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\title{Stat 311-Lecture 04, Independence}
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%\author{\Large Probability}
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Sept 15, 2005


\begin{enumerate}


\item {\em Additional problem on conditional probability.} In a bolt factory, machines
1,2 and 3 respectively produce 20 $\%$, 30$\%$ and $50\%$ of the total output.
Of their output, $5\%$, $3\%$ and $2\%$ are defective. A bolt is selected random.
(a) What is the probability that it is defective? (b) Given that it is defective, what is the probability that it was made by machine 1?\\
{\em Solution.} Let $D$ be the event that the bolt is defective and
$M_1,M_2,M_3$ be the events that the selected bolt comes from
machines 1,2 and 3 respectively. $P(M_1) = 0.2,
P(M_2)=0.3,P(M_3)=0.5$. From the law of total probability,
$$P(D) = \sum_{i=1}^3 P(D|M_i)P(M_i),$$
where $P(D|M_1)= 0.05$, $P(D|M_2)=0.03$, $P(D|M_3)=0.02$. Hence,
$P(D)= 0.029$. $P(M_1|D) = \frac{P(D \cap M_1)}{P(D)} =
\frac{P(D|M_1)P(M_1)}{P(D)} = 0.52.$


\item Two events $A$ and $B$ are {\em independent} if $P(A|B)=P(A)$.


{\em Theorem.} $A$ and $B$ are independent if $$P(A \cap B)=
P(A)P(B).$$

\item {\em Theorem.} If $A$ and $B$ are independent events, $A^c$ and $B$
are also independent.\\
{\em Solution 1.} Since $A$ and $B$ are independent, $P(A \cap B) =
P(A)P(B)$. We need show $P(A^c \cap B) = P(A^c)P(B)$. $P(B) = P(A
\cap B) + P(A^c \cap B) = P(A)P(B) + P(A^c \cap B)$. Hence $P(A^c
\cap B) = P(B) -P(A)P(B) = (1-P(A))P(B)=P(A^c)P(B)$.

{\em Solution 2.} $P(A^c|B)=\frac{P(A^c \cap B)}{P(B)} =
\frac{P(B)-P(A \cap B)}{P(B)} = \frac{P(B)-P(A)P(B)}{P(B)} = 1-
P(A)=P(A^c)$.

\item Suppose that $A \subset B$ and $P(A) > 0$ and $P(B) < 1$.
Are two events $A$ and $B$ independent?\\
{\it Solution.} Since $A \subset B$, $P(A \cap B) = P(A)$. The
condition for the independence is $P(A \cap B) = P(A)P(B)$.
$P(B)=P(A)P(B)$ iff. $P(B)=1$. Hence, if $P(B)=1$, $A$ and $B$ are
independent but if $P(B) <1$, $A$ and $B$ are not independent.


\item Suppose that $A$ and $B$ are independent events such that the probability
that neither occurs is $\frac{1}{2}$ and the probability of $B$
occurring is $\frac{1}{3}$. Determine the probability of
$A$ occurring.\\
{\em Solution.} $P ((A \cup B)^c) = 1-P(A \cup B)=1/2$. So $P(A \cup
B)=1/2$. Since $A$ and $B$ are independent $P(A \cap
B)=P(A)P(B)=P(A)/3$. By substituting terms into $P(A \cup B)  =
P(A)+P(B) - P(A \cap B)$, we have $1/2 = P(A) + 1/3 - P(A)/3$. Hence
$P(A) = 1/4$.


\item If $A$ and $B$ are independent events, with $P(A)= \frac{1}{3}$ and $P(B)=\frac{1}{4}$, find the following:\\
(a) $P(A^c \cap B^c)$\\
{\em Solution 1.} Since $A$ and $B$ are independent, $A^c$ and $B^c$ are independent. So $P(A^c \cap B^c)= P(A^c)P(B^c)=\big(1-P(A)\big)\big(1-P(B)\big) =(1-\frac{1}{3})(1-\frac{1}{4})= \frac{1}{2}$.\\
{\em Solution 2.} $P(A \cap B) = P(A)P(B) = \frac{1}{12}$. $P(A^c
\cap B^c) = 1-P(A \cup B) = 1 - \big( P(A) + P(B) - P(A \cap B)\big)
= 1 - (\frac{1}{3} + \frac{1}{4} - \frac{1}{12})= \frac{1}{2}$.

(b) $P(A^c|B)$.\\
{\em Solution.} Since $A$ and $B$ are independent, $A^c$ and $B$ are
also independent. So $P(A^c|B) = P(A^c) = 1-P(A) = \frac{2}{3}$.

\end{enumerate}



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