frac {\partial }{\partial p}}Var(X)&=&{\frac {\partial }{\partial p}}np(1-p)\\&=&n(1-p)-np\\&=&n-np-np\\&=&n-2np\\{\text{set }}n-2np&=&0\\p&=&1/2\\\\{\frac {\partial ^{2}}{\partial p^{2}}}Var(X)&=&-2n<0\end{array}}$
Since the second derivative is less than zero for all values of $p$ we know that when the first derivative is zero that we will have a maximum. The maximum is when $p=1/2$.

### Exercise 63.

* Part a. Show that $b(x;n,1-p)=b(n-x;n,p)$
    ${\begin{array}{rcl}b(x;n,1-p)&=&b(n-x;n,p)\\{\binom {n}{x}}\left(1-p\right)^{x}\left(1-(1-p)\right)^{n-x}&=&{\binom {n}{n-x}}\left(p\right)^{n-x}\left(1-p\right)^{n-(n-x)}\\{\frac {n!}{x!(n-x)!}}\left(1-p\right)^{x}\left(p\right)^{n-x}&=&{\frac {n!}{(n-x)!(n-(n-x))!}}\left(p\right)^{n-x}(1-p)^{x}\\{\frac {n!}{x!(n-x)!}}\left(p\right)^{n-x}\left(1-p\right)^{x}&=&{\frac {n!}{(n-x)!x!}}\left(p\right)^{n-x}(1-p)^{x}\end{array}}$
The functions are the same.
* Part b. Show that the CDF for a binomial with $n$ trials and success probability $1-p$, $B(x;n,1-p)$ is the same as $1-B(n-x-1;n,p)$
To do this define another random variable $Y=n-X$. The distributions are:
$X\sim {\text{Binomial}}\left(n,1-p\right)$
$Y\sim {\text{Binomial}}\left(n,p\right)$
${\begin{array}{rcl}B(x;n,1-p)&=&P(X\leq x)\\&=&1-P(X>x)\\&=&1-P(n-Y>x)\\&=&1-P(Y<n-x)\\&=&1-P(Y\leq n-x-1)\\&=&1-B(n-x-1;n,p)\end{array}}$
* Part c. The above two parts imply that in the Binomial probability tables in the back of the textbook, it is not necessary to publish the value for the success probability $p>0.50$ because you can find the needed probabilities using the failure probability $1-p$ which will be less than $0.50$ when ever $p>0.50$.

### Exercise 64.

Prove the expectation of a binomial random variable is $E(X)=np$.
${\begin{array}{rcll}E(X)&=&\sum _{x=0}^{n}xP(X=x)\\&=&\sum _{x=0}^{n}x{\binom {n}{x}}p^{x}(1-p)^{n-x}\\\\&=&\sum _{x=0}^{n}x{\frac {n!}{x!(n-x)!}}p^{x}(1-p)^{n-x}&{\text{the x=0 term in the sum will be 0,}}\\&&&{\text{so we can change the index on the sum}}\\\\&=&\sum _{x=1}^{n}x{\frac {n!}{x!(n-x)!}}p^{x}(1-p)^{n-x}\\\\&=&\sum _{x=1}^{n}{\frac {n!}{(x-1)!(n-x)!}}p^{x}(1-p)^{n-x}&{\text{move the index back to zero}}\\\\&=&\sum _{x=0}^{n-1}{\frac {n!}{x!(n-(x+1))!}}p^{x+1}(1-p)^{n-(x+1)}\\\\&=&np\sum _{x=0}^{n-1}\underbrace {{\frac {(n-1)!}{x!(n-1-x)!}}p^{x}(1-p)^{n-1-x}} _{\text{pmf of Binomial(n-1,p)}}&{\text{the sum of the pmf = l}}\\\\&=&np\end{array}}$

### Exercise 65.

* Part a. Let $X$ be the number of people who pay with a debit card.
    $X\sim {\text{Binomial}}\left(n=100,p=0.2\right)$
    The expectation and variance is:
    ${\begin{array}{rcccl}E(X)&=&np&=&20\\Var(X)&=&np(1-p)&=&16\end{array}}$
    The reasoning here, which is only in defining the random variable is the idea that a success is a customer paying with a debit card and a failure is for any other form of payment.
* Part b. Let $Y$ be the number of person who do not pay with cash. A success in this case is a customer paying with any form not cash and a failure is to paying with cash.
    $Y\sim {\text{Binomial}}\left(n=100,p=0.70\right)$
    ${\begin{array}{rcccl}E(Y)&=&np&=&70\\Var(Y)&=&np(1-p)&=&21\end{array}}$

### Exercise 66.

Let $X$ be the number of people with a reservation who show up for the limo.
$X\sim {\text{Binomial}}\left(n=6,p=0.80\right)$
* Part a. What is the probability at least one person who shows up for the limo cannot be accommodated? Since the limo only holds four people we will not be able to accommodate everyone if five or six show up.
    ${\begin{array}{rcl}P\left(X\leq 5\right)&=&P(X=5)+P(X=6)\\&=&0.3932+0.2621\\&=&0.6553\end{array}}$
* Part b. If there are six reservations made, what is the expected number of open seats in the limo? We need to define a function $h(x)$ for the number of open seats as a function of $x$ the number of people who show for the limo.
    ${\begin{array}{r|ccccccc}x&0&1&2&3&4&5&6\\\hline h(x)&4&3&2&1&0&0&0\end{array}}$
    The expectation of $h(X)$ is:
    ${\begin{array}{rcl}E(h(X))&=&\sum _{x=0}^{6}h(x)P(X=x)\\&=&(4)(0.0001)+(3)(0.0015)+(2)(0.0154)+(1)(0.0819)\\&&+(0)(0.2458+0.3932+0.2621)\\&=&0.1176\end{array}}$
* Part c. If the probability distribution for the number of reservations is:
    ${\begin{array}{r|cccc}{\text{Reservarions}}&3&4&5&6\\\hline {\text{Probability}}&0.1&0.2&0.3&0.4\\\end{array}}$
    Then the probability mass function for the number of passengers on a randomly selected trip is found as follows.
    We know that $X$ will take on the values 0, 1, 2, 3, 4. The value of 0 will happen if no one shows up for the limo for any one of the number of possible reservations totals. We use the law of total probability to find this. Let $R$ be the number of reservations made and $Y$ be the number of people who show up.
    Note that the conditional distribution of $Y|R$ is:
    $Y|R\sim {\text{Binomial}}\left(n=R,p=0.80\right)$
    ${\begin{array}{rcl}P(X=0)&=&\sum _{r=3}^{6}P(Y=0|R=r)P(R=r)\\&=&\sum _{r=3}^{6}\left({\binom {r}{0}}0.80^{x}0.20^{r}\right)P(R=r)\\&=&0.0080\times 0.1+0.0016\times 0.2+0.0003\times 0.3+0.0001\times 0.4\\&=&0.00125\\\\P(X=1)&=&\sum _{r=3}^{6}P(Y=1|R=r)P(R=r)\\&=&0.01724\\\\P(X=2)&=&\sum _{r=3}^{6}P(Y=2|R=r)P(R=r)\\&=&0.09064\\\\P(X=3)&=&\sum _{r=3}^{6}P(Y=3|R=r)P(R=r)\\&=&0.22732\\\\P(X=4)&=&1-(0.00125+0.01724+0.09064+0.22732)\\&=&0.66355\end{array}}$
The probability $P(X=4)$ is a little more work because you have to deal with the options of having five or six people show. The easy way to deal with this is to note that $P(X=4)=1-P(X\neq 4)$.

### Exercise 67.

Using Chebyshev's Inequality find the lower bounds for $k=2$ and $k=3$ for
$X\sim {\text{Binomial}}\left(n=20,p=0.5\right)$
The Chebyshev bounds are:
${\begin{array}{rcl}P\left(\left|X-\mu \right|>k\sigma \right)&\leq &{\frac {1}{k^{2}}}\\&=&{\frac {1}{4}}\quad {\text{For }}k=2\\\\&=&{\frac {1}{9}}\quad {\text{For }}k=3\end{array}}$
For the binomial distribution we have:
${\begin{array}{rcl}P\left(\left|X-\mu \right|>k\sigma \right)&=&1-P\left(\mu -k\sigma <X<\mu +k\sigma \right)\\&=&1-P\left(np-k{\sqrt {np(1-p)}}<X<np+k{\sqrt {np(1-p)}}\right)\\&=&1-P\left(5.527<X<14.472\right)\quad {\text{For }}k=2\\&=&1-P\left(6\leq X\leq 14\right)\\&=&1-0.9586105\\&=&0.0413895\\&=&1-P(3.29<X<16.708)\quad {\text{For }}k=3\\&=&1-P(4\leq X\leq 16)\\&=&1-0.9974232\\&=&0.002576828\end{array}}$
For the binomial distribution
$X\sim {\text{Binomial}}\left(n=20,p=.75\right)$
For $k=2$ the probability is 0.06523779 and for $k=3$ the probability is 0.003942142.

## Section 3.5 - Hyper geometric and Negative Binomial Distributions

For all the ISE 261 Students ;)

## Section 3.6 - The Poisson Probability Distribution

93a) Let P0(t,delta(t)) be the probability of no events (zero events) in the total time interval from 0 to t+delta(t). Let P0(t) be the probability of no events in the subinterval 0 to t. Let P0(delta(t)) be the probability of no events in the subinterval delta(t).
Part I: what needs to occur in both subintervals is for no events to occur.
Part II: P0(t) and P0(t,delta(t)) are not yet known so they are used as variables. but it is known that the probability of exactly 1 event occurring in the time frame delta(t) is alpha*delta(t). note: assume that delta(t) is so small that the occurrence of more than 1 event in delta(t) is zero i.e. o(delta(t)) is zero. so the probability of no events in the subinterval delta(t) is 1- alpha*delta(t). thus the total probability in the combined two subintervals of no events, assuming independence is
P0(t,delta(t)) = P0(t)*(1-alpha*delta(t)). This the relationship.
93b) Expand and rearrange the relationship P0(t,delta(t)) = P0(t)*(1-alpha*delta(t)) to give
(P0(t,delta(t)) - P0(t))/delta(t) = -alpha*P0(t)
This defines the derivative of P0(t) as delta(t) goes to zero.
93c) it is easy to show that the derivative of P0(t) = e^(-alpha*t) is -alpha*P0(t).
93d) left side of the equation: differentiate given Pk(t) with respect to t using product rule. but first factor out constant alpha^(k)/k! then right side of equation: use factorial identity that