A number is selected at random from first 200 natural numbers. Find the probability that it is divisible by either 6 or 8.

Explanation

To solve the problem of finding the probability that a number selected at random from the first 200 natural numbers is divisible by either 6 or 8, we can follow these steps:

### Step 1: Define the Events
Let:
- Event A = the event that a number is divisible by 6.
- Event B = the event that a number is divisible by 8.

### Step 2: Count the Total Outcomes
The total number of natural numbers we are considering is 200.

### Step 3: Count the Favorable Outcomes for Event A
To find the numbers divisible by 6:
- The multiples of 6 up to 200 are: 6, 12, 18, ..., 198.
- This forms an arithmetic sequence where:
- First term (a) = 6
- Common difference (d) = 6
- Last term (l) = 198

To find the number of terms (n) in this sequence, we can use the formula for the nth term of an arithmetic sequence:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 198 = 6 + (n-1)6 \]
\[ 198 - 6 = (n-1)6 \]
\[ 192 = (n-1)6 \]
\[ n-1 = \frac{192}{6} = 32 \]
\[ n = 32 + 1 = 33 \]
So, there are 33 numbers divisible by 6.

### Step 4: Count the Favorable Outcomes for Event B
To find the numbers divisible by 8:
- The multiples of 8 up to 200 are: 8, 16, 24, ..., 200.
- This forms another arithmetic sequence where:
- First term (a) = 8
- Common difference (d) = 8
- Last term (l) = 200

Using the same formula:
\[ 200 = 8 + (n-1)8 \]
\[ 200 - 8 = (n-1)8 \]
\[ 192 = (n-1)8 \]
\[ n-1 = \frac{192}{8} = 24 \]
\[ n = 24 + 1 = 25 \]
So, there are 25 numbers divisible by 8.

### Step 5: Count the Favorable Outcomes for A ∩ B (Intersection)
To find the numbers divisible by both 6 and 8, we need to find the least common multiple (LCM) of 6 and 8:
- LCM(6, 8) = 24.

The multiples of 24 up to 200 are: 24, 48, 72, ..., 192.
- This forms another arithmetic sequence where:
- First term (a) = 24
- Common difference (d) = 24
- Last term (l) = 192

Using the same formula:
\[ 192 = 24 + (n-1)24 \]
\[ 192 - 24 = (n-1)24 \]
\[ 168 = (n-1)24 \]
\[ n-1 = \frac{168}{24} = 7 \]
\[ n = 7 + 1 = 8 \]
So, there are 8 numbers divisible by both 6 and 8.

### Step 6: Calculate the Probability
Using the principle of inclusion-exclusion for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Where:
- \( P(A) = \frac{33}{200} \)
- \( P(B) = \frac{25}{200} \)
- \( P(A \cap B) = \frac{8}{200} \)

Now substituting these values:
\[ P(A \cup B) = \frac{33}{200} + \frac{25}{200} - \frac{8}{200} \]
\[ = \frac{33 + 25 - 8}{200} = \frac{50}{200} = \frac{1}{4} \]

### Final Answer
The probability that a number selected at random from the first 200 natural numbers is divisible by either 6 or 8 is \( \frac{1}{4} \).

---