An amateur music composer is arranging a sequence of four musical notes followed by three beats. There are 7 possible notes (labeled A to G) and 5 possible beats (labeled 1 to 5). The order of the notes and beats is important in the composition. Find the probability that a randomly chosen sequence has

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 $P(N \text{ and } B') + P(N' \text{ and } B)$:
$P(N \text{ only}) = P(N) \times (1 - P(B)) = \frac{180}{343} \times \left(1 - \frac{12}{25}\right) = \frac{180}{343} \times \frac{13}{25} = \frac{2340}{8575}$
$P(B \text{ only}) = (1 - P(N)) \times P(B) = \left(1 - \frac{180}{343}\right) \times \frac{12}{25} = \frac{163}{343} \times \frac{12}{25} = \frac{1956}{8575}$

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