In a statistical investigation of 1,003 families of Calcutta, it was found that 63 families had neither a radio nor a T.V, 794 families had a radio and 187 had a T.V. The number of families in that group having both a radio and a T.V is (a) `36` (b)`41` (c) `32` (d) None of these

Options

A. 36
B. 41
C. 32
D. None of these

Explanation

To solve the problem step by step, we can use the principle of inclusion-exclusion in set theory. Let's denote:

- \( N \) = Total number of families = 1003
- \( A \) = Families with a TV = 187
- \( B \) = Families with a radio = 794
- \( N(A \cap B) \) = Families with both a TV and a radio
- Families with neither a TV nor a radio = 63

### Step 1: Calculate the number of families that have either a TV or a radio (or both).

Since 63 families had neither a radio nor a TV, we can find the number of families that had either a TV or a radio (or both) by subtracting the families with neither from the total number of families.

\[ N(A \cup B) = N - \text{(Families with neither)} = 1003 - 63 = 940 \]

### Step 2: Use the inclusion-exclusion principle.

According to the inclusion-exclusion principle, the number of families that have either a TV or a radio (or both) can be expressed as:

\[ N(A \cup B) = N(A) + N(B) - N(A \cap B) \]

Substituting the known values:

\[ 940 = 187 + 794 - N(A \cap B) \]

### Step 3: Simplify the equation.

Now, we can simplify the equation:

\[ 940 = 981 - N(A \cap B) \]

### Step 4: Solve for \( N(A \cap B) \).

Rearranging the equation to find \( N(A \cap B) \):

\[ N(A \cap B) = 981 - 940 \]
\[ N(A \cap B) = 41 \]

### Conclusion

Thus, the number of families that have both a radio and a TV is \( 41 \).

### Final Answer
The correct option is (b) 41.

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