Probability Distribution Examples (Easy)

Q: How do I know whether to use the binomial or the normal distribution?

Use the binomial when you count successes in a fixed number of separate trials, each with the same probability, such as heads in 333 tosses. Use the normal for a continuous measurement like mass or height that clusters around a mean.

Q: What are p and q in the binomial formula?

ppp is the probability of success on a single trial and q=1−pq=1-pq=1−p is the probability of failure. They always add to 111, so once you know one you know the other.

Q: Why do I subtract from 1 for P(X>a) in the normal distribution?

The standard normal table lists the area to the left, P(Z<z)P(Z<z)P(Z<z). Since the total area is 111, the right tail is P(Z>z)=1−P(Z<z)P(Z>z)=1-P(Z<z)P(Z>z)=1−P(Z<z).

Q: Do I need to check that my binomial probabilities add to 1?

It is not required, but it is a fast, reliable check. If the probabilities for all possible values sum to 111, your individual figures are almost certainly right.

What these examples cover

These four easy Probability Distribution examples rehearse the two workhorses of the chapter: the binomial distribution for counting successes in a fixed number of trials, and the normal distribution for a continuous quantity such as mass or height. You will compute one binomial probability, find the mean, variance and standard deviation of a binomial variable, and read two straightforward normal probabilities from the standard score $Z$ .

Every number here is small and clean, so you can follow each line without wrestling the calculator. Cover the solution, attempt the question fully on paper, and only then check line by line.

Where your answer parts from ours, find the exact step, that single line is usually where the real learning sits.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and watch how the same routine, name the distribution and its parameters, write the formula, substitute, then simplify, carries every question.

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[3 marks]

A biased coin lands on heads with probability $0.4$ . It is tossed $3$ times.

If $X$ is the number of heads obtained, find $P(X=2)$ .

Show worked solution

Here $X$ follows a binomial distribution with $n=3$ trials and success probability $p=0.4$ , so $q=1-p=0.6$ . Use the binomial probability formula with $r=2$ :

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

Substitute $n=3$ , $r=2$ , $p=0.4$ and $q=0.6$ :

P(X=2)={}^{3}C_{2}(0.4)^{2}(0.6)^{1}=3\times 0.16\times 0.6

P(X=2)=3\times 0.096=0.288

Answer

$P(X=2)=0.288$ . As a check, the four probabilities $P(0)=0.216$ , $P(1)=0.432$ , $P(2)=0.288$ and $P(3)=0.064$ add to exactly $1$ , which confirms the working.

Q2[4 marks]

A type of seed germinates with probability $0.25$ . A gardener plants $16$ such seeds.

If $X$ is the number that germinate, find the mean, the variance and the standard deviation of $X$ .

Show worked solution

$X$ is binomial with $n=16$ and $p=0.25$ , so $q=0.75$ . Use $\text{mean}=np$ , $\text{variance}=npq$ and $\text{standard deviation}=\sqrt{npq}$ .

\mu=np=16\times 0.25=4

\sigma^{2}=npq=16\times 0.25\times 0.75=3

\sigma=\sqrt{3}=1.732

Answer

The mean is $4$ , the variance is $3$ , and the standard deviation is $\sqrt{3}\approx 1.732$ . The mean of $4$ matches intuition: a quarter of $16$ seeds is expected to germinate.

Q3[3 marks]

The mass of a fruit, $X$ grams, is normally distributed with mean $50$ g and standard deviation $8$ g. Find $P(X<58)$ .

Show worked solution

Standardise with $Z=\dfrac{X-\mu}{\sigma}$ , using $\mu=50$ and $\sigma=8$ . Convert $X=58$ to a $Z$ -score:

Z=\frac{58-50}{8}=\frac{8}{8}=1

So $P(X<58)=P(Z<1)$ . Read this directly from the standard normal table:

P(Z<1)=0.8413

Answer

$P(X<58)=0.8413$ . Because $58$ is exactly one standard deviation above the mean, just over $84\%$ of the fruit lie below it.

Q4[3 marks]

The height of a plant, $X$ cm, is normally distributed with mean $100$ cm and standard deviation $15$ cm. Find $P(X>130)$ .

Show worked solution

Standardise $X=130$ using $\mu=100$ and $\sigma=15$ :

Z=\frac{130-100}{15}=\frac{30}{15}=2

So $P(X>130)=P(Z>2)$ . The table gives the area to the left, so subtract from $1$ :

P(Z>2)=1-P(Z<2)=1-0.9772=0.0228

Answer

$P(X>130)=0.0228$ . A height above $130$ cm is two standard deviations above the mean, so only about $2.3\%$ of plants reach it.

Two distributions, one habit: state which distribution you are using and its parameters first, then the rest is careful substitution. That single opening line often decides whether the marks come easily or slip away.

Common slips to avoid

Two mistakes cost the most marks here. The first is mixing up the distributions, counting a fixed number of trials calls for the binomial, while a measured quantity calls for the normal.

The second is forgetting that the table reads left: for $P(X>a)$ you must subtract from $1$ . Name the distribution, write the formula, and check the direction of the tail, and both errors disappear.

Q5[3 marks]

A factory produces LED bulbs, and the probability that a bulb is defective is $0.2$ . A random sample of $4$ bulbs is selected.

If $X$ is the number of defective bulbs, find $P(X\geq 1)$ .

Show worked solution

$X$ follows a binomial distribution with $n=4$ trials and defective probability $p=0.2$ , so $q=1-p=0.8$ . Use the complement rule, since "at least one" is the opposite of "none":

P(X\geq 1)=1-P(X=0)

Substitute $n=4$ , $r=0$ , $p=0.2$ and $q=0.8$ :

P(X=0)={}^{4}C_{0}(0.2)^{0}(0.8)^{4}=0.4096\;\Rightarrow\;P(X\geq 1)=1-0.4096=0.5904

Answer

$P(X\geq 1)=0.5904$ . Finding the complement $P(X=0)$ first is far quicker than adding $P(X=1)+P(X=2)+P(X=3)+P(X=4)$ separately.

Q6[3 marks]

The time taken by students to complete a quiz, $X$ minutes, is normally distributed with mean $40$ minutes and standard deviation $5$ minutes. Find $P(35<X<45)$ .

Show worked solution

Standardise both boundaries with $Z=\dfrac{X-\mu}{\sigma}$ , using $\mu=40$ and $\sigma=5$ .

Z_{1}=\frac{35-40}{5}=-1,\qquad Z_{2}=\frac{45-40}{5}=1

So $P(35<X<45)=P(-1<Z<1)=P(Z<1)-P(Z<-1)$ . Use symmetry for the negative value: $P(Z<-1)=1-P(Z<1)=1-0.8413=0.1587$ .

P(-1<Z<1)=0.8413-0.1587=0.6826

Answer

$P(35<X<45)=0.6826$ . This matches the familiar rule that about $68\%$ of values in a normal distribution lie within one standard deviation of the mean.

Q7[3 marks]

The lifespan of a certain type of battery, $X$ hours, is normally distributed with mean $200$ hours and standard deviation $10$ hours. Given that $P(X<k)=0.9332$ , find the value of $k$ .

Show worked solution

Since $P(X<k)=0.9332$ is more than $0.5$ , $k$ must lie above the mean. From the standard normal table, the $Z$ -value with $P(Z<Z)=0.9332$ is $Z=1.5$ .

P(Z<1.5)=0.9332\;\Rightarrow\;Z=1.5

Rearrange the standardisation formula $Z=\dfrac{X-\mu}{\sigma}$ to solve for $k$ , using $\mu=200$ and $\sigma=10$ :

k=\mu+Z\sigma=200+1.5\times 10=215

Answer

$k=215$ hours. Since $Z=1.5$ is positive, $k$ should lie above the mean of $200$ , and $215$ does.

Q8[3 marks]

In a multiple-choice quiz, the probability that a student answers a question correctly by guessing is $0.3$ . If a student guesses the answers to $5$ independent questions, and $X$ is the number of correct guesses, find $P(X\leq 1)$ .

Show worked solution

$X$ follows a binomial distribution with $n=5$ , $p=0.3$ , $q=0.7$ . $P(X\leq 1)=P(X=0)+P(X=1)$ , the sum of the two lowest outcomes.

P(X\leq 1)=P(X=0)+P(X=1)

Substitute $n=5$ , $p=0.3$ , $q=0.7$ into each term:

P(X\leq 1)={}^{5}C_{0}(0.3)^{0}(0.7)^{5}+{}^{5}C_{1}(0.3)^{1}(0.7)^{4}=0.16807+0.36015=0.52822\approx 0.5282

Answer

$P(X\leq 1)\approx 0.5282$ . "At most 1" always means adding every outcome from $0$ up to that value, unlike "at least 1", which is quicker to find using the complement.

Key method points

These four examples rehearse the everyday moves that open almost every Probability Distribution question in Add Math. Keep the following points in mind as you practise more.

Name the distribution and its parameters before anything else, binomial needs $n$ and $p$ ; normal needs $\mu$ and $\sigma$ .
For a binomial probability, use $P(X=r)={}^{n}C_{r}\,p^{r}q^{n-r}$ with $q=1-p$ .
For a binomial variable, $\text{mean}=np$ , $\text{variance}=npq$ and $\text{standard deviation}=\sqrt{npq}$ .
For any normal probability, first standardise with $Z=\dfrac{X-\mu}{\sigma}$ .
The table gives the left area $P(Z<z)$ ; for a right tail use $1-P(Z<z)$ .
Because marking is analytic, a correct formula and substitution can earn method marks even if the final figure slips.

How a teacher helps

On easy Probability Distribution questions, marks are usually lost to one of two habits: reaching for the wrong distribution, or misreading the standard normal table. In a one-to-one lesson our teacher checks the very first line, the distribution and its parameters, and the exact table entry, so a small slip is corrected before it becomes routine.

Our teachers are experienced, so you learn the reasoning, not just the steps. Lessons are taught in English, while SPM papers are set in Malay and English, so the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

How do I know whether to use the binomial or the normal distribution?

Use the binomial when you count successes in a fixed number of separate trials, each with the same probability, such as heads in $3$ tosses. Use the normal for a continuous measurement like mass or height that clusters around a mean.

What are

p

and

q

in the binomial formula?

$p$ is the probability of success on a single trial and $q=1-p$ is the probability of failure. They always add to $1$ , so once you know one you know the other.

Why do I subtract from

1

for

P(X>a)

in the normal distribution?

The standard normal table lists the area to the left, $P(Z<z)$ . Since the total area is $1$ , the right tail is $P(Z>z)=1-P(Z<z)$ .

Do I need to check that my binomial probabilities add to

1

It is not required, but it is a fast, reliable check. If the probabilities for all possible values sum to $1$ , your individual figures are almost certainly right.

Source:SRC-DSKP-EN

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

Probability Distribution, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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