1. **Problem statement:** We have a stock price that varies uniformly between $16 and $25. We want to find the probability that the stock price is more than $21 given that it is already more than $18. 2. **Recall the uniform distribution properties:** For a uniform distribution between $a$ and $b$, the probability that the variable $X$ lies between $x_1$ and $x_2$ (where $a \leq x_1 < x_2 \leq b$) is given by: $$P(x_1 \leq X \leq x_2) = \frac{x_2 - x_1}{b - a}$$ 3. **Calculate the conditional probability:** We want $P(X > 21 \mid X > 18)$. By definition of conditional probability: $$P(X > 21 \mid X > 18) = \frac{P(X > 21 \cap X > 18)}{P(X > 18)} = \frac{P(X > 21)}{P(X > 18)}$$ 4. **Calculate each probability:** - $P(X > 21) = \frac{25 - 21}{25 - 16} = \frac{4}{9}$ - $P(X > 18) = \frac{25 - 18}{25 - 16} = \frac{7}{9}$ 5. **Compute the conditional probability:** $$P(X > 21 \mid X > 18) = \frac{\frac{4}{9}}{\frac{7}{9}} = \frac{4}{9} \times \frac{9}{7} = \frac{4}{7}$$ 6. **Final answer:** The probability that the stock price is more than 21 given it is more than 18 is: $$\boxed{\frac{4}{7}}$$