KBAT: Decision-Making with Probability

Q: When is a probability question binomial rather than normal?

Use the binomial model when you count successes in a fixed number of independent trials, each with the same probability, flawed or not, pass or fail, hit or miss. Use the normal model when the quantity is a continuous measurement that clusters around a mean, such as mass, length or time. Naming the right model correctly is the first mark on the page, and it decides every step that follows.

Q: Why factor out the complement instead of adding terms?

For 'at least two out of five', adding P(X=2)+P(X=3)+P(X=4)+P(X=5)P(X=2)+P(X=3)+P(X=4)+P(X=5)P(X=2)+P(X=3)+P(X=4)+P(X=5) is four terms; the complement 1−P(X=0)−P(X=1)1-P(X=0)-P(X=1)1−P(X=0)−P(X=1) is two. Both are correct, but the complement is shorter and less error-prone, so markers see it often. Choose whichever needs fewer terms, and write the complement statement clearly so the method is visible.

Q: How do I turn a decision limit into a comparison?

Read the limit as a probability. 'At most one box in ten' means the qualifying probability must be ≤0.1\le 0.1≤0.1. Compute the event probability, then compare: if it exceeds the limit the rule is too generous; if it is under the limit the rule fits. End with a sentence naming the option you choose, because the decision is what earns the final marks.

Q: What does the mean np tell me?

For X∼B(n,p)X\sim B(n,p)X∼B(n,p) the mean is npnpnp, the number of successes you expect on average. Here np=5×0.2=1np=5\times 0.2=1np=5×0.2=1, so a typical box has about one flaw. The mean is a quick sanity check: if a probability you compute clashes with that expected count, say you find 'three or more' to be very likely, you have probably slipped somewhere and should re-check.

Q: How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Naming the model, writing the correct event, substituting into the binomial formula and stating a justified decision each earn credit, so show every step and finish with a clear choice.

What makes this a KBAT question

A routine probability question tells you the model and the event: 'find $P(X=2)$ for $X\sim B(5,0.2)$ '. A decision-making KBAT question hides both inside a situation and adds a judgement at the end, should the workshop adopt this rule, is the game worth playing, does the batch pass?

That is higher-order thinking, Kemahiran Berfikir Aras Tinggi: the binomial and normal tools are familiar, but you must decide which event matters ('at least two', 'fewer than three'), compute its probability, then compare it against a stated limit and state a conclusion.

The situations change from year to year, a warranty, an acceptance test, a stall game, a screening decision, but the shape stays the same: name the model, choose the event, compute the probability, compare it with a limit, and conclude. Nothing on the page tells you which inequality to set up or where to stop.

In Add Math this rewards students who can connect a number to a real choice, not only students who can substitute into a formula and read off a value.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. It looks like a story, but every sentence is really telling you a model, an event, or a limit.

Try it before reading the solution.

Q1[8 marks]

A small workshop makes handmade candles. From long experience, each candle independently has probability $0.2$ of a minor flaw.

Candles are packed in boxes of $5$ , and $X$ is the number of flawed candles in a box. (a) State the distribution of $X$ and find the probability that a box contains exactly one flawed candle.

(b) The workshop is choosing a free-replacement rule and can afford to replace, on average, at most $1$ box in every $10$ . Rule A replaces any box with at least $2$ flawed candles; Rule B replaces any box with at least $3$ flawed candles.

Determine which rule the workshop can adopt.

Show worked solution

Understand. Each candle is flawed or not, independently, with a fixed probability $0.2$ , and there are $5$ candles in a box, so $X$ follows a binomial distribution.

'At most $1$ box in every $10$ ' is a limit written as a fraction; read as a probability it means the chance that a box qualifies for replacement must be no more than $0.1$ . So for each rule we find the probability of its replacement event and compare it with $0.1$ .

Plan. Write $X\sim B(5,\,0.2)$ and use $P(X=r)=\binom{5}{r}(0.2)^{r}(0.8)^{5-r}$ .

For Rule A find $P(X\ge 2)=1-P(X=0)-P(X=1)$ ; for Rule B find $P(X\ge 3)=1-P(X=0)-P(X=1)-P(X=2)$ . Using the complement is shorter than adding several separate terms.

Then compare each result with $0.1$ and name the rule that fits.

Execute and check. (a) The distribution is $X\sim B(5,\,0.2)$ .

For exactly one flaw, one candle is flawed and the other four are sound:

P(X=1)=\binom{5}{1}(0.2)^{1}(0.8)^{4}=5\times 0.2\times 0.4096=0.4096

(b) First the two probabilities that both rules need. A box with no flaws at all means every one of the five candles is sound:

P(X=0)=(0.8)^{5}=0.32768

P(X=2)=\binom{5}{2}(0.2)^{2}(0.8)^{3}=10\times 0.04\times 0.512=0.2048

For Rule A a box is replaced when it has at least two flaws, so we take the complement of 'zero or one flaw':

P(X\ge 2)=1-P(X=0)-P(X=1)=1-0.32768-0.4096=0.26272

Since $0.26272>0.1$ , Rule A would replace about $26\%$ of boxes, more than a quarter, far more than the workshop can afford.

For Rule B a box is replaced when it has at least three flaws:

P(X\ge 3)=1-P(X=0)-P(X=1)-P(X=2)=1-0.32768-0.4096-0.2048=0.05792

Since $0.05792<0.1$ , Rule B replaces under $6\%$ of boxes, comfortably within the limit. So the workshop can adopt Rule B.

Check. The probabilities $P(X=0)+P(X=1)+P(X=2)+P(X\ge 3)=0.32768+0.4096+0.2048+0.05792=1$ , so nothing was miscounted.

The mean number of flaws per box is $np=5\times 0.2=1$ , so most boxes have none or one flaw and three-or-more is genuinely rare, consistent with the small $0.058$ we found.

Read the two answers as the workshop owner would. Rule A sounds generous to customers, but it would send back roughly one box in four, which is expensive; Rule B triggers only when a box is unusually bad, about one in seventeen, and stays inside the budget.

The mathematics has not made the decision, it has told the owner exactly what each choice would cost, and that is what a decision question in Add Math is really testing.

Finding a sensible first step

When a probability question is wrapped in a story, the first step is always the same: name the model and define the variable in one line. Ask whether each 'trial' has just two outcomes (flaw or no flaw, pass or fail) with a fixed probability, repeated a fixed number of times, that is the binomial model, $X\sim B(n,p)$ .

If instead the quantity is a measurement that clusters around a mean (mass, length, time), it is the normal model. Once the model is named, translate the decision phrase into an inequality on $X$ : 'at least two' is $X\ge 2$ , 'fewer than three' is $X\le 2$ , 'more than one' is $X\ge 2$ .

Write the event down before you compute anything.

Here is the normal case in miniature: if bags of sugar have masses that are normally distributed with a known mean and standard deviation, 'a bag is underweight' becomes $P(X<\text{limit})$ , found by standardising with $Z=\dfrac{X-\mu}{\sigma}$ . The tool changes, but the habit does not, name the model, write the event, then compute.

Many of the marks on these questions are won or lost at this translation step, so slow down and pin the event first.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a decision-making probability question a marker looks for:

The model named and the variable defined, for example $X\sim B(5,0.2)$ , where $X$ is the number of flawed candles.
The decision phrase turned into a correct event, such as 'at least two' becoming $P(X\ge 2)$ .
The binomial formula written with the correct $\binom{n}{r}$ , power and complement, not just a final number.
Use of the complement $1-P(X=0)-P(X=1)$ where it is shorter than adding many terms.
A clear comparison against the stated limit, $0.263>0.1$ , or $0.058<0.1$ .
A final sentence that answers the actual question, such as 'adopt Rule B', not just a probability.

How a teacher helps

The place students lose marks here is not the arithmetic, a calculator handles that, it is choosing the event and reading the answer as a decision. So that is what our teachers rehearse.

In a one-to-one lesson we practise reading a situation aloud and writing the model and the event before touching the calculator, then saying the conclusion in a full sentence. We keep a running list of the phrases that flip an inequality,'at least', 'fewer than', 'no more than', 'more than', because one careless flip turns a correct method into a wrong answer.

We also build the habit of checking that the probabilities sum to one. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

When is a probability question binomial rather than normal?

Use the binomial model when you count successes in a fixed number of independent trials, each with the same probability, flawed or not, pass or fail, hit or miss. Use the normal model when the quantity is a continuous measurement that clusters around a mean, such as mass, length or time.

Naming the right model correctly is the first mark on the page, and it decides every step that follows.

Why factor out the complement instead of adding terms?

For 'at least two out of five', adding $P(X=2)+P(X=3)+P(X=4)+P(X=5)$ is four terms; the complement $1-P(X=0)-P(X=1)$ is two. Both are correct, but the complement is shorter and less error-prone, so markers see it often.

Choose whichever needs fewer terms, and write the complement statement clearly so the method is visible.

How do I turn a decision limit into a comparison?

Read the limit as a probability. 'At most one box in ten' means the qualifying probability must be $\le 0.1$ .

Compute the event probability, then compare: if it exceeds the limit the rule is too generous; if it is under the limit the rule fits. End with a sentence naming the option you choose, because the decision is what earns the final marks.

What does the mean

np

tell me?

For $X\sim B(n,p)$ the mean is $np$ , the number of successes you expect on average. Here $np=5\times 0.2=1$ , so a typical box has about one flaw.

The mean is a quick sanity check: if a probability you compute clashes with that expected count, say you find 'three or more' to be very likely, you have probably slipped somewhere and should re-check.

How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Naming the model, writing the correct event, substituting into the binomial formula and stating a justified decision each earn credit, so show every step and finish with a clear choice.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026