Because nC0=1{}^{n}C_{0}=1nC0=1 and p0=1p^{0}=1p0=1, the formula nC0 p0qn{}^{n}C_{0}\,p^{0}q^{n}nC0p0qn collapses to qnq^{n}qn, every one of the nnn trials must be a failure.

Probability Distribution Examples (Medium)

Q: When should I use the complement rule?

Whenever a question asks for 'at least one' or 'at least ...', the complement is usually shortest. For 'at least one', P(X≥1)=1−P(X=0)P(X\ge 1)=1-P(X=0)P(X≥1)=1−P(X=0) replaces a long sum with a single term.

Q: How do I handle a normal interval that crosses the mean?

Standardise both ends, then subtract the two left-areas: P(Z<z2)−P(Z<z1)P(Z<z_{2})-P(Z<z_{1})P(Z<z2)−P(Z<z1). Do not add tails; the region is a single band across the middle.

Q: How do I find a boundary from a given probability?

Reverse the process. Turn the probability into a left-area P(Z<z)P(Z<z)P(Z<z), read the matching zzz from the table, then solve z=X−μσz=\dfrac{X-\mu}{\sigma}z=σX−μ for the unknown.

What these examples cover

These three medium examples push a little past the basics. You will combine binomial probabilities using the complement rule to answer an 'at least one' question, read a normal probability across an interval that straddles the mean, and work a normal problem backwards to find an unknown boundary from a given probability.

The numbers stay clean, but each question now asks for two or three linked steps rather than a single lookup. As before, cover the solution, attempt the whole question on paper, then check line by line.

The habit that matters most here is deciding your plan, which rule, which direction, before you touch the calculator. Because the marking is analytic, writing that plan down as your first line already earns method marks, even before the arithmetic begins.

Read each question twice, once for the distribution, once for exactly which probability it wants, and the rest follows.

Worked examples

Work through all three. Attempt each fully before you read the solution, and notice how naming your plan first, complement, straddle-and-subtract, or reverse lookup, is what keeps the working short and correct.

Reading the standard normal table

The standard normal table gives the left-hand area $P(Z<z)$ for positive $z$ . Everything else is built from that one fact: a right tail is $1-P(Z<z)$ ; a negative score uses symmetry, $P(Z<-z)=1-P(Z<z)$ ; and an interval is the difference of two left-areas.

Sketching a quick bell curve and shading the region you want stops almost every sign error before it starts.

Q1[4 marks]

A random variable $X$ follows a binomial distribution with $n=6$ and $p=0.2$ . Find (a) $P(X=2)$ , and (b) $P(X\ge 1)$ .

Show worked solution

(a) With $n=6$ , $p=0.2$ and $q=0.8$ , apply the binomial formula with $r=2$ :

P(X=2)={}^{6}C_{2}(0.2)^{2}(0.8)^{4}

P(X=2)=15\times 0.04\times 0.4096=0.24576

(b) 'At least one' is easiest through the complement, since $P(X\ge 1)=1-P(X=0)$ . Compute $P(X=0)$ first:

P(X=0)={}^{6}C_{0}(0.2)^{0}(0.8)^{6}=(0.8)^{6}=0.262144

P(X\ge 1)=1-0.262144=0.737856

Both parts rest on the same three numbers $n=6$ , $p=0.2$ and $q=0.8$ ; only the value of $r$ changes, and in part (b) the decision to work through the complement.

Answer

$P(X=2)=0.24576$ and $P(X\ge 1)=0.737856$ . The complement saves work: adding $P(1)$ through $P(6)$ directly would take six terms instead of one.

Q2[4 marks]

A continuous random variable $X$ is normally distributed with mean $60$ and standard deviation $10$ . Find $P(50<X<75)$ .

Show worked solution

Standardise both boundaries with $Z=\dfrac{X-\mu}{\sigma}$ , using $\mu=60$ and $\sigma=10$ :

Z_{1}=\frac{50-60}{10}=-1,\qquad Z_{2}=\frac{75-60}{10}=1.5

The interval straddles the mean, so $P(50<X<75)=P(-1<Z<1.5)=P(Z<1.5)-P(Z<-1)$ . Read $P(Z<1.5)=0.9332$ , and use symmetry for the negative score, $P(Z<-1)=1-P(Z<1)=1-0.8413=0.1587$ :

P(50<X<75)=0.9332-0.1587=0.7745

A quick sketch makes the subtraction obvious: shade the band from $Z=-1$ to $Z=1.5$ , and its area is the larger left-area minus the smaller one.

Answer

$P(50<X<75)=0.7745$ . The two boundaries sit on opposite sides of the mean, which is why we subtract two left-areas rather than adding tails.

Q3[3 marks]

A random variable $X$ is normally distributed with mean $70$ and standard deviation $12$ . Given that $P(X>k)=0.0668$ , find the value of $k$ .

Show worked solution

First turn the probability into a $Z$ -score. Since the table gives left-areas, rewrite the right tail: $P(Z>z)=0.0668$ means $P(Z<z)=1-0.0668=0.9332$ .

From the table, $P(Z<1.5)=0.9332$ , so $z=1.5$ .

Now undo the standardisation, $z=\dfrac{k-\mu}{\sigma}$ , and solve for $k$ :

1.5=\frac{k-70}{12}

k-70=1.5\times 12=18 \;\Rightarrow\; k=88

Reversing a normal problem always follows the same order, probability, then left-area, then $z$ , then $X$ , which is exactly the forward process read from right to left.

Answer

$k=88$ . Check: $Z=\dfrac{88-70}{12}=1.5$ , and $P(Z>1.5)=1-0.9332=0.0668$ , as required.

Notice the shared decision in all three: before any arithmetic, you chose a route, complement, straddle-and-subtract, or reverse lookup. In Probability Distribution the plan is most of the marks; the calculator only finishes what the plan starts.

Common slips to avoid

Two slips recur on these medium questions. The first is adding a long string of binomial terms when the complement $1-P(X=0)$ would do it in one line.

The second is treating a straddling interval as a single tail, when the two boundaries sit on opposite sides of the mean you subtract two left-areas, and the negative score still needs the symmetry step. Decide the route first and both are avoided.

Q4[3 marks]

The probability that a seed germinates is $0.4$ . A gardener plants $n=10$ of these seeds, and $X$ is the number that germinate.

Find the mean and the variance of $X$ , and hence state its standard deviation.

Show worked solution

$X$ follows a binomial distribution with $n=10$ , $p=0.4$ and $q=0.6$ . The mean of a binomial variable is $E(X)=np$ :

E(X)=10\times 0.4=4

The variance is $Var(X)=npq$ :

Var(X)=10\times 0.4\times 0.6=2.4

The standard deviation is the positive square root of the variance:

\sigma=\sqrt{2.4}=1.549\text{ (4 s.f.)}

Answer

Mean $=4$ , variance $=2.4$ , standard deviation $\approx 1.549$ . Sense check: a mean of $4$ out of $10$ seeds matches $p=0.4$ , i.e. $40\%$ , exactly as expected.

Q5[4 marks]

The scores of students in a test are normally distributed with mean $100$ and standard deviation $15$ . Find $P(115<X<130)$ .

Show worked solution

Standardise both boundaries using $Z=\dfrac{X-\mu}{\sigma}$ with $\mu=100$ , $\sigma=15$ :

Z_{1}=\frac{115-100}{15}=1,\qquad Z_{2}=\frac{130-100}{15}=2

Both boundaries lie above the mean, so both $Z$ -scores are positive; the working is still a subtraction of two left-areas, $P(115<X<130)=P(Z<2)-P(Z<1)$ . Read $P(Z<2)=0.9772$ and $P(Z<1)=0.8413$ from the table:

P(115<X<130)=0.9772-0.8413=0.1359

Because neither boundary sits below the mean, there is no symmetry step here, only the subtraction.

Answer

$P(115<X<130)=0.1359$ . Sense check: since both scores are above the mean, the probability should be well under $0.5$ , which $0.1359$ is.

Q6[3 marks]

A basketball player takes $n=4$ free throws, each with the same probability $p$ of scoring, independent of the others. Let $X$ be the number of free throws scored.

Given that $P(X=4)=0.0625$ , find the value of $p$ .

Show worked solution

When every trial succeeds, $r=n=4$ , so the binomial formula collapses because ${}^{4}C_{4}=1$ and $q^{0}=1$ :

P(X=4)={}^{4}C_{4}p^{4}q^{0}=p^{4}

Set this equal to the given probability and solve for $p$ :

p^{4}=0.0625 \;\Rightarrow\; p=\sqrt[4]{0.0625}=0.5

Answer

$p=0.5$ . Check: $0.5^{4}=0.0625$ , matching the given probability, and a value between $0$ and $1$ as a probability must be.

Q7[3 marks]

A random variable $X$ is normally distributed with mean $50$ . Given that $P(X<65)=0.8413$ , find the standard deviation $\sigma$ .

Show worked solution

Convert the given probability to a $Z$ -score. From the table, $P(Z<1)=0.8413$ , so the boundary $X=65$ corresponds to $Z=1$ .

Now use the standardisation formula $z=\dfrac{X-\mu}{\sigma}$ with $X=65$ , $\mu=50$ , and solve for $\sigma$ :

1=\frac{65-50}{\sigma}

\sigma=\frac{15}{1}=15

Answer

$\sigma=15$ . Check: $Z=\dfrac{65-50}{15}=1$ , and $P(Z<1)=0.8413$ , as given.

Q8[4 marks]

In a school of $200$ students, the time (in minutes) each student takes to finish a puzzle is normally distributed with mean $65$ and standard deviation $8$ . Find the number of students expected to take more than $77$ minutes.

Show worked solution

Standardise the boundary using $Z=\dfrac{X-\mu}{\sigma}$ with $\mu=65$ , $\sigma=8$ :

Z=\frac{77-65}{8}=1.5

This is a right tail, so $P(X>77)=1-P(Z<1.5)$ . Reading $P(Z<1.5)=0.9332$ from the table and multiplying by the population of $200$ gives the expected headcount:

P(X>77)=1-0.9332=0.0668, \qquad 200\times 0.0668=13.36

Answer

About $13$ students ( $13.36$ rounded down to a whole student) are expected to take more than $77$ minutes. Sense check: $0.0668$ is a small tail probability, so only a small fraction of the $200$ students, not half, should exceed $77$ minutes.

Key method points

These three examples train the linked-step thinking that medium questions demand. Keep the following points in mind as you practise.

For 'at least one', use the complement: $P(X\ge 1)=1-P(X=0)$ .
$P(X=0)=q^{n}$ , because ${}^{n}C_{0}=1$ and $p^{0}=1$ .
When a normal interval straddles the mean, subtract two left-areas: $P(a<X<b)=P(Z<z_{2})-P(Z<z_{1})$ .
For a negative score use symmetry: $P(Z<-z)=1-P(Z<z)$ .
To reverse a normal problem, convert the probability to a left-area first, read the $Z$ -score, then solve $z=\dfrac{X-\mu}{\sigma}$ .
Analytic marking rewards each correct step, so write the standardisation line even if the final value slips.

How a teacher helps

Medium questions reward planning, and that is exactly what we coach. Before any figures, our teacher asks the student to say the route out loud, complement, subtract two areas, or reverse the standardisation, because naming the plan prevents the most common error, solving the wrong quantity.

Working one-to-one, we can pause at the precise line where a sign or a table entry goes astray and fix it on the spot. Our teachers are experienced.

Lessons are taught in English, while SPM papers are set in Malay and English, so the symbols stay familiar in both.

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Frequently asked questions

When should I use the complement rule?

Whenever a question asks for 'at least one' or 'at least ...', the complement is usually shortest. For 'at least one', $P(X\ge 1)=1-P(X=0)$ replaces a long sum with a single term.

Why is

P(X=0)=q^{n}

Because ${}^{n}C_{0}=1$ and $p^{0}=1$ , the formula ${}^{n}C_{0}\,p^{0}q^{n}$ collapses to $q^{n}$ , every one of the $n$ trials must be a failure.

How do I handle a normal interval that crosses the mean?

Standardise both ends, then subtract the two left-areas: $P(Z<z_{2})-P(Z<z_{1})$ . Do not add tails; the region is a single band across the middle.

How do I find a boundary from a given probability?

Reverse the process. Turn the probability into a left-area $P(Z<z)$ , read the matching $z$ from the table, then solve $z=\dfrac{X-\mu}{\sigma}$ for the unknown.

Source:SRC-DSKP-EN

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

Probability Distribution, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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