(18 The school's star basketball team is about to be formed, and reserve players need to be selected from each class. It is stipulated that players with a shooting score of $A$ level can be considered as candidates. During the selection process, each person can shoot up to 5 times. If they hit 3 times, they are determined to be $B$ level. If they hit more than 4 times, they can be determined to be $A$ level. It is known that in Class 1 of Senior 3, Ah Ming has a probability of $\frac{1}{2}$ of hitting each shot.
(1) Find the probability that Ah Ming is determined to be $B$ level after shooting 4 times;
(2) If he misses two consecutive shots, he stops shooting. Find the probability that Ah Ming cannot be a candidate.

)
18 (1) Find the probability that Ah Ming is determined to be level $B$ after 4 attempts, which means the probability of hitting exactly 2 out of the first 3 attempts and definitely hitting on the 4th attempt. The probability $p=\mathrm{C}_{3}^{2} \cdot\left(\frac{1}{2}\right)^{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{3}{16}$.
(2) If Ah Ming misses two consecutive attempts, he stops and is not selected. This event can be divided into the following categories:
$\left(1^{\circ}\right)$ 3 hits out of 5 attempts have $\mathrm{C}_{4}^{2}$ possible ways of shooting, with a probability of
$$
p(3)=\mathrm{C}_{4}^{2}\left(\frac{1}{2}\right)^{5}=\frac{3}{16} ;
$$
( $2^{\circ}$ ) 2 hits, which can be categorized as “hit-hit-miss-miss”, “miss-hit-miss-miss”, “miss-hit-hit-miss-miss”, “miss-hit-miss-hit-miss”, with a probability of
$$
p(2)=\left(\frac{1}{2}\right)^{4}+3 \times\left(\frac{1}{2}\right)^{5}=\frac{5}{32} ;
$$
( $3^{\circ}$ ) 1 hit, which can be categorized as “hit-miss-miss”, “miss-hit-miss-miss”, with a probability of
$$
p(1)=\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{4}=\frac{3}{16} ;
$$
( $4^{\circ}$ ) 0 hits, which is only “miss-miss”, with a probability of
$$
p(0)=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \text {. }
$$

Therefore,
$$
p=\frac{3}{16}+\frac{5}{32}+\frac{3}{16}+\frac{1}{4}=\frac{25}{32} .
$$