You're on a roll — Step 1: Define the probabilities for the first game. Let $P(W_1)$ be the probability of winning the first game. Let $P(L_1)$ be the probability of losing the first game. Given $P(W_1) = \frac{1}{4}$. The probability of losing the first game is $P(L_1) = 1 - P(W_1) = 1 - \frac{1}{4} = \frac{3}{4}$. Step 2: Determine the conditional probabilities for winning the second game. Let $P(W_2 | W_1)$ be the probability of winning the second game given that the first game was won. "If it wins, the probability of winning the next game increases by 10%." This means the probability value itself increases by $0.10$. $$P(W_2 | W_1) = P(W_1) + 0.10 = \frac{1}{4} + 0.10 = 0.25 + 0.10 = 0.35$$ Let $P(W_2 | L_1)$ be the probability of winning the second game given that the first game was lost. "If it loses, the probability of winning the next game increases by 20%." This means the probability value itself increases by $0.20$. $$P(W_2 | L_1) = P(W_1) + 0.20 = \frac{1}{4} + 0.20 = 0.25 + 0.20 = 0.45$$ Step 3: Calculate the probability that the team wins in the second round, $P(W_2)$, using the law of total probability. $$P(W_2) = P(W_2 | W_1) P(W_1) + P(W_2 | L_1) P(L_1)$$ Substitute the values: $$P(W_2) = (0.35) \times \left(\frac{1}{4}\right) + (0.45) \times \left(\frac{3}{4}\right)$$ Convert the decimals to fractions: $0.35 = \frac{35}{100} = \frac{7}{20}$ and $0.45 = \frac{45}{100} = \frac{9}{20}$. $$P(W_2) = \left(\frac{7}{20}\right