18. (15 points) Clothing retailers A and B wish to deal in clothing produced by a certain brand's manufacturing company. The design department of the company, without any information about A and B's sales, randomly provides them with $n$ different designs. A and B independently select the designs they approve of. What is the probability that at least one design is approved by both A and B?


18. Let the set of $n$ styles be denoted as $V$, and the choices of Party A and Party B be denoted as
$$
P_{\text {A }} \subseteq V, P_{\text{B}} \subseteq V .
$$

The combination of Party A's and Party B's choices is referred to as a selection plan, denoted as $\left(P_{\text {A }}, P_{\text {B }}\right)$.

First, we prove that the probability of any selection plan $\left(P_{\text {A}}, P_{\text{B}}\right)$ occurring is $\left(\frac{1}{4}\right)^{n}$.

In fact, since the design department has no information about the sales situation of Party A and Party B, the probability of each style being recognized or rejected by Party A or Party B is $\frac{1}{2}$.

If Party A selects $k(k=0,1, \cdots, n)$ styles and simultaneously rejects the remaining $n-k$ styles, the probability of this choice occurring is
$$
\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}=\left(\frac{1}{2}\right)^{n} \text {. }
$$

The same applies to Party B.
Since the choices of Party A and Party B are independent, the probability of any selection plan $\left(P_{\text {A }}, P_{\text{B}}\right)$ occurring is
$$
\left(\frac{1}{2}\right)^{n}\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{4}\right)^{n} \text {. }
$$

Let $\bar{P}$ denote the probability of all selection plans $\left(P_{\text {A }}, P_{\text{B}}\right)$ where $P_{\text {A }} \cap P_{\text {B }}=\varnothing$. Then the required probability is
$$
P=1-\bar{P} \text {. }
$$

To calculate $\bar{P}$, we need to calculate the number of all selection plans $S$ that satisfy $P_{\text {A }} \cap P_{\text {B }}=\varnothing$.
We classify these based on the number of elements in $P_{\text {A}} \cup P_{\text{B}}$, denoted as $\left|P_{\text {A}} \cup P_{\text{B}}\right|$.

If $\left|P_{\text {A }} \cup P_{\text {B}}\right|=i(i=0,1, \cdots, n)$, then $P_{\text {A}}$ is any subset of this $i$-element set, and the corresponding $P_{\text{B}}$ is its complement.

Thus, when $\left|P_{\text {A}} \cup P_{\text{B}}\right|=i$, the number of possible selection plans is $2^{i} \mathrm{C}_{n}^{i}$.

Therefore, by the principle of addition, when $P_{\text {A }} \cap P_{\text {B }} \neq \varnothing$, the number of all possible selection plans is
$$
S=\sum_{i=0}^{n} 2^{i} \mathrm{C}_{n}^{i}=(1+2)^{n}=3^{n} \text {. }
$$

Hence, $\bar{P}=\left(\frac{1}{4}\right)^{n} \times 3^{n}=\left(\frac{3}{4}\right)^{n}, P=1-\left(\frac{3}{4}\right)^{n}$.