To find the probability that a randomly picked ball from a box containing 200 balls (numbered from 1 to 200) has a number that is divisible by either 3 or 5, we can follow these steps:
### Step 1: Determine the total number of balls
The total number of balls in the box is 200.
### Step 2: Find the count of numbers divisible by 3
To find how many numbers between 1 and 200 are divisible by 3:
- The first number divisible by 3 is 3.
- The last number divisible by 3 within 200 is 198 (since 3 × 66 = 198).
- The sequence of numbers divisible by 3 is: 3, 6, 9, ..., 198.
Using the formula for the nth term of an arithmetic series:
\[ a_n = a + (n-1) \cdot d \]
where:
- \( a = 3 \) (the first term),
- \( d = 3 \) (the common difference),
- \( a_n = 198 \) (the last term).
Setting up the equation:
\[ 198 = 3 + (n-1) \cdot 3 \]
\[ 198 - 3 = (n-1) \cdot 3 \]
\[ 195 = (n-1) \cdot 3 \]
\[ n-1 = 65 \]
\[ n = 66 \]
So, there are 66 numbers divisible by 3.
### Step 3: Find the count of numbers divisible by 5
To find how many numbers between 1 and 200 are divisible by 5:
- The first number divisible by 5 is 5.
- The last number divisible by 5 within 200 is 200 (since 5 × 40 = 200).
- The sequence of numbers divisible by 5 is: 5, 10, 15, ..., 200.
Using the same formula:
\[ a_n = a + (n-1) \cdot d \]
where:
- \( a = 5 \),
- \( d = 5 \),
- \( a_n = 200 \).
Setting up the equation:
\[ 200 = 5 + (n-1) \cdot 5 \]
\[ 200 - 5 = (n-1) \cdot 5 \]
\[ 195 = (n-1) \cdot 5 \]
\[ n-1 = 39 \]
\[ n = 40 \]
So, there are 40 numbers divisible by 5.
### Step 4: Find the count of numbers divisible by both 3 and 5 (i.e., divisible by 15)
To find how many numbers between 1 and 200 are divisible by 15:
- The first number divisible by 15 is 15.
- The last number divisible by 15 within 200 is 195 (since 15 × 13 = 195).
- The sequence of numbers divisible by 15 is: 15, 30, 45, ..., 195.
Using the formula:
\[ a_n = a + (n-1) \cdot d \]
where:
- \( a = 15 \),
- \( d = 15 \),
- \( a_n = 195 \).
Setting up the equation:
\[ 195 = 15 + (n-1) \cdot 15 \]
\[ 195 - 15 = (n-1) \cdot 15 \]
\[ 180 = (n-1) \cdot 15 \]
\[ n-1 = 12 \]
\[ n = 13 \]
So, there are 13 numbers divisible by both 3 and 5.
### Step 5: Apply the principle of inclusion-exclusion
To find the total count of numbers divisible by either 3 or 5:
\[ \text{Count} = (\text{Count divisible by 3}) + (\text{Count divisible by 5}) - (\text{Count divisible by both 3 and 5}) \]
\[ \text{Count} = 66 + 40 - 13 = 93 \]
### Step 6: Calculate the probability
The probability \( P \) that a randomly picked ball has a number divisible by either 3 or 5 is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} \]
\[ P = \frac{93}{200} \]
### Conclusion
The probability that the ball picked has a number which is divisible by either 3 or 5 is \( \frac{93}{200} \).
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