List the sample space of equally likely outcomes and calculate the theoretical probability of an event using $P(E)=\dfrac{n(E)}{n(S)}$, the number of favourable outcomes divided by the total number of outcomes

Part (a): the sample space is $\{1, 2, 3, 4, 5, 6\}$ with $n(S) = 6$. The numbers that are even AND greater than $3$ are $4$ and $6$, so $n(E) = 2$ and $P(E) = \tfrac{2}{6} = \tfrac{1}{3}$.

Part (b): the factors of $6$ are $1, 2, 3, 6$, so $n(E) = 4$ and $P(E) = \tfrac{4}{6} = \tfrac{2}{3}$.

Markers award one mark for a correct sample space or count of favourable outcomes, and one mark for each correct probability written in simplest form. A candidate who lists $4, 6$ but forgets to simplify $\tfrac{2}{6}$ still earns the count mark; the simplest-form fraction secures full marks.

Part (a): each die has $6$ outcomes, and pairing each of the $6$ on the first die with each of the $6$ on the second gives $6 \times 6 = 36$ equally likely ordered outcomes. One mark for the clear multiplication reason.

Part (b): the ordered pairs summing to $8$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$, which is $5$ pairs, so $P(\text{sum} = 8) = \tfrac{5}{36}$. One mark.

Part (c): "at least $10$" means a sum of $10$, $11$ or $12$. The pairs are $(4,6), (5,5), (6,4)$ for $10$, $(5,6), (6,5)$ for $11$, and $(6,6)$ for $12$, a total of $6$ pairs, so $P(\text{sum} \ge 10) = \tfrac{6}{36} = \tfrac{1}{6}$. Markers reward correctly interpreting "at least" as $\ge 10$ and counting all six pairs; a candidate who omits the $(4,4)$-style pairs or miscounts loses the final mark.

Part (a): $n(S) = 12$ and there are $5$ milk chocolates, so $P(\text{milk}) = \tfrac{5}{12}$. One mark.

Part (b): the chocolates that are not white are the $4$ dark and $5$ milk, that is $9$, so $P(\text{not white}) = \tfrac{9}{12} = \tfrac{3}{4}$. Equivalently, $3$ of $12$ are white so $P(\text{not white}) = 1 - \tfrac{3}{12} = \tfrac{3}{4}$. One mark, either route accepted.

Part (c): $\tfrac{3}{4} = 0.75$. One mark. Markers accept the complement method in (b) and reward a simplified fraction; the decimal mark requires the exact value $0.75$.

Part (a) - list every possible outcome. A standard die has faces numbered $1$ to $6$, and each is a single possible result of one roll. Listing them as a set:

Part (b) - count the outcomes. There are $6$ faces, and each is equally likely on a fair die, so there are $n(S) = 6$ equally likely outcomes. (Check: the faces are distinct and nothing is left out or repeated, so the count of $6$ matches the number of items listed.)

Set up the sample space. The outcomes are $S = \{1, 2, 3, 4, 5, 6\}$, so $n(S) = 6$ equally likely outcomes.

Part (a) - probability of a $4$. Only one face shows a $4$, so $n(E) = 1$:

Part (b) - probability of an even number. The even faces are $2, 4, 6$, so $n(E) = 3$:

(Check: $\tfrac{1}{2}$ matches the intuition that exactly half the faces are even.)

Find the total number of outcomes. Each marble is one equally likely outcome, so

Part (a) - probability of blue. There are $3$ blue marbles, so $n(E) = 3$:

Part (b) - probability of not green. The marbles that are not green are the $5$ red and $3$ blue, giving $n(E) = 8$:

(Check: there are $2$ green out of $10$, that is $\tfrac{2}{10}=\tfrac{1}{5}$, and $\tfrac{1}{5}+\tfrac{4}{5}=1$, which is the whole bag.)

Set up the sample space. A drawn card is equally likely to be any of the $52$ cards, so $n(S) = 52$.

Part (a) - a heart. There are $13$ hearts, so $n(E) = 13$:

Part (b) - a king. There are $4$ kings, one per suit, so $n(E) = 4$:

Part (c) - a red face card. The face cards are jack, queen and king, that is $3$ per suit. The red suits are hearts and diamonds, so the red face cards number $3 \times 2 = 6$, giving $n(E) = 6$:

(Check: each fraction is below $1$ and the heart probability $\tfrac14$ matches there being four equal suits.)

Part (a) - total outcomes. The red die can show any of $6$ numbers, and for each of those the blue die can show any of $6$, so by listing the ordered pairs in a $6 \times 6$ grid:

Part (b) - a double. The doubles are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$, which is $n(E) = 6$:

Part (c) - a sum of $7$. The ordered pairs adding to $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, which is $n(E) = 6$:

(Check: both events have $6$ favourable pairs out of $36$, and a sum of $7$ is well known to be the most likely total, consistent with $\tfrac16$.)

Set up the sample space for one spinner. The sectors are $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$, all equally likely, so $n(S) = 8$.

Part (a) - a multiple of $3$. The multiples of $3$ in the list are $3$ and $6$, so $n(E) = 2$:

Part (b) - a prime number. The primes from $1$ to $8$ are $2, 3, 5, 7$ (recall $1$ is not prime), so $n(E) = 4$:

Part (c) - why $64$ outcomes. Each spin of the first spinner gives one of $8$ equally likely results, and for every one of those the second spinner independently gives one of $8$ results. Pairing each of the $8$ with each of the $8$ gives

equally likely ordered outcomes, which is the same multiplication that builds the $6 \times 6$ grid for two dice. (Check: $\tfrac14$ and $\tfrac12$ are both sensible, since multiples of $3$ are rarer than primes in this range.)

Set up the sample space. Any of the $200$ tickets is equally likely, so $n(S) = 200$.

Part (a) - a multiple of $10$. The multiples of $10$ from $1$ to $200$ are $10, 20, \ldots, 200$, which is $200 \div 10 = 20$ tickets, so $n(E) = 20$:

Part (b) - less than $50$. The numbers $1, 2, \ldots, 49$ are less than $50$, which is $49$ tickets, so $n(E) = 49$:

Part (c) - multiple of $10$ or less than $50$. List the favourable numbers without double-counting. The multiples of $10$ below $50$ are $10, 20, 30, 40$, and these are already counted among the $49$ numbers below $50$. So the new multiples of $10$ to add are $50, 60, 70, \ldots, 200$, which is $20 - 4 = 16$ extra tickets. The total favourable count is

so

Part (d) - why not just add. Adding (a) and (b) would give $\tfrac{20}{200} + \tfrac{49}{200} = \tfrac{69}{200}$, but that counts the four tickets $10, 20, 30, 40$ twice, since each is both a multiple of $10$ and less than $50$. Removing the $4$ double-counted tickets gives $69 - 4 = 65$, which matches part (c). (Check: $\tfrac{13}{40} = 0.325$, comfortably between the two separate probabilities $0.10$ and $0.245$.)