Drill 1: $p(\text{red}) \approx \tfrac{22}{50} = 0.44$, $p(\text{blue}) \approx \tfrac{18}{50} = 0.36$, $p(\text{green}) \approx \tfrac{10}{50} = 0.20$. Sum $= 0.44 + 0.36 + 0.20 = 1.00$ ✓
Drill 2: $0.2 + 0.35 + a + 0.15 = 1$, so $a = 0.30$. It must be non-negative because a probability can never be less than 0, and a negative value would make the list invalid regardless of the sum.
Drill 3: $p(0) = \tfrac{1}{8} = 0.125$, $p(1) = \tfrac{3}{8} = 0.375$, $p(2) = \tfrac{3}{8} = 0.375$, $p(3) = \tfrac{1}{8} = 0.125$. There are 8 equally likely sequences, and 3 of them contain exactly one head.
Drill 4: Each trial produces exactly one of the possible values, so the frequencies account for all $n$ trials and total $n$. Dividing every frequency by $n$ therefore gives numbers totalling $\tfrac{n}{n} = 1$.
Drill 5: $k(1 + 4 + 9) = 1$, so $14k = 1$ and $k = \tfrac{1}{14}$. The bars have heights $\tfrac{1}{14}$, $\tfrac{4}{14}$ and $\tfrac{9}{14}$, increasing sharply, so the sketch rises steeply from left to right.
Q1 (3 marks): (a) $p(1) \approx \tfrac{88}{400} = 0.22$, $p(2) \approx \tfrac{112}{400} = 0.28$, $p(3) \approx \tfrac{96}{400} = 0.24$, $p(4) \approx \tfrac{104}{400} = 0.26$ [1]. (b) Every spin lands on exactly one of the four numbers, so the four frequencies account for all 400 spins and sum to 400; dividing each by 400 gives values summing to $\tfrac{400}{400} = 1$ [1]. (c) A fair spinner gives $0.25$ for each. The largest departure is $0.03$ (values 1 and 2), which is small relative to the variation expected in 400 spins, so the data do not give strong evidence of bias [1].
Q2 (3 marks): (a) $X \in \{2, 3, 4, 5, 6, 7, 8\}$ [0.5]. (b) With 16 equally likely ordered outcomes: $p(2) = \tfrac{1}{16}$, $p(3) = \tfrac{2}{16}$, $p(4) = \tfrac{3}{16}$, $p(5) = \tfrac{4}{16}$, $p(6) = \tfrac{3}{16}$, $p(7) = \tfrac{2}{16}$, $p(8) = \tfrac{1}{16}$ [1.5]. (c) Every $p(x)$ lies in $[0, 1]$, and the sum is $\tfrac{1 + 2 + 3 + 4 + 3 + 2 + 1}{16} = \tfrac{16}{16} = 1$ ✓. The most likely value is $X = 5$, with probability $\tfrac{4}{16} = 0.25$ [1].
Q3 (3 marks): (a) Each value lies in $[0, 1]$, and $0.25 + 0.40 + 0.20 + 0.15 = 1.00$, so it is a valid distribution [1]. (b) First, the sample of 20 is very small, so each estimate carries substantial uncertainty; a single different response shifts a probability by $0.05$. Second, the sample is not representative of the school: one class is not a random sample of all students, and classmates may share year group, and possibly family characteristics, that relate to sibling numbers. A third acceptable reason: the distribution is truncated at 3, so any student with 4 or more siblings has been excluded or misrecorded [1]. (c) Take a larger random sample drawn from across the whole school rather than one class. Increasing $n$ reduces the variability of each relative frequency, and randomising across year groups removes the selection bias, so the estimates converge on the true school-wide probabilities [1].