A book contains 32 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is :
(i) divisible by 2
(ii) divisible by 3.
(iii) divisible by 5
(iv) divisible by 2 or 3 or 5
(v) divisible by 2, 3 and 5 ?

Answer

To solve the problem, we will calculate the probability that the sum of the digits on a randomly chosen page from a book containing 32 pages is divisible by 2, 3, 5, or combinations of these numbers.

### Step-by-Step Solution:

1. **Identify the Total Pages**:
The book contains 32 pages, numbered from 1 to 32.

2. **Calculate the Sum of Digits for Each Page**:
We need to find the sum of the digits for each page number:
- For pages 1 to 9, the sum is simply the page number itself.
- For pages 10 to 32, the sum is calculated as follows:
- Page 10: 1 + 0 = 1
- Page 11: 1 + 1 = 2
- Page 12: 1 + 2 = 3
- Page 13: 1 + 3 = 4
- Page 14: 1 + 4 = 5
- Page 15: 1 + 5 = 6
- Page 16: 1 + 6 = 7
- Page 17: 1 + 7 = 8
- Page 18: 1 + 8 = 9
- Page 19: 1 + 9 = 10
- Page 20: 2 + 0 = 2
- Page 21: 2 + 1 = 3
- Page 22: 2 + 2 = 4
- Page 23: 2 + 3 = 5
- Page 24: 2 + 4 = 6
- Page 25: 2 + 5 = 7
- Page 26: 2 + 6 = 8
- Page 27: 2 + 7 = 9
- Page 28: 2 + 8 = 10
- Page 29: 2 + 9 = 11
- Page 30: 3 + 0 = 3
- Page 31: 3 + 1 = 4
- Page 32: 3 + 2 = 5

3. **Count the Pages with Sums Divisible by 2**:
The sums that are divisible by 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32.
- Total pages: 16 pages (2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32).

**Probability = Favorable outcomes / Total outcomes = 16 / 32 = 1/2**.

4. **Count the Pages with Sums Divisible by 3**:
The sums that are divisible by 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
- Total pages: 10 pages (3, 6, 9, 12, 15, 18, 21, 24, 27, 30).

**Probability = 10 / 32 = 5/16**.

5. **Count the Pages with Sums Divisible by 5**:
The sums that are divisible by 5 are: 5, 10, 15, 20, 25, 30.
- Total pages: 6 pages (5, 10, 15, 20, 25, 30).

**Probability = 6 / 32 = 3/16**.

6. **Count the Pages with Sums Divisible by 2 or 3 or 5**:
We can use the principle of inclusion-exclusion:
- Let A be the set of pages where the sum is divisible by 2 (16 pages).
- Let B be the set of pages where the sum is divisible by 3 (10 pages).
- Let C be the set of pages where the sum is divisible by 5 (6 pages).
- We need to find |A ∪ B ∪ C|.

Using inclusion-exclusion:
\[ |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| \]
- |A ∩ B| (divisible by 6): 6 pages (6, 12, 18, 24, 30).
- |A ∩ C| (divisible by 10): 3 pages (10, 20, 30).
- |B ∩ C| (divisible by 15): 2 pages (15, 30).
- |A ∩ B ∩ C| (divisible by 30): 1 page (30).

Plugging in the values:
\[ |A ∪ B ∪ C| = 16 + 10 + 6 - 6 - 3 - 2 + 1 = 22 \]
**Probability = 22 / 32 = 11/16**.

7. **Count the Pages with Sums Divisible by 2, 3, and 5**:
This is the same as finding the pages where the sum is divisible by 30.
- Only 1 page (30) meets this criterion.

**Probability = 1 / 32**.

### Final Answers:
(i) Probability divisible by 2: **1/2**
(ii) Probability divisible by 3: **5/16**
(iii) Probability divisible by 5: **3/16**
(iv) Probability divisible by 2 or 3 or 5: **11/16**
(v) Probability divisible by 2, 3, and 5: **1/32**