In a class of 80 students, 32 play chess, 20 play football and 8 play both…

Step 1. The two probabilities. $P(\text{chess}) = \dfrac{32}{80} = 0.4$ and $P(\text{football}) = \dfrac{20}{80} = 0.25$.

Step 2. The product. $0.4 \times 0.25 = 0.1$.

Step 3. The overlap. $P(\text{chess} \cap \text{football}) = \dfrac{8}{80} = 0.1$. It equals the product, so the events are independent.

Step 4. Chess given football. $P(\text{chess} \mid \text{football}) = \dfrac{8}{20} = 0.4$, which is the same as $P(\text{chess})$.