一个密码由 10 个字符组成。 密码的前 5 个字符由从集合 $\{A, B, C, D, E, F, G, H\}$ 中选择的 5 个字母组成，密码的后 5 个字符由从集合 $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 0\}$ 中选择的 5 个数字组成。密码允许字母 and/or 数字重复。

1. Calculate the number of ways to arrange the first 5 characters (letters) with repetition

Since there are 8 letters and repetitions are allowed, the number of ways is $8^5$.

2. Calculate the number of ways to arrange the last 5 characters (digits) with repetition

Since there are 10 digits and repetitions are allowed, the number of ways is $10^5$.

3. Find the total number of different codes using the multiplication principle

Total number of codes $= 8^5 \times 10^5$
$= 32\,768 \times 100\,000$
$= 3\,276\,800\,000$

1. Calculate the probability that the letter 'E' appears exactly 3 times in the first 5 characters.

$P(\text{exactly 3 'E's}) = \binom{5}{3} \left(\frac{1}{8}\right)^3 \left(\frac{7}{8}\right)^2 = 10 \times \frac{1}{512} \times \frac{49}{64} = \frac{490}{32768}$

2. Calculate the probability that the digit '5' appears exactly once in the last 5 characters.

$P(\text{exactly one '5'}) = \binom{5}{1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^4 = 5 \times \frac{1}{10} \times \frac{6561}{10000} = \frac{32805}{100000}$

3. Multiply the probabilities since the conditions apply to separate parts of the code and are independent.

$\text{Overall Probability} = \frac{490}{32768} \times \frac{32805}{100000} = \frac{16074450}{3276800000} = \frac{321489}{65536000} \approx 0.00491 \text{ (to 3 s.f.)}$

1. Find the probabilities of the relevant independent events for the first and tenth characters.

Let $A$ be the event that the first character is E, and $B$ be the event that the tenth character is 4.
Since characters are chosen with replacement from their respective sets of $8$ letters and $10$ digits:
$P(A) = \frac{1}{8}, \quad P(A') = \frac{7}{8}$
$P(B) = \frac{1}{10}, \quad P(B') = \frac{9}{10}$

2. Calculate the probability that exactly one of the two events occurs.

The choices for the different positions in the code are independent. The probability of having E as the first character or 4 as the tenth, but not both, is given by $P(A \text{ and not } B) + P(\text{not } A \text{ and } B)$:
$P(A \cap B') + P(A' \cap B)$
$= \left( \frac{1}{8} \times \frac{9}{10} \right) + \left( \frac{7}{8} \times \frac{1}{10} \right)$
$= \frac{9}{80} + \frac{7}{80}$
$= \frac{16}{80} = \frac{1}{5}$