1. **Problem Statement:** We have a color wheel with four colors: red, orange, yellow, and green. The wheel is spun twice, selecting two colors. Define the random variable $X$ as the number of times the color orange appears in the two spins. 2. **Goal:** Construct the probability distribution of $X$, i.e., find $P(X=0)$, $P(X=1)$, and $P(X=2)$. 3. **Total possible outcomes:** Since each spin can result in any of the 4 colors independently, the total number of outcomes is: $$4 \times 4 = 16$$ 4. **Possible values of $X$:** - $X=0$: orange appears zero times. - $X=1$: orange appears exactly once. - $X=2$: orange appears twice. 5. **Calculate $P(X=0)$:** Number of outcomes with no orange in either spin. Each spin can be red, yellow, or green (3 options), so: $$3 \times 3 = 9$$ Probability: $$P(X=0) = \frac{9}{16}$$ 6. **Calculate $P(X=1)$:** Number of outcomes with exactly one orange. This can happen in two ways: - First spin orange, second spin not orange: $1 \times 3 = 3$ - First spin not orange, second spin orange: $3 \times 1 = 3$ Total: $$3 + 3 = 6$$ Probability: $$P(X=1) = \frac{6}{16} = \frac{3}{8}$$ 7. **Calculate $P(X=2)$:** Number of outcomes with orange in both spins: $$1 \times 1 = 1$$ Probability: $$P(X=2) = \frac{1}{16}$$ 8. **Summary of the probability distribution:** | $X$ | 0 | 1 | 2 | |-----|---|---|---| | $P(X)$ | $\frac{9}{16}$ | $\frac{6}{16}$ | $\frac{1}{16}$ | 9. **Check sum:** $$\frac{9}{16} + \frac{6}{16} + \frac{1}{16} = \frac{16}{16} = 1$$ This confirms the distribution is valid. **Final answer:** $$\boxed{P(X=0) = \frac{9}{16}, \quad P(X=1) = \frac{6}{16}, \quad P(X=2) = \frac{1}{16}}$$