I don't know how to enter these values into a calculator, can you please help me finish the calculations.

Certainly! I'll complete the calculations for each probability using the relevant formulas.

### 1. Probability that the digit 5 appears in a 5-digit number
\[
P(\text{at least one 5}) = 1 - \frac{8^5}{9^5}
\]
First, let's calculate \(8^5\) and \(9^5\):
\[
8^5 = 32768
\]
\[
9^5 = 59049
\]
Now, the probability:
\[
P(\text{at least one 5}) = 1 - \frac{32768}{59049} \approx 1 - 0.5547 = 0.4453
\]
So, the probability is approximately **0.4453** or **44.53%**.

### 2. Probability that the digit 5 appears **twice** in any position in a 5-digit number
\[
P(\text{exactly two 5s}) = \frac{\binom{5}{2} \times 8^3}{9^5}
\]
First, calculate the binomial coefficient \( \binom{5}{2} \):
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
Next, calculate \(8^3\):
\[
8^3 = 512
\]
Now, calculate the probability:
\[
P(\text{exactly two 5s}) = \frac{10 \times 512}{59049} = \frac{5120}{59049} \approx 0.0867
\]
So, the probability is approximately **0.0867** or **8.67%**.

### 3. Probability that the digit 5 appears **three times** in any position in a 5-digit number
\[
P(\text{exactly three 5s}) = \frac{\binom{5}{3} \times 8^2}{9^5}
\]
Calculate the binomial coefficient \( \binom{5}{3} \):
\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
\]
Next, calculate \(8^2\):
\[
8^2 = 64
\]
Now, calculate the probability:
\[
P(\text{exactly three 5s}) = \frac{10 \times 64}{59049} = \frac{640}{59049} \approx 0.0108
\]
So, the probability is approximately **0.0108** or **1.08%**.

### 4. Probability that the digit 5 appears **next to each other** twice or thrice in a 5-digit number
For two consecutive 5s:
\[
\text{Number of favorable outcomes} = 4 \times 8^3 = 4 \times 512 = 2048
\]
For three consecutive 5s:
\[
\text{Number of favorable outcomes} = 3 \times 8^2 = 3 \times 64 = 192
\]
Now, calculate the total probability:
\[
P(\text{two or three consecutive 5s}) = \frac{2048 + 192}{59049} = \frac{2240}{59049} \approx 0.0379
\]
So, the probability is approximately **0.0379** or **3.79%**.

### 5. Probability that both the digits 5 and 4 appear in any position in a 5-digit number
We first need to subtract the probability of neither digit 5 nor 4 appearing:
\[
\text{Total without 5 or 4} = 7^5 = 16807
\]
The probability of at least one 5 and one 4 is:
\[
P(\text{at least one 5 and one 4}) = 1 - \frac{16807}{59049} \approx 1 - 0.2846 = 0.7154
\]
So, the probability is approximately **0.7154** or **71.54%**.

### 6. Probability that there are **2 occurrences of the digit 5** and **one occurrence of the digit 4** in a 5-digit number
\[
P(\text{2 fives and 1 four}) = \frac{\binom{5}{2} \times \binom{3}{1} \times 7^2}{9^5}
\]
First, calculate the binomial coefficients:
\[
\binom{5}{2} = 10, \quad \binom{3}{1} = 3
\]
Next, calculate \(7^2\):
\[
7^2 = 49
\]
Now, calculate the probability:
\[
P(\text{2 fives and 1 four}) = \frac{10 \times 3 \times 49}{59049} = \frac{1470}{59049} \approx 0.0249
\]
So, the probability is approximately **0.0249** or **2.49%**.

These are the probabilities for each of your scenarios! Do you have any further questions or need clarification on any of these steps?

Explore how to calculate probabilities involving the appearance of the digit 5 in 5-digit numbers. This detailed guide covers scenarios such as at least one 5, exactly two 5s, three consecutive 5s, and more. Suitable for high school students and beyond.

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