一位业余作曲家正在编排一段由四个音符和接下来的三个节拍组成的序列。有 7 种可能的音符（标记为 A 到 G）和 5 种可能的节拍（标记为 1 到 5）。音符和节拍的顺序在创作中很重要。求随机选择的序列满足以下条件的概率：

1. Identify the total number of outcomes for the relevant part of the sequence.

Since each position in the sequence is independently generated (repetition is allowed as indicated by the context of later parts), the condition only restricts the 2nd and 3rd beats. We can focus solely on these two positions, as the choices for the notes and the 1st beat cancel out in the probability calculation.
With 5 possible beats, the total possible pairs $(b_2, b_3)$ for the 2nd and 3rd beats is:
$5 \times 5 = 25$

2. Determine the number of favorable outcomes where the third beat is strictly greater than the second beat.

For the 3rd beat to be a higher number than the 2nd beat ($b_3 > b_2$), we can just select 2 distinct beats out of the 5 available. Each chosen pair of distinct beats has exactly 1 valid arrangement (in increasing order).
Number of favorable pairs $= \binom{5}{2} = 10$.

(Alternatively, by listing possibilities for $b_2$: if $b_2=1$, there are 4 options for $b_3$; if $b_2=2$, 3 options; if $b_2=3$, 2 options; if $b_2=4$, 1 option. Total $= 4+3+2+1=10$)

3. Calculate the required probability.

$\text{Probability} = \frac{\text{Number of favorable pairs}}{\text{Total number of pairs}} = \frac{10}{25} = \frac{2}{5}$

1. Find the probability that a sequence has exactly two notes the same.

Let $N$ be the event that exactly two notes are the same.
The total number of possible arrangements for the 4 notes is $7^4 = 2401$.
To form a sequence with exactly one pair of identical notes (e.g., AABC):

• Choose the note to be duplicated: $\binom{7}{1} = 7$ ways
• Choose the 2 positions for this duplicated note: $\binom{4}{2} = 6$ ways
• Choose the notes for the 2 remaining positions from the other 6 notes: $6 \times 5 = 30$ ways

The number of favorable note arrangements is $7 \times 6 \times 30 = 1260$.
$P(N) = \frac{1260}{2401} = \frac{180}{343}$

2. Find the probability that a sequence has exactly two beats the same.

Let $B$ be the event that exactly two beats are the same.
The total number of possible arrangements for the 3 beats is $5^3 = 125$.
To form a sequence with exactly one pair of identical beats (e.g., 112):

• Choose the beat to be duplicated: $\binom{5}{1} = 5$ ways
• Choose the 2 positions for this beat: $\binom{3}{2} = 3$ ways
• Choose the remaining beat from the other 4 beats: $4$ ways

The number of favorable beat arrangements is $5 \times 3 \times 4 = 60$.
$P(B) = \frac{60}{125} = \frac{12}{25}$

3. Calculate the probability that exactly one of these events occurs (but not both).

Since the choices of notes and beats are independent, we calculate $P(N \text{ and } B') + P(N' \text{ and } B)$:
$P(N \text{ only}) = P(N) \times (1 - P(B)) = \frac{180}{343} \times \left(1 - \frac{12}{25}\right) = \frac{180}{343} \times \frac{13}{25} = \frac{2340}{8575}$
$P(B \text{ only}) = (1 - P(N)) \times P(B) = \left(1 - \frac{180}{343}\right) \times \frac{12}{25} = \frac{163}{343} \times \frac{12}{25} = \frac{1956}{8575}$

$P(\text{exactly one}) = \frac{2340}{8575} + \frac{1956}{8575} = \frac{4296}{8575}$