17. Use Math Tools Each of the spinners at the right is spun once. Use a tree diagram to find the following probabilities. a. P(at least one 2) b. P(at least one 3)

We assume that each spinner has three equally likely outcomes: 1, 2, and 3. When each spinner is spun once, the total number of outcomes is \[ 3 \times 3 = 9. \] A tree diagram for the two spins is constructed as follows. Let the first spin be represented by the first branching and the second spin by the subsequent branches: - First spin: outcomes 1, 2, 3. - If the first spinner shows 1, the second spinner can show 1, 2, or 3: \((1,1),\ (1,2),\ (1,3)\). - If the first spinner shows 2, the second spinner can show 1, 2, or 3: \((2,1),\ (2,2),\ (2,3)\). - If the first spinner shows 3, the second spinner can show 1, 2, or 3: \((3,1),\ (3,2),\ (3,3)\). Thus, the sample space is \[ \{(1,1),\ (1,2),\ (1,3),\ (2,1),\ (2,2),\ (2,3),\ (3,1),\ (3,2),\ (3,3)\}. \] --- **(a) Probability of at least one 2** We need to count all outcomes that include at least one 2. These outcomes are: - Outcomes with a 2 from the first spin: \((2,1),\ (2,2),\ (2,3)\) - Outcomes with a 2 from the second spin (but not double-counting those already listed): \((1,2),\ (3,2)\) The total number of outcomes with at least one 2 is \[ 3 + 2 = 5. \] Thus, the probability is \[ P(\text{at least one }2) = \frac{5}{9}. \] --- **(b) Probability of at least one 3** Similarly, we list the outcomes that include at least one 3: - Outcomes with a 3 from the first spinner: \((3,1),\ (3,2),\ (3,3)\) - Outcomes with a 3 from the second spinner (avoiding double-counting): \((1,3),\ (2,3)\) The total number of outcomes with at least one 3 is \[ 3 + 2 = 5. \] Thus, the probability is \[ P(\text{at least one }3) = \frac{5}{9}. \] --- Final answers: (a) \( \frac{5}{9} \) (b) \( \frac{5}{9} \)