1.4.5 problem 47

(a)

Consider the experiment of flipping a fair coin twice. The sample space $S$ is $\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}.$ 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 $\mathrm{HH}$.

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

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

(b)

$A_1$ 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.

(c)

$$\begin{array}{cccccccc}P(A\cup B)& =P\left(A\right)+P\left(B\right)-P(A\cap B)& & & & & & \\ & =P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)& & & & & & \\ & =P\left(A\right)(1-P(B\left)\right)+P\left(B\right)& & & & & & \\ & =P\left(A\right)P\left(B^c\right)+P\left(B\right)& & & & & & \\ & =P\left(A\right)P\left(B^c\right)+1-P\left(B^c\right)& & & & & & \\ & =1+P\left(B^c\right)\left(P\right(A)-1)& & & & & & \\ & =1-P\left(B^c\right)P\left(A^c\right)& & & & & & \end{array}$$