Find the probability for the function $Z = \min\{X,Y\}$ [closed]

Let X and Y be two independent random variables with the cumulative distribution functions

$F_X(x)= 1−(2/3)^x, x=1,2,3,···;$ and

$G_Y(y)= 1−(3/4)^y, y=1,2,3,\ldots;$ respectively.

Let $Z = \min\{X, Y \}$. Then, the probability $P (Z ≥ 6)$ is $\frac{1}{32}$. I am getting $\frac{63}{64}$.

Solution 1:

\begin{align} P(\min(X,Y)\geq 6) & =P(X\geq 6,Y\geq 6) \\ & =P(X\geq 6)P(Y\geq6) \\ & = [1-P(X\leq 5)][1-P(Y\leq 5)] \\ & = [1-F_{X}(5)][1-G_{Y}(5)] \\ & = \left(\frac{2}{3}\right)^{5}\left(\frac{3}{4}\right)^{5} \\ & = \frac{1}{2^{5}} \\ & = \frac{1}{32} \end{align}