# Probability that length of shortest nonzero vector is less than a number

Let $$\Lambda\subset \mathbb{Z}^n$$ be an $$n-$$ dimensional lattice with determinant $$d$$. We know that the probability that a uniformly random integer vector $$x$$ is a point in $$\Lambda$$ is given by $$\dfrac{1}{d}$$. Denote $$\lambda_1$$ be the length of the shortest nonzero vector in $$\Lambda$$ and we know by Minkowski that if the $$n-$$dimensional ball $$B:=B_R(\mathbf{0}$$) has the property that $$\textbf{vol}(B)\geq 2^n\cdot d$$, then $$B$$ contains a nonzero lattice point in $$\Lambda$$ and also that $$\lambda_1\leq \sqrt{n}d^{1/n}$$.

Now, suppose that $$\Lambda$$ is a uniformly random chosen $$n-$$dimensional lattice with determinant $$d$$ and let $$\delta\in (0,1)$$. I want to get a nice upper bound for $$\Pr[\lambda_1(\Lambda)\leq\delta\sqrt{n}d^{1 / n}]$$, in other words, would like to make it as negligible as it can be with the proper choice of $$\delta$$.

My attempt is that $$\Pr[\lambda_1(\Lambda)\leq\delta\sqrt{n}d^{1 / n}]$$ is equal to $$\Pr[\text{at least one nonzero vector in } \Lambda \text{ has norm }\leq \delta\sqrt{n}d^{1 / n}]$$. If I take $$R=\delta\sqrt{n}d^{1 / n}$$, then the above probability is equivalent to $$\Pr[\exists\mathbf{0}\neq\mathbf{x}\in\Lambda \cap B_R(\mathbf{0})]$$. So this seems that I have to use a union bound, that is, $$\Pr[\exists\mathbf{0}\neq\mathbf{x}\in\Lambda \cap B_R(\mathbf{0})]\leq \displaystyle\sum \Pr[\mathbf{x}\in\Lambda \cap B_R(\mathbf{0})]$$, where the sum is over all nonzero integer vectors $$\mathbf{x}\in\mathbb{Z}^n$$.

So my questions are:

1. Is the above inequality that I did using the union bound correct? Are the description of events I made correct?
2. If the answers to my first question are in the affirmative, how do I calculate $$\displaystyle\sum \Pr[\mathbf{x}\in\Lambda \cap B_R(\mathbf{0})]$$? Is it equal to $$\dfrac{\textbf{vol}(B_R(\mathbf{0}))}{d}$$?