To find the probability that a ticket drawn from a set of 200 tickets numbered from 1 to 200 is divisible by 2, 3, or 5, we can use the principle of inclusion-exclusion. Here’s a step-by-step solution:
### Step 1: Define the Sets
Let:
- \( A \) = the set of numbers divisible by 2
- \( B \) = the set of numbers divisible by 3
- \( C \) = the set of numbers divisible by 5
### Step 2: Count the Elements in Each Set
1. **Count of numbers divisible by 2**:
- The numbers divisible by 2 from 1 to 200 are: 2, 4, 6, ..., 200.
- This forms an arithmetic series where:
- First term \( a = 2 \)
- Last term \( l = 200 \)
- Common difference \( d = 2 \)
- The number of terms \( n \) can be calculated using the formula:
\[
n = \frac{l - a}{d} + 1 = \frac{200 - 2}{2} + 1 = 100
\]
- Thus, \( |A| = 100 \).
2. **Count of numbers divisible by 3**:
- The numbers divisible by 3 from 1 to 200 are: 3, 6, 9, ..., 198.
- This is also an arithmetic series where:
- First term \( a = 3 \)
- Last term \( l = 198 \)
- Common difference \( d = 3 \)
- The number of terms \( n \) is:
\[
n = \frac{198 - 3}{3} + 1 = 66
\]
- Thus, \( |B| = 66 \).
3. **Count of numbers divisible by 5**:
- The numbers divisible by 5 from 1 to 200 are: 5, 10, 15, ..., 200.
- This is an arithmetic series where:
- First term \( a = 5 \)
- Last term \( l = 200 \)
- Common difference \( d = 5 \)
- The number of terms \( n \) is:
\[
n = \frac{200 - 5}{5} + 1 = 40
\]
- Thus, \( |C| = 40 \).
### Step 3: Count the Elements in the Intersections of Sets
1. **Count of numbers divisible by both 2 and 3 (i.e., divisible by 6)**:
- The numbers divisible by 6 from 1 to 200 are: 6, 12, 18, ..., 198.
- The number of terms \( n \) is:
\[
n = \frac{198 - 6}{6} + 1 = 33
\]
- Thus, \( |A \cap B| = 33 \).
2. **Count of numbers divisible by both 3 and 5 (i.e., divisible by 15)**:
- The numbers divisible by 15 from 1 to 200 are: 15, 30, 45, ..., 195.
- The number of terms \( n \) is:
\[
n = \frac{195 - 15}{15} + 1 = 13
\]
- Thus, \( |B \cap C| = 13 \).
3. **Count of numbers divisible by both 2 and 5 (i.e., divisible by 10)**:
- The numbers divisible by 10 from 1 to 200 are: 10, 20, 30, ..., 200.
- The number of terms \( n \) is:
\[
n = \frac{200 - 10}{10} + 1 = 20
\]
- Thus, \( |A \cap C| = 20 \).
4. **Count of numbers divisible by 2, 3, and 5 (i.e., divisible by 30)**:
- The numbers divisible by 30 from 1 to 200 are: 30, 60, 90, ..., 180.
- The number of terms \( n \) is:
\[
n = \frac{180 - 30}{30} + 1 = 6
\]
- Thus, \( |A \cap B \cap C| = 6 \).
### Step 4: Apply the Inclusion-Exclusion Principle
Using the formula for the union of three sets:
\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|
\]
Substituting the values:
\[
|A \cup B \cup C| = 100 + 66 + 40 - 33 - 13 - 20 + 6 = 146
\]
### Step 5: Calculate the Probability
The total number of tickets is 200. Therefore, the probability \( P \) that a ticket drawn is divisible by 2, 3, or 5 is:
\[
P = \frac{|A \cup B \cup C|}{\text{Total Outcomes}} = \frac{146}{200} = \frac{73}{100}
\]
### Final Answer
The probability that the number is divisible by 2, 3, or 5 is \( \frac{73}{100} \).
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