Exercise 1.47

Question

Events $A$ and $B$ are independent if $P(A \cap B)=P(A) P(B)$ (independence is explored in detail in the next chapter).
ewline
(a) Give an example of independent events $A$ and $B$ in a finite sample space $S$ (with neither equal to $\emptyset$ or $S$ ), and illustrate it with a Pebble World diagram.
ewline
(b) Consider the experiment of picking a random point in the rectangle
$$
R=\{(x, y): 0<x<1,0<y<1\}
$$
where the probability of the point being in any particular region contained within $R$ is the area of that region. Let $A_{1}$ and $B_{1}$ be rectangles contained within $R$, with areas not equal to 0 or 1 . Let $A$ be the event that the random point is in $A_{1}$, and $B$ be the event that the random point is in $B_{1}$. Give a geometric description of when it is true that $A$ and $B$ are independent. Also, give an example where they are independent and another example where they are not independent.
(c) Show that if $A$ and $B$ are independent, then
$$
P(A \cup B)=P(A)+P(B)-P(A) P(B)=1-P\left(A^{c}ight) P\left(B^{c}ight)
$$

fifthist x · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item Consider the experiment of flipping a fair coin twice. 
The sample space $S$ is $\{HH, HT, TH, TT\}.$ 
Let $A$ be the event that the first flip lands heads and $B$ be the event
that the second flip lands heads. 
$P(A \cap B) = \frac{1}{4}$ since $A \cap B$ corresponds to the outcome $HH$.

On the other hand, $A$ corresponds to the outcomes $\{HH, HT\}$ 
and $B$ corresponds to the outcomes $\{HH, TH\}$. Thus, 
$P(A) = P(B) = \frac{1}{2}.$

Since $P(A \cap B) = P(A)P(B),$ $A$ and $B$ are independent events.

\item $A_{1}$ and $B_{1}$ should intersect such that the ratio of 
the area of $A_{1} \cap B_{1}$ to the area of $A_{1}$ equals the ratio of 
the area of $B_{1}$ to the area of $R$.

As a simple, extreme case, if $A_{1} = B_{1},$ then $A$ and $B$ are dependent,
since the condition above is violated.

\item 
\begin{flalign}
P(A \cup B) & = P(A) + P(B) - P(A \cap B) 
onumber && \
 & = P(A) + P(B) - P(A)P(B) 
onumber && \
 & = P(A)(1 - P(B)) + P(B) 
onumber && \
 & = P(A)P(B^{c}) + P(B) 
onumber && \
 & = P(A)P(B^{c}) + 1 - P(B^{c}) 
onumber && \
 & = 1 + P(B^{c})(P(A) - 1) 
onumber && \
 & = 1 - P(B^{c})P(A^{c}) 
onumber
\end{flalign}
\end{enumerate}

Exercise 1.47

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Exercise 1.47

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