Making probability distributions make sense

In earlier chapters, a variable such as $x$ simply stood in for a number you were solving for. In this chapter, a random variable stands for the numerical outcome of a chance experiment, something you cannot know in advance, only describe the possible values of.

Toss three coins and count the heads, and the outcome could be 0, 1, 2 or 3; that count is a random variable, usually written $X$.

Add Math splits random variables into two families. A discrete random variable can only take separate, countable values, you can list them with commas, like $X = \{x : x = 0, 1, 2, 3\}$ for the coin example, or the number of defective items in a batch, or the number of correct guesses on a quiz.

A continuous random variable is measured on a scale rather than counted, and its possible values fill an interval, like $X = \{x : x \text{ is the height of a student in cm}, 140 < x < 180\}$. Heights, times, and masses are continuous; anything you count on your fingers is discrete.

Once you have a random variable, a probability distribution is just the pairing of every value it can take with the probability of that value happening, shown as a table, or as a graph with a bar or point for each value. Whatever form it takes, the probabilities across all possible values must add up to exactly 1, because something in the sample space always happens.

That single check, do the probabilities sum to 1, is worth running on almost every distribution question you meet.

A binomial distribution appears whenever four conditions hold together: a fixed number of trials $n$, each trial has only two outcomes (call them success and failure), the trials are independent of each other, and the probability of success $p$ stays the same on every trial. Guessing answers on a multiple-choice quiz, checking items off a production line for a fixed batch size, or shooting a fixed number of free throws with a constant success rate, all of these fit the binomial pattern, with $X$ counting the number of successes.

$$P(X=r) = {}^{n}C_{r}\, p^{r} q^{\,n-r}$$

Here $q = 1-p$ is the probability of failure, and ${}^{n}C_{r}$ counts the number of different orders the $r$ successes could land in among the $n$ trials. Say a student guesses on 5 multiple-choice questions, each with 4 options, so $p=\frac{1}{4}$ and $q=\frac{3}{4}$.

The chance of getting exactly 2 correct is $P(X=2) = {}^{5}C_{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3} = 10 \times \frac{1}{16} \times \frac{27}{64} \approx 0.264$.

Because it's a distribution, the mean, variance, and standard deviation are also worth knowing: mean $=np$, variance $=npq$, standard deviation $=\sqrt{npq}$. These aren't in the given formula list, so they're worth committing to memory.

The mean is best read as an expected average, not a certain outcome, if a fair coin is tossed 20 times, the mean number of heads is 10, but any individual set of 20 tosses could easily land on 8, 11, or 13.

$$\text{mean} = np,\qquad \text{variance} = npq$$

The normal distribution describes continuous data that clusters symmetrically around a central value, exam scores, heights in a large population, or measurement errors often behave this way. Plotted as a graph, it forms the familiar bell shape: a peak at the mean $\mu$, falling away evenly on both sides, with the spread controlled by the standard deviation $\sigma$.

A small $\sigma$ gives a tall, narrow bell; a large $\sigma$ gives a wide, flat one.

Because $X$ is continuous here, you can't list separate values and add up their probabilities the way you did for the binomial distribution. Instead, probability is read as an area under the curve, the area between two values of $X$ is the probability that the outcome falls between them, and the total area under the whole curve is exactly 1, just like the discrete case.

Every normal distribution has its own mean and standard deviation, which would mean a different area calculation for every single question, except for one shortcut. Any normal distribution can be converted into the standard normal distribution, which always has mean 0 and standard deviation 1, using the standard score:

$$Z = \dfrac{X-\mu}{\sigma}$$

$Z$ tells you how many standard deviations $X$ sits above or below the mean. Suppose exam scores are normally distributed with mean $\mu=65$ and standard deviation $\sigma=8$, and you want $P(X>75)$.

First standardise: $Z = \frac{75-65}{8}=1.25$. Then use your calculator's normal-distribution function (or a standard normal table, if the question supplies one) to read off $P(Z>1.25)$, which comes out to roughly $0.106$, about a 10.6% chance of scoring above 75.

The reverse direction matters just as often: you're given a probability and asked to find the corresponding $Z$, then work back to find $X$, $\mu$, or $\sigma$. Rearranging $Z=\frac{X-\mu}{\sigma}$ into $X = \mu + Z\sigma$ handles that direction.

Whichever way the question runs, sketching a quick bell curve and shading the region you need, even roughly, makes it far easier to see whether you should be adding, subtracting, or using the complement $1-P(\cdot)$.

It helps to see the reading and the calculation work together in one question. Suppose a quiz has 8 multiple-choice questions, each with 4 options, and a student guesses every answer.

What is the probability of getting at least 2 correct? This is binomial with $n=8$, $p=\frac{1}{4}$, and $q=\frac{3}{4}$, and "at least 2" means $X \ge 2$.

Adding $P(X=2)$ all the way to $P(X=8)$ would be seven separate calculations. The complement is far shorter: $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.

Work out the two easy terms first, $P(X=0) = \left(\frac{3}{4}\right)^{8} \approx 0.1001$, and $P(X=1) = {}^{8}C_{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{7} \approx 0.2670$.

$$P(X \ge 2) = 1 - 0.1001 - 0.2670 \approx 0.633$$

So a pure guesser has about a 63% chance of getting at least two right, a result that feels reasonable once you see it, and a good example of why the complement rule is usually the quicker road. Notice the discipline: name the distribution, write the complement expression in full, then reach for the calculator.

Even if a decimal slips, the method line has already earned its marks.

The same care pays off in the other direction. If the question instead told you that only the top 15% of students scored above a certain mark and asked for that mark, you would read the probability first, find the matching $Z$ from your calculator (here $Z \approx 1.04$), and only then substitute into $X = \mu + Z\sigma$.

Reading before calculating is the habit that keeps these questions from unravelling.

Distribution questions are as much about reading carefully as calculating. A handful of phrases decide which inequality you need:

• "At least" 2 means $X \ge 2$, which includes 2 itself.
• "More than" 2 means $X > 2$, which does not include 2.
• "At most" or "no more than" 2 means $X \le 2$.
• "Exactly" 2 means $X = 2$, a single value, not a range.

The complement rule is worth reaching for often: $P(X \ge 1) = 1 - P(X=0)$ is almost always faster than adding up $P(X=1)$ through $P(X=n)$ individually, especially for a large $n$.

Because SPM Add Math is marked analytically, a correctly set-up expression earns method marks even if a later decimal comes out slightly off. Writing $P(X \ge 2) = 1 - P(X=0) - P(X=1)$ clearly, before you touch a calculator, protects most of the marks even on a bad arithmetic day.

Probability distribution is a chapter where the maths is rarely the hard part, most students can substitute into a formula. What trips people up is the reading: discrete or continuous, at least or more than, and whether the calculator is set up to give a normal-distribution area rather than a single point value.

A teacher watching you work through a problem live can catch exactly which of those you're missing, far faster than a mark scheme alone can explain it. Our teachers are experienced, and lessons run online in English, which also builds comfort with the English terms used throughout this chapter, while SPM papers themselves are set bilingually in Bahasa Melayu and English.

If you'd like to try a session, the first lesson is a one-hour paid class at the teacher's rate, from RM50 an hour, depending on the teacher's experience, quoted on WhatsApp. We meet you wherever you are in the chapter, whether that's the discrete/continuous split or standardising a normal distribution, and build outward at your own pace.