1. **Problem Statement:** We have a uniform distribution for the average number of donuts eaten by a nine-year-old child per month, ranging from 1 to 6 donuts inclusive. We want to find: (i) The probability that a child eats less than 4 donuts. (ii) The probability that a child eats between 2 and 5 donuts, given that the child eats more than 1.5 donuts. 2. **Formula and Important Rules:** For a continuous uniform distribution on the interval $[a,b]$, the probability density function (pdf) is: $$f(x) = \frac{1}{b - a}$$ The probability that $X$ lies between $x_1$ and $x_2$ (where $a \leq x_1 < x_2 \leq b$) is: $$P(x_1 \leq X \leq x_2) = \frac{x_2 - x_1}{b - a}$$ For conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$. 3. **Given:** - $a = 1$ - $b = 6$ 4. **Calculations:** (i) Probability that $X < 4$: $$P(X < 4) = P(1 \leq X < 4) = \frac{4 - 1}{6 - 1} = \frac{3}{5} = 0.6$$ (ii) Probability that $2 \leq X \leq 5$ given $X > 1.5$: First, find $P(2 \leq X \leq 5)$: $$P(2 \leq X \leq 5) = \frac{5 - 2}{6 - 1} = \frac{3}{5} = 0.6$$ Next, find $P(X > 1.5)$: $$P(X > 1.5) = P(1.5 < X \leq 6) = \frac{6 - 1.5}{6 - 1} = \frac{4.5}{5} = 0.9$$ Now, the conditional probability: $$P(2 \leq X \leq 5 \mid X > 1.5) = \frac{P(2 \leq X \leq 5)}{P(X > 1.5)} = \frac{0.6}{0.9} = \frac{2}{3} \approx 0.6667$$ 5. **Final Answers:** - (i) $P(X < 4) = 0.6$ - (ii) $P(2 \leq X \leq 5 \mid X > 1.5) = \frac{2}{3} \approx 0.6667$ These results mean there is a 60% chance a child eats less than 4 donuts, and about a 66.67% chance a child eats between 2 and 5 donuts given they eat more than 1.5 donuts.