1. **Stating the problem:** We want to find the probability that Sean does **not** win either the chess game or the pool game. 2. **Understanding the problem:** - Chess outcomes: Win with probability $\frac{2}{10}$, Lose with $\frac{7}{10}$, Draw with $\frac{1}{10}$. - Pool outcomes (independent of chess result): Win with probability $\frac{5}{7}$, Lose with $\frac{2}{7}$. 3. **Key rule:** The probability of independent events both happening is the product of their probabilities. 4. **Calculate probability of not winning chess:** Not winning chess means either Lose or Draw: $$P(\text{not win chess}) = \frac{7}{10} + \frac{1}{10} = \frac{8}{10} = \frac{4}{5}$$ 5. **Calculate probability of not winning pool:** $$P(\text{not win pool}) = \frac{2}{7}$$ 6. **Calculate probability of not winning either game:** Since chess and pool outcomes are independent, $$P(\text{not win either}) = P(\text{not win chess}) \times P(\text{not win pool}) = \frac{4}{5} \times \frac{2}{7}$$ 7. **Simplify the product:** $$\frac{4}{5} \times \frac{2}{7} = \frac{4 \times 2}{5 \times 7} = \frac{8}{35}$$ **Final answer:** $$\boxed{\frac{8}{35}}$$ This means there is an $\frac{8}{35}$ chance Sean does not win either game.