Question

Fifty college teachers are surveyed as to their possession of colour TV, VCR and tape recorder. Of them, 22 own colour TV, 15 own VCR and 14 own tape recorders. Nine of these college teachers own exactly two items out of colour TV, VCR and tape recorder; and, one college teacher owns all three. How many of the 50 teachers own none of the three, colour TV, VCR or tape recorder?

• A. 4
• B. 9
• C. 10
• D. 11

Hint:

When given “exactly two” counts in set problems, first convert them into pairwise intersection counts by adding back those with all three items before applying the inclusion–exclusion principle.

Answer 1

The Correct Option is B

Step 1: Define sets and given data Let: $C =$ set of teachers who own a Colour TV, $|C| = 22$
$V =$ set of teachers who own a VCR, $|V| = 15$
$T =$ set of teachers who own a Tape Recorder, $|T| = 14$
Number of teachers who own exactly two items = $9$
Number of teachers who own all three = $1$
Total teachers = $50$ Step 2: Understanding “exactly two” data If $x = |C \cap V|$, $y = |V \cap T|$, $z = |T \cap C|$, then: - $x-1$ = number owning $C$ and $V$ only (subtract those who also have $T$)
- $y-1$ = number owning $V$ and $T$ only
- $z-1$ = number owning $T$ and $C$ only Given: $(x-1) + (y-1) + (z-1) = 9$ $\Rightarrow x + y + z - 3 = 9$ $\Rightarrow x + y + z = 12$ Step 3: Inclusion–exclusion principle for at least one item The total owning at least one = \[ |C \cup V \cup T| = |C| + |V| + |T| - (x + y + z) + |C \cap V \cap T| \] Substitute values: \[ |C \cup V \cup T| = 22 + 15 + 14 - 12 + 1 \] \[ = 51 - 12 + 1 = 40 \] Step 4: Find those with none Number owning none = Total - At least one
$= 50 - 40 = 9$ \[ \boxed{9} \]