Recognise that probabilities of events range from 0 to 1, identify the complement of an event and use the relationship that the probability of an event and its complement sum to 1, so the probability of 'not E' equals 1 minus the probability of E

Part (a): the bus being not late is the complement of being late, so $P(\text{not late}) = 1 - 0.18 = 0.82$. One mark for the correct complement.

Part (b): the largest value is $1$ (a certain event) and the smallest is $0$ (an impossible event), so every probability satisfies $0 \le P \le 1$. One mark for both bounds.

Markers reward the explicit subtraction from $1$ for the complement and both correct bounds. A common loss is writing $0.82$ without showing $1 - 0.18$, or giving only one of the two bounds.

Part (a): the eight equally likely outcomes range from $HHH$ to $TTT$; only $TTT$ has no heads, so $P(\text{no heads}) = \tfrac{1}{8}$. Equivalently $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$. One mark.

Part (b): "at least one head" is the complement of "no heads", so $P(\text{at least one head}) = 1 - \tfrac{1}{8} = \tfrac{7}{8}$. One mark.

Part (c): "at least one head" covers exactly one, two or three heads (seven of the eight outcomes), so the complement replaces three separate counts with the single case $TTT$. One mark for a clear reason.

Markers reward the word "complement" and the subtraction $1 - \tfrac{1}{8}$. The standard error is trying to add the cases directly and miscounting, or forgetting that "at least one" includes the all-heads case.

Part (a): $2$ of the $25$ globes are faulty, so $P(\text{faulty}) = \tfrac{2}{25} = 0.08$. One mark.

Part (b): "not faulty" is the complement, so $P(\text{not faulty}) = 1 - 0.08 = 0.92$ (equivalently $\tfrac{23}{25}$). One mark.

Part (c): a probability and its complement must add to $1$, but $0.08 + 0.82 = 0.90 \ne 1$. The error is the complement: it should be $1 - 0.08 = 0.92$, not $0.82$. One mark for identifying that the pair fails to sum to $1$ and giving the correct $0.92$.

Markers reward checking that the two probabilities sum to $1$ as the test for the error, and the corrected value $0.92$.

Count the sample space. There are $3 + 5 + 2 = 10$ marbles in all.

Part (a) - probability of blue. Five of the $10$ marbles are blue, so

Part (b) - probability of not blue, by the complement. "Not blue" is the complement of "blue", so subtract from $1$:

(Check by counting directly: $3 + 2 = 5$ marbles are not blue, and $\tfrac{5}{10} = \tfrac{1}{2}$, which agrees.)

Recognise the complement. "It does not rain" is the complement of "it rains", and the two probabilities must add to $1$.

Subtract from $1$.

so the probability of no rain is $0.65$. (Sanity check: $0.35 + 0.65 = 1$, as a probability and its complement must.)

List the sample space. The faces are $1, 2, 3, 4, 5, 6$, so $n(S) = 6$ equally likely outcomes.

Part (a) - greater than $4$. The favourable faces are $5$ and $6$, so

Part (b) - not greater than $4$, by the complement. Subtract from $1$:

(Check directly: the faces $1, 2, 3, 4$ are not greater than $4$, giving $\tfrac{4}{6} = \tfrac{2}{3}$, which matches.)

Part (a) - the total must be $1$. For one spin exactly one colour occurs, so the probabilities of all the outcomes must add to $1$. The three given values total

which is less than $1$, so the remaining probability of $0.35$ must belong to green; the spinner cannot be described by the first three alone.

Part (b) - find $P(\text{green})$ as the complement of the other three. Green is "none of red, blue or yellow", so

(Check: $0.2 + 0.3 + 0.15 + 0.35 = 1$, so the four probabilities form a complete set.)

Part (a) - the sample space. Writing $H$ for heads and $T$ for tails, the four equally likely outcomes are

so $n(S) = 4$.

Part (b) - no heads. The only outcome with no heads is $TT$, so

Part (c) - at least one head, by the complement. "At least one head" is the complement of "no heads", so

(Check by listing: $HH, HT, TH$ each contain a head, so $\tfrac{3}{4}$ agrees - and the complement saved counting all of them.)

Part (a) - none faulty. Each item is not faulty with probability $1 - 0.05 = 0.95$, and the three items are independent, so multiply:

Part (b) - at least one faulty, by the complement. "At least one faulty" is the complement of "none faulty", so

Correct to four decimal places, $P(\text{at least one faulty}) \approx 0.1426$.

Part (c) - why the complement wins. "At least one" covers three separate cases (exactly one, exactly two, or exactly three faulty), each needing its own calculation, then a sum. The complement "none faulty" is a single case, so $1 - P(\text{none})$ replaces three calculations with one. (Check: the direct cases would still total $0.142625$, but the complement reaches it in one line.)

Part (a) - Mia wins. Mia holds $8$ of the $200$ equally likely tickets, so

Part (b) - neither Mia nor Sam wins. Together they hold $8 + 5 = 13$ tickets, so a win for one of them has probability $\tfrac{13}{200}$. "Neither wins" is the complement of "Mia or Sam wins":

Part (c) - someone other than Mia wins. This is the complement of "Mia wins":

Part (d) - the error. A probability and its complement must add to exactly $1$. The friend's pair gives $0.04 + 0.94 = 0.98$, not $1$, so it is impossible. The correct complement of $0.04$ is $1 - 0.04 = 0.96$, matching part (c).