Consider a sequence of independently and identically distributed random variables $X_1,X_2,...,X_n$, where $X_i \in \mathcal{X} = \{ 0,1\}$ and $P[X_i = 1]=q$ for all $i \in \{ 1,...,n \}$. Let $\delta$ be a real-valued constant, $\delta \in (0,1]$. Let us define the random variable: $Z_n = \prod_{i=1}^{n} X_i$.
Is the following statement true? If $q \leq \frac{1}{2}$, then $Pr\{Z_8 \geq \delta^8 \} \leq 10^{-2}$
Given that $0 < \delta \leq 1$, then $Z_n \geq \delta^n$ if $X_1,...,X_n$ are all 1's, as $Z_n = \prod_{i = 1}^{n} X_i$ and a number multiplied by zero is zero. So, if one the RV $X_i$ is zero, then $Z_n=0$, if not, then $Z_n = 1$.
Therefore, $$P[Z_n \geq \delta^n] = P_{Z_n}(1)=\prod_{i=1}^{n}P(X_i=1)=q^n$$ as the RV's $X_1,...,X_n$ are independent.
The statement says that $P[Z_8 \geq \delta^8] \leq 10^{-2}$, therefore, $$P[Z_8 \geq \delta^8] = q^8 \leq 10^{-2}$$
In other words, $$q \leq (10^{-16})^{\frac{1}{8}} = 10^{-\frac{1}{4}} \approx 0.5 = \frac{1}{2}$$
In conclusion, I have found that the statement is false but apparently, it is true. Is there any mistake in my calculations?
Thanks.