(i)A survey of 500 TV viewers produced the following information: 280 watch football, 150 watch hockey, 200 watch basketball, 70 watch football and basketball, 90 watch football and hockey, 80 watch hockey and basketball and 50 watch all the three games.Then,
(i) How many viewers watch atleast one of three games?

Answer

To solve the problem of how many viewers watch at least one of the three games (football, hockey, or basketball), we can use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Define the Sets:**
- Let \( F \) be the set of viewers who watch football.
- Let \( H \) be the set of viewers who watch hockey.
- Let \( B \) be the set of viewers who watch basketball.

From the problem, we have:
- \( |F| = 280 \) (viewers who watch football)
- \( |H| = 150 \) (viewers who watch hockey)
- \( |B| = 200 \) (viewers who watch basketball)
- \( |F \cap B| = 70 \) (viewers who watch both football and basketball)
- \( |F \cap H| = 90 \) (viewers who watch both football and hockey)
- \( |H \cap B| = 80 \) (viewers who watch both hockey and basketball)
- \( |F \cap H \cap B| = 50 \) (viewers who watch all three games)

2. **Apply the Inclusion-Exclusion Principle:**
The formula for the union of three sets is given by:
\[ |F \cup H \cup B| = |F| + |H| + |B| - |F \cap H| - |F \cap B| - |H \cap B| + |F \cap H \cap B| \]

3. **Substitute the Values:**
Now, substituting the values we have:
\[ |F \cup H \cup B| = 280 + 150 + 200 - 90 - 70 - 80 + 50 \]

4. **Calculate Step-by-Step:**
- First, calculate the sum of individual viewers:
\[ 280 + 150 + 200 = 630 \]
- Next, calculate the sum of the intersections of two sets:
\[ 90 + 70 + 80 = 240 \]
- Now, substitute these values into the equation:
\[ |F \cup H \cup B| = 630 - 240 + 50 \]
- Simplify this:
\[ |F \cup H \cup B| = 630 - 240 = 390 \]
\[ |F \cup H \cup B| = 390 + 50 = 440 \]

5. **Final Result:**
Therefore, the number of viewers who watch at least one of the three games is \( 440 \).