A mail-order computer business has six telephone lines. Let $X$ denote the number of lines in use at a specified time. Suppose the pmf of $X$ is as given in the accompanying table.

$x$	0	1	2	3	4	5	6
$p(x)$	.10	.15	.20	.25	.20	.06	.04

 

Calculate the probability of each of the following events.

• (a) $\{$ at most 3 lines are in use $\}$
• (b) $\{$ fewer than 3 lines are in use $\}$
• (c) $\{$ at least 3 lines are in use $\}$
• (d) $\{$ between 2 and 5 lines, inclusive, are in use $\}$
• (e) $\{$ between 2 and 4 lines, inclusive, are not in use $\}$
• (f) $\{$ at least 4 lines are not in use $\}$
   

An automobile service facility specializing in engine tune-ups knows that $45\%$ of all tune-ups are done on four-cylinder automobiles, $40\%$ on six-cylinder automobiles, and $15\%$ on eight-cylinder automobiles. Let $X=$ the number of cylinders on the next car to be tuned.

• (a) What is the pmf of $X$?
• (b) Draw both a line graph and a probability histogram for the pmf of part (a).

 

Remark. Here pmf stands for probability mass function.

Introduction:

Two types of random variables:

• discrete random variables
• continuous random variables

 

[Def] For a given sample space $\mathcal{S}$ of some experiment, a random variable is any rule that associates a number with each outcome in $\mathcal{S}$.

$X(s)=x$ means that $x$ is the value associated with the outcomes by the rv $X$.

Here we use the abbreviation "rv" to stand for random variable. 

 

[Def] Any random variable whose only possible values are 0 and 1 is called Bernoulli random variable. (伯努利随机变量)

 

[Def] A discrete random variable is an rv whose possible values consitute a countable set.

[Def] A random variable is continuous if its set of possible values consist of an entire interval on the number line.

 

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Section 3.2 Probability Distributions for Discrete Random Variables （离散型随机变量的概率分布）

The probability distribution of $X$ says how the total probability of 1 is distributed among (allocated to) the various possible $X$ values.

 

[Def] The probability distribution or probability mass function (pmf) of a discrete rv is defined for every number $x$ by $p(x)=P(X=x)=P(\text{all} s\in\mathcal{S}; X(s)=x)$.

Here $P(X=x)$ is read "the probability that the rv $X$ assumes the value $x$".

In words, for every possible value $x$ of the random variable, the pmf specifies the probability of observing that value when the experiment is performed. The conditions $p(x)\geqslant 0$ and $\sum_{\text{all possible } x}p(x)=1$ are required of any pmf.

 

[Def] Suppose $p(x)$ depends on a quantity that can be assigned any one of a number of possible values, with each different value determining a different probability distribution. Such a quantity is called a parameter of the distribution. 

The collection of all probability distributions for different values of the parameter is called a family of probability distributions.

 

The Cumulative Distribution Function (CDF) （累积分布函数）

For some fixed value $x$, we often wish to compute the probability that the observed value of $X$ will be at most $x$.

When $X$ is a discete random variable and $x$ is a possible value of $X$, we have

\[
P(X < x) < P(X\leqslant x)
\]

 

[Def] The cumulative distribution function (cdf) $F(x)$ of a discrete rv $X$ with pmf $p(x)$ is defined for every number $x$ by

\[
F(x)=P(X\leqslant x)=\sum_{y:\ y\leqslant x}p(y)
\]

For any number $x$, $F(x)$ is the probability that the observed value of $X$ will be at most $x$.

 

The cdf has been derived from the pmf. It is possible to reverse this procedure and obtain the pmf from the cdf whenever the later function is available.

[Prop] For any two numbers $a$ and $b$ with $a\leqslant b$, 

\[
P(a\leqslant x\leqslant b)=F(b)-F(a-)
\]

where "$a-$" represents the largest possible $X$ value that is strictly less than $a$.

In particular, if the only possible values are integers, and if $a$ and $b$ are integers, then

\[
\begin{split}
P(a\leqslant x\leqslant b)&=P(X=a \text{or} a+1 \text{or} \cdots \text{or} b)\\
&=F(b)-F(a-1)
\end{split}
\]

Taking $b=a$, yields

\[
P(X=a)=F(a)-F(a-1)
\]

in this case.

 

Subsection 3.3 Expected Values of Discrete Random Variables （离散型随机变量的期望）

[Def] Let $X$ be  a discrete rv with set of possible values $D$ and pmf $p(x)$. The expected value or mean value of $X$, denoted by $E(X)$ or $\mu_X$, is

\[
E(X)=\mu_X=\sum_{x\in D}x\cdot p(x)
\]

When it is clear to which $X$ the expected value refers, $\mu$ rather $\mu_X$ is often used.

 

---

Remark:

These notes are copied for the following book:

Jay L. Devore, Probability and Statistics For Engineering and The Sciences (Fifth Edition)

A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let $p$ denote the probability that the flaw is detected during any one fixation (this model is discussed in "Human Performance in Sampling Inspection", Human Factors, 1979: 99--105).

 

(a) Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? (Hint: $\{\text{flaw detected in at most two fixations}\}$=$\{\text{flaw detected on the first fixation}\}$ $\cup$ $\{\text{flaw undetected on the first and detected on the second}\}$.)

(b) Give an expression for the probability that a flaw will be detected by the end of the $n$th fixation.

(c) If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection?

(d) Suppose $10\%$ of all items contain a flaw [$P$(randomly chosen item is flawed)$=.1$]. With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)?

(e) Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for $p=.5$.

 

---

Remark: The exercise is copied from the reference book Devore2017B.

 

Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works iff either 1 or 2 works; since 3 and 4 are connected in series, that subsystem works iff both 3 and 4 work. If components work independently of one another and $P(\text{component works})=.9$, calculate $P(\text{system works})$.

 

 

 

---

Remark: The exercise is copied from the reference book Devore2017B.

For customers purchasing a full set of tires at a particular tire store, consider the events
\[
A=\{\text{tires purchased were made in the United States}\}
\]
\[
B=\{\text{purchaser has tires balanced immediately}\}
\]
\[
C=\{\text{purchaser requests front-end alignment}\}
\]
along with $A^c$, $B^c$, and $C^c$. Assume the following unconditional and conditional probabilities:
\[
P(A)=.75\qquad P(B|A)=.9\qquad P(B|A^c)=.8
\]
\[
P(C|A\cap B)=.8\qquad P(C|A\cap B^c)=.6
\]
\[
P(C|A^c\cap B)=.7\qquad P(C|A^c\cap B^c)=.3
\]

1.  Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.
2.  Compute $P(A\cap B\cap C)$.
3.  Compute $P(B\cap C)$.
4.  Compute $P(C)$.
5.  Compute $P(A|B\cap C)$, the probability of a purchase of U.S. tires given that both balancing and an alignment were requested.
    

 

A certain sports car comes equipped with either an automatic or a manual transmission, and the car is available in one of four colors. Relevant probabilities for various combinations of transmission type and color are given in the accompanying table.

		White	Blue	Black	Red
Transmission Type	A	.15	.10	.10	.10
Transmission Type	M	.15	.05	.15	.20

 

Let $A=\{\text{automatic transmission}\}$, $B=\{\text{black}\}$, and $C=\{\text{white}\}$.

1.   Calculate $P(A)$, $P(B)$ and $P(A\cap B)$.
2.   Calculate both $P(A|B)$ and $P(B|A)$, and explain in context what each of these probabilities represents.
3.   Calculate and interpret $P(A|C)$ and $P(A|C^c)$.
    

 

---

 

References:

Book P83, Exer 45.

Axioms of Probabilies:

Suppose $\Omega$ is the sample space, $A$ and $B$ are events. The probabilities should obey the following rules:

(1) $P(A)\geqslant 0$,

(2) $P(\Omega)=1$;

(3) If $A,B\subset\Omega$ and $A\cap B=\emptyset$, then $P(A\cup B)=P(A)+P(B)$.

 

---

Thm. $P(A)\leqslant 1$, $\forall\ A\subset\Omega$.

Proof. By axioms, $1=P(\Omega)=P(A\cup A^c)=P(A)+P(A^c)$, then we have

\[P(A)=1-P(A^c)\leqslant 1\]

 

---

Cor. If $A,B,C\subset\Omega$, and $A,B,C$ are disjoint sets. Then

\[
P(A\cup B\cup C)=P(A)+P(B)+P(C)
\]

Proof. Since $A,B,C$ are disjoint sets, $A$ and $B\cup C$ are disjoint. Thus

\[
P(A\cup B\cup C)=P(A\cup (B\cup C))=P(A)+P(B\cup C),
\]

and by Axiom(3), $P(B\cup C)=P(B)+P(C)$, substitute it in the above, we get

\[
P(A\cup B\cup C)=P(A)+P(B)+P(C).
\]

---

It is easy to infer the following corollary.

Cor. If $A_1,A_2,\ldots,A_n$ are disjoint sets in $\Omega$, then

\[
P(A_1\cup A_2\cdots A_n)=P(A_1)+P(A_2)+\cdots+P(A_n).
\]

Proof. By mathematical induction.

 

---

 

Cor. If $s_1,\ldots,s_n$ are $n$ distinct points(or outcomes) in $\Omega$ then

\[
P(\{s_1,\ldots,s_n\})=P(\{s_1\})+\cdots+P(\{s_n\})=P(s_1)+\cdots+P(s_n).
\]

Here, we use $P(s_i)$ to denote the probability $P(\{s_i\})$.

---

 

Suppose the sample space have $N$ points, and event $A$ has $n$ points. Here $\Omega$ is a finite set, i.e., $N < \infty$.  If all outcomes are equally likely, then 

\[
P(A)=\frac{\text{number of elements of}\ A}{\text{total number of sample points}}=\frac{\# A}{\#\Omega}
\]

 

---

Example. If $A,B$ are two subsets of $\Omega$. Let $C=A-B=A\cap B^c$, $D=B-A=B\cap A^c$, $E=A\cap B$.

Then we have $P(C\cup D)=P(A)+P(B)-2P(E)$.

Solution. 

Note that $C$ and $D$ are disjoint, hence we have

\[
P(C\cup D)=P(C)+P(D).\tag{1}
\]

On the other hand, $A=C\cup E$, $B=D\cup E$, and $C\cap E=\emptyset$, $D\cap E=\emptyset$. Thus

\[
\begin{aligned}
P(A)&=P(C)+P(E),\\
P(B)&=P(D)+P(E).
\end{aligned}
\]

It infers that $P(A)+P(B)=P(C)+P(D)+2P(E)$. By using (1), we have

\[
P(C\cup D)=P(A)+P(B)-2P(E).
\]

---

 

Prop. Suppose $A,B\subset\Omega$, we have

\[
P(A\cup B)=P(A)+P(B)-P(A\cap B).
\]

Proof. 

\[
P(A\cup B)=P(A\cup(B-A))=P(A\cup(B\cap A^c))=P(A)+P(B\cap A^c).
\]

Note that $P(B)=P(B\cap A^c)+P(B\cap A)$, we substitute it in the above formula, and get

\[
P(A\cup B)=P(A)+P(B)-P(A\cap B).
\]

---

 

Prop. Suppose $A,B,C\subset\Omega$, we have

\[
P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).
\]

 

Exercise. Using the axioms to show that if one event $A$ is contained in another event $B$ (i.e. $A$ is a subset of $B$), then $P(A)\leqslant P(B)$.

For general $A$ and $B$, what does this imply about the relationship among $P(A\cap B)$, $P(A)$ and $P(A\cup B)$ ?

Solution. Since $A\subset B$, $B=A\cup(B\cap A^c)$, here $A$ and $B\cap A^c$ are disjoint. Thus by axioms,

\[
P(B)=P(A)+P(B\cap A^c)\geqslant P(A).
\]

For general $A$ and $B$, we have $A\cap B\subset A\subset A\cup B$, thus we have

\[
P(A\cap B)\leqslant P(A)\leqslant P(A\cup B).
\]

Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2-3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2-3 subsystem.

The experiment consists of determining the condition of each component [$S$ (success) for a functioning component and $F$ (failure) for a nonfunctioning component].

A. What outcomes are contained in the event $A$ that exactly two out of the three components function?
B. What outcomes are contained in the event $B$ that at least two of the components function?
C. What outcomes are contained in the event $C$ that the system functions?
D. List outcomes in $C^{c}$, $A\cup C$, $A\cap C$, $B\cup C$, and $B\cap C$.
 

 

---

Reference:

Jay L. Devore, Probability and Statistics, For Engineering and The Sciences (Fifth Edtion)

Four universities -- 1,2,3 and 4 -- are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4. Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4 in first round games, and then 1 beats 3 and 2 beats 4).

A. List all outcomes in $\mathcal{S}$.
B. Let $A$ denote the event that 1 wins the tournament. List outcomes in $A$.
C. Let $B$ denote the event that 2 gets into the championship game. List outcomes in $B$.
D. What are the outcomes in $A\cup B$ and in $A\cap B$? What are the outcomes in $A^{c}$?
 

 

 

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Reference:

Jay L. Devore, Probability and Statistics, For Engineering and The Sciences (Fifth Edtion)