How to find a joint confidence region? 
I need some help with this problem. They are constructed using different confidence interval, then how to combine them together? And how does knowing the fact that $Pr\{A\text{ or } B\}=Pr\{A\}+Pr\{B\}-Pr\{A \text{ and } B\}$ help? 
 A: To illustrate the solution, consider the following plot. Given independence of the two confidence intervals, we want to know the probability of both estimates being in their respective bounds.
Let $A$ be $\hat{\theta}_{1}\in[\hat{L}_{1},\hat{U}_{1}]$ and $B$ be $\hat{\theta}_{2}\in[\hat{L}_{1},\hat{U}_{2}]$.
So it should be straightforward from here:
$$\begin{align}
\text{Pr}(A)&=95\%\\
\text{Pr}(B)&=99\%\\
\text{Pr}(A\cap B)&=95\%\times 99\%=94.05\%\\
\text{Pr}(A\cup B)&=\text{Pr}(A)+\text{Pr}(B)-\text{Pr}(A\cup B)\\
&=95\%+99\%-94.05\%\\
&=99.95\%
\end{align}$$
Therefore, the answer is true.

A: The rectangle represents an area where $\hat\theta_1$ falls within $[\hat L_1; \hat U_1]$, and $\hat\theta_2$ falls within $[\hat L_2; \hat U_2]$. (Try drawing it if you don't understand why.)
Now you only need to calculate the probability of this happening, and check if it matches the number suggested in the exercise. The formula hint might come in handy, but it seems possible to do it without it as well.
A: The probability that the ''true'' $\theta_1$ is outside the interval $[\hat{l}_1,\hat{u}_1]$ is 0.05, the probability that the ''true'' $\theta_2$ is outside the interval $[\hat{l}_2,\hat{u}_2]$ is 0.01. 
The probability that the true $\theta_1$ is outside $[\hat{l}_1,\hat{u}_1]$ OR that the true $\theta_2$ is outside $[\hat{l}_2,\hat{u}_2]$ is therefore $0.01+0.05-0.01 \times 0.05=0.0595$ (see the rule you mention). 
So the probability that $(\theta_1,\theta_2)$ fall inside the rectangular region is $1-0.0595=0.9405=94.05\%$. 
