Describe the likelihood of an event using the language of chance and the probability scale from 0 to 1, distinguishing fair from biased and equally likely outcomes

Part (a): not equally likely, because the sectors are different sizes - red covers half while blue and green each cover only a quarter - so red is more likely than blue or green. The justification (unequal sector sizes) is required for the mark, not just "no".

Part (b): blue fills one quarter, so $P(\text{blue}) = \tfrac{1}{4} = 0.25$, which is below a half, so landing on blue is unlikely. Markers want both the value $0.25$ and the word "unlikely".

Part (c): red plus blue covers one half plus one quarter $= \tfrac{3}{4} = 0.75$, so red-or-blue is likely. Award the final mark for $0.75$ with the word "likely". A common loss is calling blue "even chance" by confusing it with the fifty-fifty red sector.

Part (a): fair means all six faces are equally likely, because the cube is symmetric so no face is favoured. The phrase "equally likely" is the key idea markers look for.

Part (b): the faces greater than $1$ are $2, 3, 4, 5, 6$, which is $5$ of the $6$ faces, so $P = \tfrac{5}{6} \approx 0.83$. As this is well above a half, the event is likely (almost certain). The reason - five of the six faces succeed - earns the second mark.

Part (c): only one face shows a $2$, so $P(2) = \tfrac{1}{6} \approx 0.17$, which is below a half, so it is unlikely. Again the count "one face out of six" is the required reason.

Part (d): biased, because the outcomes are not equally likely - one face (the $6$) always comes up while the others never do. Markers reward the verdict "biased" plus the equally-likely justification; "biased" alone is not enough for full marks.

Part (a) - check every outcome. An ordinary die shows $1, 2, 3, 4, 5$ or $6$, and every one of these is less than $7$. The event always happens, so it is certain (a probability of $1$).

Part (b) - count the equally likely outcomes. A fair coin has two equally likely outcomes, a head or a tail, and exactly one of them is a head. A head happens on half of all tosses, so a head is an even chance (a probability of $0.5$). (Sanity check: "certain" sits at the right-hand end of the likelihood scale and "even chance" sits in the middle, which matches a guaranteed event and a fifty-fifty event.)

Use the likelihood scale

Probabilities run from $0$ (impossible) to $1$ (certain), with the words spread evenly across the scale.

Part (a) impossible

An impossible event never happens, so its probability is $0$.

Part (b) even chance

An even chance is fifty-fifty, halfway along the scale, so its probability is $0.5$.

Part (c) certain

A certain event always happens, so its probability is $1$. (The two left-over values, $0.25$ and $0.75$, are "unlikely" and "likely" respectively, which is worth remembering for the next question.)

Part (a) equal sectors mean equal chance. The four sectors are the same size, so the pointer has the same chance of stopping on each one. The outcomes are therefore equally likely.

Part (b) one favourable outcome out of four. Red is one of four equally likely results, so it happens on one quarter of the spins. A probability of $\tfrac{1}{4} = 0.25$ sits a quarter of the way along the scale, which is described as unlikely. (Check: four equally likely colours must share the whole certainty of $1$ between them, and $4 \times 0.25 = 1$, so each colour having a probability of $0.25$ is consistent.)

Fair means every outcome is equally likely; biased means it is not

Part (a) the die

A standard die is a symmetric cube, so each face has the same chance of landing up. The outcomes are equally likely, so the die is fair.

Part (b) the drawing pin

A drawing pin is not symmetric: its heavy flat head and light point make point-up and point-down land at different rates. The two outcomes are not equally likely, so this is biased.

Part (c) the raffle

Each ticket is equally likely to be drawn, so at the level of single tickets the draw is fair. But the people are not equally likely to win: the person holding $50$ of the $100$ tickets has a $\tfrac{50}{100} = 0.5$ chance, far more than someone with one ticket. So while the draw itself is fair, the chances between people are biased toward the big buyer. (The key test in every part is the same: ask whether the outcomes you are comparing are equally likely.)

There are six equally likely outcomes. The faces are $1, 2, 3, 4, 5, 6$, so each probability is the count of favourable faces divided by $6$.

Part (a) an even number. The even faces are $2, 4, 6$, which is $3$ faces:

A probability of $\tfrac{1}{2}$ is an even chance.

Part (b) greater than $4$. The faces greater than $4$ are $5$ and $6$, which is $2$ faces:

A probability of about $0.33$ is below a half, so this event is unlikely.

Part (c) a $7$. No face shows a $7$, so it cannot happen:

which is impossible. (Check: $\tfrac{1}{3}$ lies between $0$ and $\tfrac{1}{2}$, so "unlikely" is the right word, sitting left of the midpoint of the scale.)

Part (a) what "at random" means. Drawing at random means every marble has the same chance of being selected, so the outcomes are equally likely at the level of individual marbles. The colours are not equally likely, because there are different numbers of each.

Part (b) probability of green. There are $12$ green marbles out of $20$:

Part (c) green in words. A probability of $0.6$ is more than a half but not certain, so drawing a green marble is likely.

Part (d) blue in words. There are $3$ blue marbles out of $20$:

which is well below a half, so drawing a blue marble is unlikely.

Part (e) red compared with an even chance. There are $5$ red marbles out of $20$:

Since $0.25 < 0.5$, drawing a red marble is less likely than an even chance. (Check: the three probabilities are $0.6 + 0.15 + 0.25 = 1$, as they must be, because every draw lands on red, blue or green.)

Part (a) a fair coin. A fair coin has two equally likely outcomes, so a head has an even chance, a probability of

Part (b) compare with the prediction

For a fair coin, $200$ tosses would give roughly $200 \times 0.5 = 100$ heads. The class recorded $176$ heads, which is far more than $100$, and only $24$ tails instead of about $100$.

Part (c) fair or biased

The results are nowhere near the fifty-fifty split a fair coin produces over many tosses, so the evidence suggests the coin is biased toward heads. (One run can wander a little from the prediction, but $176$ out of $200$ is far too lopsided to be ordinary variation.)

Part (d) on the scale

The claimed probability of a head is $0.5$, an even chance, sitting at the midpoint of the scale. The observed proportion of heads is

which sits near the right-hand end, well past "likely" and close to "certain". The gap between $0.5$ and $0.88$ on the scale is the picture of the bias. (Check: $0.88 + \tfrac{24}{200} = 0.88 + 0.12 = 1$, the head and tail proportions adding to $1$ as they must.)