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\date{January 27, 2005}

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\title{Stat 324: Lecture 04\\
Conditional Probability}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}

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\begin{enumerate}

\item For two events $A$ and $B$ with $P(B) > 0$, the {\em
conditional probability} of A given that B has occurred is defined
as $$P(A|B) = \frac{P(A \cap B)}{P(B)}.$$

\item The Smiths have two children. At least one of them is a boy. What is the probability that both children are boys?\\
{\em Solution.} Let $X$ be the number of boys. Then
\bq P(X=2 | X\geq 1)= \frac{P(X=2,X \geq 1)}{P(X \geq 1)}\\
=\frac{P(X=2)}{1-P(X=0)}=\frac{1/4}{1-1/4}.\eq



\item {\em The law of total probability.} For disjoint
events $A_1$ and $A_2$ with $S=A_1 \cup A_2$,
$$P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2).$$
\item {\em Bayes' theorem.}
$$P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)}.$$



\item 1/10 of men and 1/7 of women are color-blind. A person is chosen at
random and that person is color-blind. What is the probability that
the
person is male. Assume males and females to be in equal numbers. \\
{\it Solution.} Let M=male,F=female,C=color-blind. Then \bq P(M|C)
&=& \frac{P(M \cap C)}{P(C)}\\&=&
\frac{P(C|M)P(M)}{P(C|M)P(M)+P(C|F)P(F)}\\&=&
\frac{\frac{1}{10}\cdot \frac{1}{2}}{\frac{1}{10}\cdot \frac{1}{2} +
\frac{1}{7} \cdot \frac{1}{2}}.\eq



\item A box contains $w$ white balls, $b$ black balls or $r$ red balls. A ball is chosen at random and if it is either black or red then it is replaced by a white ball and if it is white then it is replaced by a red ball. Now again draw a ball. What is the probability that the second ball drawn is red when the first ball drawn is red ? What is the probability that the second ball drawn is white? \\
{\it Solution.} Let $W_i,B_i,R_i$ be the event that $i-$th draw is a
white, black and red ball respectively. The sample space is given by
$$S=W_1 \cup B_1 \cup C_1 = W_2 \cup B_2 \cup C_2.$$
$$P(R_2|R_1)  = \frac{r-1}{w+b+r}.$$
\bq P(W_2) &=& P(W_2|W_1)P(W_1)\\
&+& P(W_2|B_1)P(B_1)+P(W_2|R_1)P(R_1) \\
&=& \frac{w-1}{w+b+r}\frac{w}{w+b+r} \\
&+& \frac{w+1}{w+b+r}\frac{b}{w+b+r}\\
& +& \frac{w+1}{w+b+r}\frac{r}{w+b+r}. \eq

\item Two events $A$ and $B$ are {\em independent} if $P(A|B)=P(A)$.
Obviously $A$ and $B$ are independent if $P(A \cap B)= P(A)P(B)$.
\item Suppose that $A \subset B$ and $P(A) > 0$ and $P(B) > 0$.
Are two events $A$ and $B$ independent?\\
{\it Solution.} Since $A \subset B$, $P(A \cap B) = P(B)$. 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.




\end{enumerate}

{\bf Assigned Problems} Exercise 2.60, Exercise 2.72\\
Read Chapter 3. {\bf First MIDTERM EXAM  will be on FEB 22.
9:30-10:45AM. There will be 4 problems in the midterm}.


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