There are 20 tickets numbered 1 to 20. A ticket is drawn. Find the probability of drawing a ticket whose number is:
(i) a multiple of 3
(ii) a multiple of 5
(iii) a multiple of 3 or 5.

Explanation

To solve the problem step by step, we will calculate the probabilities for each part of the question.

### Step 1: Total Tickets
We know that there are 20 tickets numbered from 1 to 20. Therefore, the total number of tickets, \( N \), is:
\[ N = 20 \]

### Step 2: Probability of Drawing a Ticket that is a Multiple of 3
1. Identify the multiples of 3 within the range of 1 to 20:
- The multiples of 3 are: 3, 6, 9, 12, 15, and 18.
- Count the multiples: There are 6 multiples of 3.

2. Calculate the probability:
\[ P(\text{Multiple of 3}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{20} = \frac{3}{10} \]

### Step 3: Probability of Drawing a Ticket that is a Multiple of 5
1. Identify the multiples of 5 within the range of 1 to 20:
- The multiples of 5 are: 5, 10, 15, and 20.
- Count the multiples: There are 4 multiples of 5.

2. Calculate the probability:
\[ P(\text{Multiple of 5}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{20} = \frac{1}{5} \]

### Step 4: Probability of Drawing a Ticket that is a Multiple of 3 or 5
1. We need to find the intersection of the events (multiples of both 3 and 5):
- The least common multiple of 3 and 5 is 15.
- The only multiple of both 3 and 5 in the range is 15, so there is 1 ticket.

2. Use the formula for the probability of the union of two events:
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
where:
- \( P(E) = P(\text{Multiple of 3}) = \frac{3}{10} \)
- \( P(F) = P(\text{Multiple of 5}) = \frac{1}{5} = \frac{2}{10} \)
- \( P(E \cap F) = P(\text{Multiple of both 3 and 5}) = \frac{1}{20} \)

3. Substitute the values into the formula:
\[ P(E \cup F) = \frac{3}{10} + \frac{2}{10} - \frac{1}{20} \]
To perform the addition and subtraction, convert \(\frac{1}{20}\) to a common denominator of 20:
\[ P(E \cup F) = \frac{6}{20} + \frac{4}{20} - \frac{1}{20} = \frac{9}{20} \]

### Final Results:
- (i) The probability of drawing a ticket that is a multiple of 3 is \( \frac{3}{10} \).
- (ii) The probability of drawing a ticket that is a multiple of 5 is \( \frac{1}{5} \).
- (iii) The probability of drawing a ticket that is a multiple of 3 or 5 is \( \frac{9}{20} \).