1. **State the problem:** Casey plays two games of football. The probability that Casey scores a goal in each game is 0.3. 2. **Complete the tree diagram probabilities:** - For the first game: - Scores: $P(\text{Scores 1st}) = 0.3$ - Does not score: $P(\text{No Score 1st}) = 1 - 0.3 = 0.7$ - For the second game (given the outcome of first): - If Casey scores in first game: - Scores: $0.3$ - Does not score: $0.7$ - If Casey does not score in first game: - Scores: $0.3$ - Does not score: $0.7$ 3. **Calculate the probability Casey scores in only one game:** The possible ways are: - Scores first game and does not score second game: $$P(\text{Score 1st and No Score 2nd})=0.3 \times 0.7 = 0.21$$ - Does not score first game and scores second game: $$P(\text{No Score 1st and Score 2nd})=0.7 \times 0.3 = 0.21$$ 4. **Add these probabilities to get the total:** $$P(\text{scores exactly one game}) = 0.21 + 0.21 = 0.42$$ **Final answer:** The probability that Casey scores in only one of the games is $0.42$.