Question

In a statistical investigation of 1003 families of Calcutta, it was found that 63 families have neither a radio nor a TV, 794 families have a radio, and 187 have a TV. The number of families having both a radio and a TV is:

• A. \( 32 \)
• B. \( 41 \)
• C. None of these
• D. None of these

Hint:

Use set operations to solve logical counting problems.

Answer 1

The Correct Option is B

Using set theory: \[ n(R) = 794, \quad n(T) = 187, \quad n(R \cup T)' = 63 \] \[ n(Total) = n(R \cup T) + n(R \cup T)' \Rightarrow 1003 = n(R \cup T) + 63 \] \[ n(R \cup T) = 940 \] Using formula: \[ n(R \cup T) = n(R) + n(T) - n(R \cap T) \] \[ 940 = 794 + 187 - n(R \cap T) \] \[ n(R \cap T) = 41 \]

Answer 2

Step 1: Defining the sets
Let \( R \) be the set of families having a radio, and \( T \) be the set of families having a TV.

We are given the following information: - Total number of families = 1003 - Families with neither a radio nor a TV = 63 - Families with a radio = 794 - Families with a TV = 187

Step 2: Using the principle of inclusion-exclusion
We can use the principle of inclusion-exclusion to find the number of families that have both a radio and a TV.
The formula for the principle of inclusion-exclusion is: \[ |R \cup T| = |R| + |T| - |R \cap T| \] where: - \( |R \cup T| \) is the total number of families that have either a radio or a TV (or both). - \( |R| \) is the number of families with a radio. - \( |T| \) is the number of families with a TV. - \( |R \cap T| \) is the number of families that have both a radio and a TV.

Step 3: Substituting the known values
The total number of families that have either a radio or a TV (or both) is: \[ |R \cup T| = 1003 - 63 = 940 \] Now, substitute the values into the inclusion-exclusion formula: \[ 940 = 794 + 187 - |R \cap T| \] \[ 940 = 981 - |R \cap T| \] \[ |R \cap T| = 981 - 940 = 41 \] So, 41 families have both a radio and a TV. However, since there seems to be an error in calculation, let's recheck and evaluate properly. In fact, we arrived earlier