Probability of a Single Event

The total number of balls in the bag is 3 + 7 = 10. Probability is calculated as the number of favourable outcomes divided by the total number of equally likely outcomes, so P(red) = 3/10. The denominator must be 10 — the total — not 7.

The value \( \frac{3}{7} \) would mean “3 reds out of 7 blues” — but we are not choosing from just the blue balls. We are choosing from all 10 balls. A useful check: if there were 3 red and 3 blue (6 total), the probability of red should be \( \frac{3}{6} = \frac{1}{2} \), not \( \frac{3}{3} = 1 \). The “divide by the other colour” method gives absurd results.

The probability is only \( \frac{1}{3} \) if all three sections are equal in size. If the red section takes up half the spinner and blue and green share the other half, P(red) = \( \frac{1}{2} \), not \( \frac{1}{3} \). The number of colours on a spinner does not determine the probability — the sizes of the sections do.

An analogy: imagine a spinner where red covers 99% of the area and blue and green are tiny slivers. Nobody would say there’s a \( \frac{1}{3} \) chance of each. Probability depends on how much of the total area each section occupies, not simply on how many differently-coloured sections exist.

These are three different ways of writing the same number. Converting: \( \frac{1}{4} = 1 \div 4 = 0.25 \), and \( 0.25 \times 100 = 25\% \). They all mean “one quarter of the time” or equivalently “25 times out of every 100.” Probability can be expressed in any of these forms — it does not change the likelihood of the event.

A common confusion is thinking fractions, decimals, and percentages are somehow different “types” of probability. But just as £0.50, 50p, and “half a pound” all describe the same amount of money, these three notations all describe the same position on the probability scale.

An event either happens or it doesn’t — those are the only two possibilities, and together they cover every possible outcome. So P(event happens) + P(event does not happen) = 1. If P(winning) = \( \frac{3}{5} \), then P(not winning) = \( 1 – \frac{3}{5} = \frac{2}{5} \).

You don’t need extra information to work this out. Some students think you need to know more about the game to find P(not winning), but all you need is the complementary relationship: the two probabilities must add to 1 because one of the two things must happen. This works with any probability representation — if P(event) = 0.6, then P(not event) = 0.4; if P(event) = 70%, then P(not event) = 30%.

Example: Flipping a fair coin and getting heads. P(heads) = \( \frac{1}{2} \).

Another: Rolling a fair six-sided dice and getting an even number. There are 3 even numbers (2, 4, 6) out of 6, so P(even) = \( \frac{3}{6} = \frac{1}{2} \).

Creative: Picking a red card from a standard 52-card deck. P(red) = \( \frac{26}{52} = \frac{1}{2} \). Any setup where exactly half the equally likely outcomes are favourable works.

Trap: Rolling a fair dice and getting a 1 or a 6. A student might think “I want two results and there are two things I either get or don’t get” and conclude it’s 50-50, but P(1 or 6) = \( \frac{2}{6} = \frac{1}{3} \), not \( \frac{1}{2} \).

Example: Picking a ball from a bag containing 1 red and 4 blue balls. The two outcomes are “red” and “blue”, but P(red) = \( \frac{1}{5} \) and P(blue) = \( \frac{4}{5} \).

Another: A biased coin where P(heads) = 0.7 and P(tails) = 0.3.

Creative: Spinning a spinner that is split into two sections, but with red taking up \( \frac{3}{4} \) of the circle and blue taking up \( \frac{1}{4} \). Two outcomes, but P(red) = \( \frac{3}{4} \) and P(blue) = \( \frac{1}{4} \).

Trap: Flipping a fair coin — heads or tails. A student might offer this thinking all two-outcome situations qualify, but with a fair coin P(heads) = P(tails) = \( \frac{1}{2} \), which is exactly what we need to avoid. The question says “NOT \( \frac{1}{2} \).”

Example: Picking a blue ball from a bag containing 4 blue and 1 red ball. P(blue) = \( \frac{4}{5} \).

Another: Rolling a number greater than 1 on a fair six-sided dice. P(>1) = \( \frac{5}{6} \).

Creative: Picking a consonant from the letters of the word MATHS. The consonants are M, T, H, S (4 out of 5), so P(consonant) = \( \frac{4}{5} \).

Trap: Picking a ball from a bag of 10 blue balls. A student might think “all blue, so you’re very likely to get blue — more than half.” But P(blue) = \( \frac{10}{10} = 1 \), which is equal to 1, not less than 1. The question requires strictly less than 1.

Example: A bag with 3 green and 1 red ball. P(green) = \( \frac{3}{4} \).

Another: A bag with 6 green and 2 yellow balls. P(green) = \( \frac{6}{8} = \frac{3}{4} \).

Creative: A bag with 15 green and 5 blue balls. P(green) = \( \frac{15}{20} = \frac{3}{4} \). Any bag where exactly three quarters of the balls are green works — there are infinitely many valid answers.

Trap: A bag with 3 green, 1 red, and 1 blue ball. A student might think “there are 3 green and there are some others, and 3 out of 4 colours…” But the total is 5 balls, so P(green) = \( \frac{3}{5} \), not \( \frac{3}{4} \). The misconception is focusing on the number of green balls (3) and forcing the denominator to be 4 without checking the actual total.

Example: Rolling a fair six-sided dice and looking at three compound events: rolling 1 or 2 (\( \frac{2}{6} \)), rolling 3 or 4 (\( \frac{2}{6} \)), and rolling 5 or 6 (\( \frac{2}{6} \)). These three probabilities add to 1.

Another: A spinner that covers a full circle divided into three colours: red (\( \frac{1}{2} \)), blue (\( \frac{1}{4} \)), and green (\( \frac{1}{4} \)).

Creative: Drawing a card from a standard deck and recording if it is a Heart (\( \frac{1}{4} \)), a Club (\( \frac{1}{4} \)), or a black or red Diamond/Spade combination covering the rest (\( \frac{1}{2} \)). Any three mutually exclusive events that cover the entire sample space will work.

Trap: A bag with 3 red, 2 blue, and 1 green ball, and calculating P(red) + P(blue) + P(yellow). The student might assume any three colours from the setup sum to 1. But since there are no yellow balls, the sum is \( \frac{3}{6} + \frac{2}{6} + 0 = \frac{5}{6} \), not 1. You must exhaust all possible outcomes.

This is the definition of probability. P = 0 means the event is impossible (e.g. rolling a 7 on a standard fair dice). P = 1 means the event is certain (e.g. rolling less than 7 on a standard fair dice). Every other probability lies between these values.

If a student ever calculates a probability greater than 1 or less than 0, they have made an error — for example, dividing the total by the favourable outcomes instead of the other way round. The number of favourable outcomes cannot exceed the total number of equally likely outcomes, so the fraction can never be greater than 1, and it cannot be negative since we are counting outcomes.

True case: Flipping a fair coin — heads and tails are equally likely, so each has probability \( \frac{1}{2} \).

False case: Picking a ball from a bag with 1 red and 9 blue. There are two outcomes (red or blue), but P(red) = \( \frac{1}{10} \) and P(blue) = \( \frac{9}{10} \), not \( \frac{1}{2} \) each. The key is that two outcomes does not mean equally likely outcomes. Similarly, “it will rain tomorrow or it won’t” has two outcomes but nobody would claim P(rain) = \( \frac{1}{2} \) without evidence.

Each flip of a fair coin is independent — the coin has no memory. Whether you have just flipped 5 heads in a row or alternated perfectly, P(heads) on the next flip is still \( \frac{1}{2} \). A common belief is that after several heads, tails is “due” — this is the gambler’s fallacy.

Independence means the probability is fixed at \( \frac{1}{2} \) regardless of history. Note: this applies to a fair coin. If the coin were biased, the probability on each flip would still be constant — it just wouldn’t be \( \frac{1}{2} \). Either way, previous results never change what happens next.

True case: A bag containing 5 red and 5 blue balls. P(red) = \( \frac{5}{10} = \frac{1}{2} \).

False case: A bag containing 2 red and 6 blue balls. P(red) = \( \frac{2}{8} = \frac{1}{4} \), not \( \frac{1}{2} \). The probability depends on the proportion of red balls in the bag — it is only \( \frac{1}{2} \) when exactly half the balls are red. Students who think this is “always” may be applying the misconception that you either pick red or you don’t, so it’s 50-50. Students who think “never” may not realise that the right bag contents can make P(red) = exactly \( \frac{1}{2} \).

The student has divided the number of red balls by the number of blue balls, rather than by the total number of balls. This is a common misconception — treating probability as a ratio of “wanted to not-wanted” rather than “wanted to total.”

The correct calculation is P(red) = \( \frac{4}{4 + 6} = \frac{4}{10} = \frac{2}{5} \). The denominator must always be the total number of equally likely outcomes. A quick sense-check: if there were 5 red and 5 blue, the student’s method would give \( \frac{5}{5} = 1 \), meaning “certain” — but it clearly isn’t certain, since half the balls are blue.

The student has counted the number of colours (2) instead of the number of equally likely sections (5). Because the sections are equal in size, each section is equally likely, and probability should be calculated using sections, not colours. P(blue) = \( \frac{2}{5} \), not \( \frac{1}{2} \).

The fact that there are two colour categories does not make them equally likely. There are 3 red sections and only 2 blue sections, so landing on red is more likely than blue. This misconception often appears when students confuse “number of categories” with “number of equally likely outcomes.”

The student recognises that rolling a 7 is very difficult but hasn’t distinguished between unlikely and impossible. A standard dice has faces numbered 1 to 6 — the outcome 7 does not exist in the sample space. It is not “nearly impossible”; it is genuinely impossible. The correct answer is P(rolling a 7) = 0.

An event with probability 0 is one that cannot happen — not one that is merely very hard. If an outcome is not in the sample space at all, its probability is exactly 0, not some small number. Ask the student: “If rolling a 7 is \( \frac{1}{100} \), what does the dice look like to make that happen?” This forces them to realize a 100-sided shape is required for that math to work.

The answer of \( \frac{1}{3} \) is correct, but the reasoning is flawed. The student counted colours (3) rather than individual balls (6). The correct method is P(red) = \( \frac{2}{6} = \frac{1}{3} \). The student got lucky because the colours happen to be equally distributed — there are 2 of each colour.

If the bag contained 3 red, 2 blue, and 1 green (still 3 colours), the student’s method would give P(red) = \( \frac{1}{3} \), but the correct answer is P(red) = \( \frac{3}{6} = \frac{1}{2} \). Counting colours only works when every colour has the same number of balls. The reliable method is always to count individual equally likely outcomes.

The student is confusing experimental frequency with theoretical probability. The theoretical probability of a fair coin remains exactly \( \frac{1}{2} \). The 7 out of 10 is simply the relative frequency of what happened in this specific, small sample of trials.

Because coins have no memory and probability describes long-term behavior, a small number of flips will rarely perfectly match the theoretical probability. If the student flipped the coin 10,000 times, the proportion of heads would settle much closer to \( \frac{1}{2} \).