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MA181 INTRODUCTION TO STATISTICAL MODELLING
ELEMENTS OF PROBABILITY
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\begin{description}
\item[1.]
An elementary result of an experiment is called an
\textit{outcome}.
\item[2.]
The totality of the outcomes of an experiment constitutes the
\textit{sample space S}.
\item[3.]
A set of outcomes is called an \textit{event}. Thus an event is a
subset of $S$. Events are often denoted by $A,B,C,\ldots$, or by
$A_1,A_2,A_3,\ldots$.
\item[Example - Birth months:]
Suppose an experiment consists of recording the birth months of my
niece's first-born child. Then the sample space $S$ consists of
the twelve outcomes \lq\lq born in January, \lq\lq born in
February'',\ldots,\lq\lq born in December''. The event $A$,
defined by the set \{\lq\lq born in March'', \lq\lq born in
April'', \lq\lq born in May''\}, may be defined as the event
\lq\lq born in Spring''.
\item[4.]
A function $P(.)$ defined on the sample space $S$ is called a
\textit{probability function} if
\begin{description}
\item[(i)]
$P(S)=1$,
\item[(ii)]
$0\leq P(A)\leq1$ for any event $A$,
\item[(iii)]
$P(A$ or $B)=P(A)+P(B)$ if $A$ and $B$ are mutually exclusive
(i.e. they have no outcomes in common).
\end{description}
Note that the event \lq\lq$A$ or $B$'' is the set of outcomes
contained in $A$ or $B$ or both, and may also be written as $A\cup
B$. Similarly $A\cap B$ or \lq\lq$A$ and $B$''. is the set of
outcomes common to both $A$ and $B$. So, if $A$ and $B$ are
mutually exclusive, then $A\cap B=\emptyset$.
If event $A$ consists of the outcomes $w_1,w_2,\ldots,w_k$, then,
by (iii),
$$P(A)=P(w_1)+P(w_2)+\ldots+P(w_k).$$
If, moreover, the outcomes of a finite sample space are all
equally probable, then the probability of an event $A$ is given by
$$P(A)=\frac{\textrm{Number of outcomes favourable to
}A}{\textrm{Total number of outcomes}}.$$
An example is provided by a perfectly symmetrical die with faces
numbered 1,2,3,4,5,6. All faces (outcomes) are equally probable
with each having probability $\frac{1}{6}$ of being observed. Such
a die is known as a fair die. If, further, $A$ is the event \lq\lq
odd number''. then $P(A)=\frac{3}{6}=\frac{1}{2}$.
\item[Example - Birth months revisited:]
Suppose the months have the following probabilities of being the
birth month:
\begin{tabular}{lll}
$P$(Jan)$=\frac{1}{12}-0.04,$&$P$(Feb)$=\frac{1}{12}-0.02,$
&$P$(Mar)$=\frac{1}{12}$\\ $P$(Apr)$=\frac{1}{12}+0.01,$&
$P$(May)$=\frac{1}{12}+0.02,$&$P$(Jun)$=\frac{1}{12}+0.02,$\\
$P$(Jul)=$\frac{1}{12}+0.02,$&$P$(Aug)$=\frac{1}{12}+0.03,$&
$P$(Sep)$=\frac{1}{12}+0.01,$\\ $P$(Oct)$=\frac{1}{12},$
&$P$(Nov)$=\frac{1}{12}-0.02,$&$P$(Dec)$=\frac{1}{12}-0.03.$
\end{tabular}
Then $P$(born in Spring)$=\frac{3}{12}+0.01+0.02=0.28$.
\item[Theorem 1]
If $A$ is an event in $S$, then $P(\overline{A})=1-P(A)$.
\item[Proof]
Now $A\cup\overline{A}=S$ while $A\cap\overline{A}=\phi$. Hence,
by (iii) above,
$$1=P(S)=P(A\cup\overline{A})=P(A)+P(\overline{A}),$$
so that $P(\overline{A})=1-P(A)$.
\item[Theorem 2]
If $A$ and $B$ are two events in $S$, then $P(A)=P(A\cap
B)+P(A\cap\overline{B})$.
\item[Proof]
In the diagram, $A\cap B$ is hatched horizontally, while
$A\cap\overline{B}$ is hatched vertically. Now
$$A=(A\cap B)\cup(A\cap \overline{B})\textrm{ and }(A\cap
B)\cap(A\cap\overline{B})=\emptyset.$$
Hence, by (iii) above,
$$P(A)=P(A\cap B)+P(A\cap\overline{B}).$$
\item[5.]
\begin{description}
\item[Theorem 3 (Addition rule)]
For any two events $A$ and $B$ in $S$,
$$P(A\textrm{ or }B)=P(A)+P(B)=P(A\textrm{ and }B).$$
\item[Proof]
In the diagram, event $A$ is hatched diagonally while
$\overline{A}\cap B$ is hatched horizontally. Now
$$A\cup B=A\cup(\overline{A}\cap B)\textrm{ and
}A\cap(\overline{A}\cap B)=\emptyset.$$
Hence
\begin{eqnarray*}
P(A\cup B)&=&P(A)+P(\overline{A}\cap B),\textrm{ by (iii),}\\
&=&P(A)_+P(B)=P(A\cap B),\textrm{ by Theorem 2.} \end{eqnarray*}
\end{description}
\item[Example]
Let $A$ be the event \lq\lq even number'' and $B$ the event \lq\lq
number$>3$'' when a fair die is thrown. Then
$P(A)=P(B)=\frac{3}{6}$ and $P(A\textrm{ and }B)=\frac{2}{6}$.
Hence $P(A\textrm{ or
}B)=\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}$, i.e.
$P(\{2,4,5,6\})=\frac{2}{3}$.
\item[6. Multiplication rule]
Let $A$ and $B$ be two events in $S$. Then the \textit{conditional
probability} of $A$ given $B$ is defined as $P(A|B)=P(A\textrm{
and }B)/P(B)$ so long as $P(B)>0$.
Similarly $P(B|A)+P(A\textrm{ and }B)/P(A)$, so long as $P(A)>0/$
Note that $P(A|B)$ represents the probability of $A$ in the
reduced sample space defined by $B$.
When these definitions are written in the form
$$P(A\textrm{ and }B)=P(A)P(B|A)=P(B)P(A|B),$$
this is know as the \textit{multiplication rule}.
\item[Example:]
A bag contains three black balls and three white balls. I draw two
balls from the bag without replacement. The probability that I
draw two white balls is
$$P(\textrm{first ball is white})P(\textrm{second ball is
white|first is white})=\frac{3}{6}\times\frac{2}{5}=\frac{1}{5}.$$
\item[Example:]
I shake a fair die. Then
\begin{eqnarray*}P(\textrm{number$>$3|even number})&=&\frac{P(\textrm{number$>$3
and even number})}{P(\textrm{even number})}\\
&=&\frac{P(\{4,6\})}{P(\{2,4,6\})}=\frac{2}{3}.
\end{eqnarray*}
\item[Example - Birth months revisited:]
Now
\begin{eqnarray*}
P(\textrm{Born in may|born in Spring})&=&\frac{P(\textrm{born in
May and born in Spring})}{P(\textrm{born in Spring})}\\
&=&\frac{\frac{1}{12}+0.02}{\frac{1}{4}+0.03}\\
&=&\frac{31}{84}=0.369 \end{eqnarray*}
\item[Example:]
Suppose a person is chosen at random from the population, that $F$
is the event that he or she speaks French and $G$ is the event
that he or she speaks German. Suppose, moreover, that
$P(F)=\frac{1}{100},\ P(G)=\frac{1}{500}$ and
$P(F|G)=\frac{2}{3}$. Then
$$P(G|F)=\frac{P(F \textrm{ and
}G)}{P(F)}=\frac{P(F|G)P(G)}{P(F)}=\frac{\frac{2}{3}\cdot\frac{1}{5
00}}{\frac{1}{100}}=\frac{2}{15}.$$
\item[Theorem 4]
For any two events $A$ and $B$ in $SW$,
$$P(A)=P(A|B)P(B)+P(A|\overline{B})P(\overline{B}).$$
[This is a simple case of the theorem of total probability.]
\item[Proof]
From Theorem 2,
$$P(A)=P(A\cap B)+P(A\cap \overline{B}).$$
However,
$$P(A\cap B)=P(B)P(A|B)\textrm{ and
}P(A\cap\overline{B}=P(\overline{B})P(A|\overline{B}).$$
Consequently, after substitution, we have the result.
\item[7.]
Events $A$ and $B$ are said to be \textit{independent} if
$$P(A\textrm{ and }B)=P(A)P(B).$$
Thus $P(A|B)=P(A)$, so long as $P(B)>0$, and $P(B|A)=P(B)$, as
long as $P(A)>0$.
\item[Example:]
On a fair die, let $A$ be the event \lq\lq number$\leq$2'' and $B$
the event \lq\lq even number''. Then
$$P(A)=\frac{1}{3},\ P(B)=\frac{1}{2},\ P(A\textrm{ and
}B)=\frac{1}{6}.$$
Hence $A$ and $B$ are independent.
\item[Example:]
Two dice are shaken together. If $A$ is an event on one of them
and $B$ an event on the other, then $A$ and $B$ are assumed to be
independent. Consequently the probability of two sixes is
$\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$.
\item[Example:]
I shake two fair dice. What is the probability that the total
shown is six?
$P$(total is 6)=$P$(first die shows 1 and second 5)+$P$(first die
shows 2 and second 4)+\ldots+$P$(first die shows 5 and second 1)
=$P$(first die shows1).P(second shows 5)+\ldots+P(first die shows
5).$P$(second shows 1)
$=\left(\frac{1}{6}\cdots\frac{1}{6}\right)+\left(\frac{1}{6}\cdot
\frac{1}{6}\right)+\ldots+\left(\frac{1}{6}\cdot\frac{1}{6}\right)=\frac{5}{36}.$
\item[Example:]
A machine produces metal rods for cars, of which $\frac{1}{100}$
are too long while $\frac{1}{60}$ are dent, these two faults
occurring independently. What is the probability of a faulty rod?
A rod is faulty if it is too long or bent (or both). Hence
$$P(\textrm{faulty
rod})=\frac{1}{100}+\frac{1}{60}-\left(\frac{1}{100}\times\frac{1}{60}\right)=\frac{53}{2000}=0.0265.$$
\item[8.]
The above results can be extended to more than two events. So, for
three events $A,B$ and $C$ in $S$,
\begin{description}
\item[(i)]
$P(A\textrm{ or $B$ or}C)=P(A)+P(B)+BP(C)-P(A\textrm{ and
}B)-P(A\textrm{ and }C)-P(B\textrm{ and }C)+P(A\textrm{ and $B$
and }C)$,
\item[(ii)]
$P(A\textrm{ and $B$ and }C)=P(A)P(B|A)P(C|A\textrm{ and }B)$ etc.
\item[(iii)]
$A