INa room of 50 people How many people in the room have a mobile where either of the last two digits is 1 or 2?

6. How many people in the room have a mobile where the last two digits are 1 and 2 in any

order?

Let's break down the problem.

### 1. People with mobile numbers where either of the last two digits is **1 or 2**:
We are assuming each person’s phone number ends in two digits, and those digits are random from 00 to 99, giving us a total of **100 possible combinations** for the last two digits.

Now, we want to find how many numbers end with either of the last two digits being **1 or 2**.

#### Case 1: First digit is 1 or 2
- If the **first digit** (tens place) is **1**, the second digit (units place) can be anything from 0 to 9. So, there are **10 numbers** like: 10, 11, 12, 13, ..., 19.
- If the **first digit** is **2**, the same reasoning applies, giving us **10 more numbers**: 20, 21, 22, ..., 29.

Thus, the number of combinations where the **first digit is either 1 or 2** is \( 10 + 10 = 20 \).

#### Case 2: Second digit is 1 or 2
- If the **second digit** (units place) is **1**, the first digit (tens place) can be anything from 0 to 9. So, there are **10 more numbers**: 01, 11, 21, ..., 91.
- If the **second digit** is **2**, the same reasoning gives **10 more numbers**: 02, 12, 22, ..., 92.

Thus, the number of combinations where the **second digit is either 1 or 2** is \( 10 + 10 = 20 \).

#### Total number of valid combinations:
The numbers where either the first or second digit is **1 or 2** are given by the union of both cases. However, the numbers where **both digits** are **either 1 or 2** have been counted twice, so we need to subtract the overlap:
- The overlap consists of the four numbers: 11, 12, 21, 22.

Thus, the total number of valid numbers is:
\[
20 + 20 - 4 = 36
\]
Out of 100 possible last-two-digit combinations, **36 numbers** have at least one digit being 1 or 2.

#### Probability:
The probability that a random phone number ends with one of these 36 numbers is \( \frac{36}{100} = 0.36 \).

Therefore, in a room of 50 people, the expected number of people with such numbers is:
\[
50 \times 0.36 = 18
\]

### 2. People with mobile numbers where the last two digits are **1 and 2 in any order**:
The possible combinations where the last two digits are **1 and 2 in any order** are:
- 12
- 21

Thus, there are **2 numbers** where the last two digits are **1 and 2**.

#### Probability:
The probability of a random phone number ending with one of these two numbers is \( \frac{2}{100} = 0.02 \).

Thus, the expected number of people in the room with these numbers is:
\[
50 \times 0.02 = 1
\]

### Final Answers:
1. **18 people** are expected to have a mobile number where either of the last two digits is 1 or 2.
2. **1 person** is expected to have a mobile number where the last two digits are 1 and 2 in any order.

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Would you like further details or have any questions?

Here are 5 related questions:
1. What is the probability of having neither of the last two digits as 1 or 2?
2. How many people would have a number where both digits are the same (like 11, 22, 33)?
3. What if the room had 100 people instead of 50? How would the results change?
4. How does this calculation change if we want the last three digits instead of two?
5. Can you calculate the number of people who have at least one digit being a specific digit (e.g., 3)?

**Tip:** In problems involving probabilities, it's useful to remember the principle of inclusion and exclusion when calculating overlaps!

In a room of 50 people, how many people in the room have a mobile where either of the last two digits is 1 or 2? How many people in the room have a mobile where the last two digits are 1 and 2 in any order?

Explore how to calculate the probability of people having mobile numbers ending with 1 or 2 as the last two digits in a room of 50. This problem uses combinatorics and the inclusion-exclusion principle, providing clear steps and expected results for both cases. Ideal for high school students (Grades 10-12).

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