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


To determine how many people in the room of 50 have a mobile number where either of the last two digits is 1 or 2, we can follow this process:

### Step 1: Probability that the last two digits contain 1 or 2
Each digit in a phone number (last two digits in this case) can be any of the 10 digits: 0 through 9. So, the total number of possibilities for each of the last two digits is 10.

- The probability that the first of the last two digits is either 1 or 2 is \(\frac{2}{10} = 0.2\).
- Similarly, the probability that the second of the last two digits is either 1 or 2 is \(\frac{2}{10} = 0.2\).

### Step 2: Calculating the probability that at least one of the last two digits is either 1 or 2
Now we need to calculate the probability that **at least one** of the last two digits is 1 or 2. We can use the complement rule:

1. The probability that **neither** of the last two digits is 1 or 2 is:
   \[
   \left(\frac{8}{10}\right) \times \left(\frac{8}{10}\right) = 0.64
   \]
2. So, the probability that at least one of the last two digits is 1 or 2 is:
   \[
   1 - 0.64 = 0.36
   \]

### Step 3: Expected number of people with this condition
The expected number of people in the room who have a mobile number where either of the last two digits is 1 or 2 is:
\[
0.36 \times 50 = 18
\]

Thus, **approximately 18 people** in the room have a mobile number where either of the last two digits is 1 or 2.

Would you like further details or have any questions?

Here are 5 questions to expand on this:
1. How would the result change if the number of people in the room increased to 100?
2. What is the probability that both of the last two digits are 1 or 2?
3. Can we generalize this method to any two specific digits, not just 1 and 2?
4. What if we considered the last three digits instead of two? How would that affect the outcome?
5. What is the probability that neither of the last two digits is a 1 or 2?

**Tip:** When calculating probabilities for "at least one" events, using the complement rule often simplifies the computation.

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?

Explore how to calculate the number of people in a room of 50 whose mobile numbers have either of the last two digits as 1 or 2. This problem uses the complement rule in probability and is suitable for students in Grades 9-11.

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