Out of 1000 boys in a school, 300 played cricket, 380 played hockey and 420 played basketball. Of the total, 120 played both basketball and hockey,100 played cricket and basketball, 70 played cricket and hockey and 56 played all the three games. If the probability of the number of boys who did not play any game is k, then 200k is equal to

Answer

To solve the problem, we will use the principle of inclusion-exclusion along with the given data about the boys playing different sports.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Total number of boys = 1000
- Boys playing cricket (C) = 300
- Boys playing hockey (H) = 380
- Boys playing basketball (B) = 420
- Boys playing both basketball and hockey (H ∩ B) = 120
- Boys playing both cricket and basketball (C ∩ B) = 100
- Boys playing both cricket and hockey (C ∩ H) = 70
- Boys playing all three games (C ∩ H ∩ B) = 56

2. **Use the Inclusion-Exclusion Principle**:
To find the total number of boys who played at least one game, we can use the formula:
\[ |C \cup H \cup B| = |C| + |H| + |B| - |C \cap H| - |H \cap B| - |C \cap B| + |C \cap H \cap B| \]

Substituting the values:
\[ |C \cup H \cup B| = 300 + 380 + 420 - 70 - 120 - 100 + 56 \]

3. **Calculate the Total**:
\[ |C \cup H \cup B| = 300 + 380 + 420 - 70 - 120 - 100 + 56 \]
\[ = 300 + 380 + 420 - 70 - 120 - 100 + 56 \]
\[ = 300 + 380 + 420 = 1100 \]
\[ = 1100 - 70 - 120 - 100 + 56 \]
\[ = 1100 - 290 + 56 = 866 \]

4. **Find the Number of Boys Who Did Not Play Any Game**:
\[ \text{Boys who did not play any game} = \text{Total boys} - |C \cup H \cup B| \]
\[ = 1000 - 866 = 134 \]

5. **Calculate the Probability (k)**:
The probability \( k \) of boys who did not play any game is given by:
\[ k = \frac{\text{Number of boys who did not play any game}}{\text{Total number of boys}} = \frac{134}{1000} \]

6. **Calculate \( 200k \)**:
\[ 200k = 200 \times \frac{134}{1000} = \frac{26800}{1000} = 26.8 \]

### Final Answer:
\[ 200k = 26.8 \]