An integer is chosen between 0 and 100. What is the probability that it is:
(i) divisible by 9? (ii) not divisible by 9?

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the total number of integers between 0 and 100.
The integers between 0 and 100 are 0, 1, 2, ..., 99. Therefore, the total number of integers is:
\[ 100 \]

### Step 2: Identify the integers between 0 and 100 that are divisible by 9.
The integers divisible by 9 in this range can be found by listing them:
- The multiples of 9 are: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

Counting these, we find there are:
\[ 12 \text{ integers (0 to 99) divisible by 9} \]

### Step 3: Calculate the probability that a randomly chosen integer is divisible by 9.
The probability \( P(E1) \) that an integer is divisible by 9 is given by the ratio of the number of favorable outcomes to the total outcomes:
\[ P(E1) = \frac{\text{Number of integers divisible by 9}}{\text{Total number of integers}} = \frac{12}{100} = \frac{3}{25} \]

### Step 4: Calculate the probability that a randomly chosen integer is not divisible by 9.
The probability \( P(E2) \) that an integer is not divisible by 9 can be found using the complement rule:
\[ P(E2) = 1 - P(E1) \]
Substituting the value we found for \( P(E1) \):
\[ P(E2) = 1 - \frac{3}{25} = \frac{25 - 3}{25} = \frac{22}{25} \]

### Final Answers:
(i) The probability that the integer is divisible by 9 is \( \frac{3}{25} \).

(ii) The probability that the integer is not divisible by 9 is \( \frac{22}{25} \).

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