Question

In a survey of $500$ TV viewers: $285$ watch football (F), $195$ hockey (H), $115$ basketball (B); $45$ watch F&B, $70$ watch F&H, $50$ watch H&B, and $50$ watch none. How many watch exactly one of the three games? 

• A. 325
• B. 405
• C. 310
• D. 372

Hint:

For "exactly one", compute each group as:  
\[\text{Only }F = F - (F \cap H + F \cap B) + t\]  and sum; find \(t\) via inclusion–exclusion.
 

Answer 1

The Correct Option is A

Step 1: At least one game: \( 500-50=450 \). Using inclusion-exclusion with triple overlap \( t \): \[ 450=285+195+115-(45+70+50)+t\ \Rightarrow\ t=20. \]

Step 2: Only-football \( =285-70-45+20=190 \), only-hockey \( =195-70-50+20=95 \), only-basketball \( =115-45-50+20=40 \).

Step 3: Adding these, exactly one \( =190+95+40=325 \).
\[ \boxed{325} \]