Probability Distribution Examples (KBAT)

Q: How do I get n and p from the mean and variance?

Divide: variancemean=npqnp=q\dfrac{\text{variance}}{\text{mean}}=\dfrac{npq}{np}=qmeanvariance=npnpq=q. Then p=1−qp=1-qp=1−q, and n=meanpn=\dfrac{\text{mean}}{p}n=pmean. Two short steps recover both parameters.

Q: Why subtract the two equations in the normal problem?

Both equations contain −μ-\mu−μ. Subtracting one from the other cancels μ\muμ and leaves a single equation in σ\sigmaσ, which you solve first, then back-substitute for μ\muμ.

Q: Why does the inequality flip when I take logarithms?

lg⁡0.6\lg 0.6lg0.6 is negative. Dividing or multiplying an inequality by a negative number reverses its direction, so n lg⁡0.6<lg⁡0.1n\,\lg 0.6<\lg 0.1nlg0.6<lg0.1 becomes n>lg⁡0.1lg⁡0.6n>\dfrac{\lg 0.1}{\lg 0.6}n>lg0.6lg0.1.

Q: Once I get n>4.51, why is the answer 5 and not 4?

nnn counts trials, so it must be a whole number. The smallest whole number greater than 4.514.514.51 is 555; checking n=4n=4n=4 gives only 0.87040.87040.8704, which fails the requirement.

What these examples cover

These three hard examples ask you to run the ideas in reverse and to join them together. In the first you are given the mean and variance of a binomial variable and must recover $n$ and $p$ before finding a tail probability.

In the second, two normal probabilities give you two equations, which you solve simultaneously for the mean and standard deviation. The third is a KBAT twist: you combine the binomial with logarithms to find the smallest number of trials that makes a success almost certain.

The numbers are still clean, but the thinking is longer, plan the whole route before you start, and keep every line so the method marks are secure.

Worked examples

Work through all three. Attempt each fully before you read the solution, and notice how every question is a familiar tool used in reverse or in combination, nothing new, only deeper.

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.

When a probability is given and the score is unknown, run the same reading backwards: convert to a left-area, then find the matching $z$ .

Q1[5 marks]

A random variable $X$ follows a binomial distribution with mean $2.4$ and variance $1.44$ . (a) Find the value of $n$ and of $p$ .

(b) Hence find $P(X\ge 5)$ .

Show worked solution

(a) For a binomial variable, $\text{mean}=np$ and $\text{variance}=npq$ . Divide the variance by the mean to isolate $q$ :

\frac{npq}{np}=\frac{1.44}{2.4} \;\Rightarrow\; q=0.6

So $p=1-q=0.4$ . Substitute back into $np=2.4$ to find $n$ :

n\times 0.4=2.4 \;\Rightarrow\; n=6

(b) With $n=6$ , $p=0.4$ and $q=0.6$ , $P(X\ge 5)=P(X=5)+P(X=6)$ :

P(X=5)={}^{6}C_{5}(0.4)^{5}(0.6)^{1}=6\times 0.01024\times 0.6=0.036864

P(X=6)={}^{6}C_{6}(0.4)^{6}=0.004096

P(X\ge 5)=0.036864+0.004096=0.04096

Answer

$n=6$ , $p=0.4$ , and $P(X\ge 5)=0.04096$ . Check the parameters: $np=6(0.4)=2.4$ and $npq=6(0.4)(0.6)=1.44$ , both as given.

Q2[5 marks]

A random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ . Given that $P(X>65)=0.0668$ and $P(X<30)=0.0228$ , find $\mu$ and $\sigma$ .

Show worked solution

Turn each probability into a $Z$ -score. For the upper value, $P(Z>z)=0.0668$ gives $P(Z<z)=0.9332$ , so $z=1.5$ :

\frac{65-\mu}{\sigma}=1.5 \;\Rightarrow\; 65-\mu=1.5\sigma \quad (1)

For the lower value, $P(Z<z)=0.0228$ matches $z=-2$ , since $P(Z<2)=0.9772$ and the score is negative:

\frac{30-\mu}{\sigma}=-2 \;\Rightarrow\; 30-\mu=-2\sigma \quad (2)

Subtract equation $(2)$ from equation $(1)$ to eliminate $\mu$ :

(65-\mu)-(30-\mu)=1.5\sigma-(-2\sigma) \;\Rightarrow\; 35=3.5\sigma \;\Rightarrow\; \sigma=10

Substitute $\sigma=10$ into equation $(1)$ :

65-\mu=1.5(10)=15 \;\Rightarrow\; \mu=50

Answer

$\mu=50$ and $\sigma=10$ . Check the lower condition: $\dfrac{30-50}{10}=-2$ , and $P(Z<-2)=0.0228$ , exactly as given.

Q3[4 marks]

In a repeated trial the probability of success each time is $0.4$ , independently. Find the smallest number of trials $n$ so that the probability of at least one success exceeds $0.9$ .

Show worked solution

Let $X$ be the number of successes in $n$ trials, so $X$ is binomial with $p=0.4$ and $q=0.6$ . 'At least one success' uses the complement:

P(X\ge 1)=1-P(X=0)=1-(0.6)^{n}

Require this to exceed $0.9$ and rearrange:

1-(0.6)^{n}>0.9 \;\Rightarrow\; (0.6)^{n}<0.1

Take logarithms of both sides. Because $\lg 0.6$ is negative, dividing by it reverses the inequality:

n\,\lg 0.6<\lg 0.1 \;\Rightarrow\; n>\frac{\lg 0.1}{\lg 0.6}=\frac{-1}{-0.2218}=4.51

Since $n$ must be a whole number greater than $4.51$ , the smallest value is $n=5$ .

Answer

$n=5$ . Check: with $n=5$ , $1-(0.6)^{5}=1-0.07776=0.92224>0.9$ ; with $n=4$ , $1-(0.6)^{4}=1-0.1296=0.8704<0.9$ , so $5$ is indeed the smallest.

Every hard question here is a familiar tool used in reverse or in combination, parameters from a mean and variance, a mean and deviation from two probabilities, a trial count from a complement and a logarithm. Read the demand slowly, set up the equations honestly, and each one unwinds to a clean answer.

Common slips to avoid

Two slips recur on these harder questions. The first is dividing mean by variance instead of variance by mean when recovering $q$ , the ratio must be $\dfrac{npq}{np}$ .

The second is forgetting to flip the inequality after taking logs, because $\lg 0.6$ is negative. Writing the ratio the right way up and checking the sign of the logarithm keeps both under control.

Q4[5 marks]

The number of goals scored by a school futsal team in a match is a discrete random variable $X$ with probability distribution $P(X=0)=0.1$ , $P(X=1)=0.3$ , $P(X=2)=k$ and $P(X=3)=0.2$ . (a) Find the value of $k$ .

(b) Find $E(X)$ . (c) Find $\text{Var}(X)$ .

Show worked solution

(a) The probabilities of a discrete random variable must sum to $1$ . Add the four given probabilities and set the total equal to $1$ :

0.1+0.3+k+0.2=1

So $k=0.4$ . (b) Use $E(X)=\sum xP(X=x)$ , multiplying each value of $X$ by its probability:

E(X)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=0+0.3+0.8+0.6=1.7

E(X^{2})=0^{2}(0.1)+1^{2}(0.3)+2^{2}(0.4)+3^{2}(0.2)=0+0.3+1.6+1.8=3.7

\text{Var}(X)=3.7-(1.7)^{2}=3.7-2.89=0.81

Answer

$k=0.4$ , $E(X)=1.7$ and $\text{Var}(X)=0.81$ . Check the table: $0.1+0.3+0.4+0.2=1$ , so the probabilities are valid.

Q5[6 marks]

In an archery practice session, a player shoots $5$ arrows independently at a target, with probability $p$ of hitting the target on each shot. Let $X$ be the number of hits out of the $5$ shots.

It is given that $P(X=3)=1.5\,P(X=2)$ . (a) Find the value of $p$ .

(b) Hence, find $P(X\ge 4)$ .

Show worked solution

(a) Write both probabilities using the binomial formula and form their ratio. Since ${}^{5}C_{3}={}^{5}C_{2}=10$ , the binomial coefficients cancel:

\frac{P(X=3)}{P(X=2)}=\frac{{}^{5}C_{3}p^{3}q^{2}}{{}^{5}C_{2}p^{2}q^{3}}=\frac{p}{q}=1.5

So $p=1.5q$ . Substitute into $p+q=1$ :

1.5q+q=1 \;\Rightarrow\; 2.5q=1 \;\Rightarrow\; q=0.4 \;\Rightarrow\; p=0.6

(b) With $n=5$ , $p=0.6$ and $q=0.4$ , $P(X\ge 4)=P(X=4)+P(X=5)$ :

P(X=4)={}^{5}C_{4}(0.6)^{4}(0.4)^{1}=5\times 0.1296\times 0.4=0.2592

P(X=5)={}^{5}C_{5}(0.6)^{5}=0.07776

P(X\ge 4)=0.2592+0.07776=0.33696

Answer

$p=0.6$ and $P(X\ge 4)=0.33696$ . Check: $P(X=3)={}^{5}C_{3}(0.6)^{3}(0.4)^{2}=0.3456$ and $P(X=2)={}^{5}C_{2}(0.6)^{2}(0.4)^{3}=0.2304$ ; indeed $0.3456=1.5\times 0.2304$ .

Q6[5 marks]

The time taken, in minutes, by students in a class to complete a puzzle is normally distributed with mean $40$ and standard deviation $\sigma$ . It is given that $30.85\%$ of the students take more than $46$ minutes to complete the puzzle.

(a) Find $\sigma$ . (b) Hence, find the probability that a randomly chosen student takes less than $28$ minutes to complete the puzzle.

Show worked solution

(a) Convert the given percentage to a $Z$ -score. Since $P(Z<0.5)=0.6915$ , the right tail is $P(Z>0.5)=0.3085$ , which matches the given $30.85\%$ :

\frac{46-40}{\sigma}=0.5 \;\Rightarrow\; \sigma=\frac{6}{0.5}=12

(b) With $\sigma=12$ , standardise $28$ and use symmetry, since the score is negative:

\frac{28-40}{12}=-1 \;\Rightarrow\; P(Z<-1)=1-P(Z<1)=1-0.8413=0.1587

Answer

$\sigma=12$ and $P(X<28)=0.1587$ . Check the sizes: $28$ is $12$ minutes below the mean ( $1\sigma$ ) while $46$ is only $6$ minutes above it ( $0.5\sigma$ ); being further from the mean, $28$ should be less likely, and indeed $0.1587<0.3085$ .

Q7[6 marks]

The diameter, in mm, of ball bearings produced by a machine is normally distributed with mean $25$ and standard deviation $2$ . A ball bearing is classified as Grade A if its diameter exceeds $27$ mm.

(a) Find the probability that a randomly selected ball bearing is Grade A. (b) Five ball bearings are selected at random.

Find the probability that exactly $2$ of them are Grade A.

Show worked solution

(a) Standardise $27$ and read the right tail:

\frac{27-25}{2}=1 \;\Rightarrow\; P(X>27)=P(Z>1)=1-0.8413=0.1587

(b) Treat this probability as a fixed chance of success in $5$ independent trials, so the number of Grade A bearings $Y$ is binomial with $n=5$ , $p=0.1587$ and $q=0.8413$ :

P(Y=2)={}^{5}C_{2}(0.1587)^{2}(0.8413)^{3}=10\times 0.02519\times 0.59546\approx 0.1500

Answer

$P(\text{Grade A})=0.1587$ and $P(Y=2)\approx 0.1500$ . A normal-distribution probability can feed straight into a binomial $p$ once you treat each bearing as an independent trial with the same two outcomes, Grade A or not.

Q8[5 marks]

In a certain school, the scores of a Mathematics test are normally distributed with mean $65$ and standard deviation $5$ . Only the top $0.13\%$ of students are awarded a Best Student certificate.

(a) Find the minimum score needed to receive the certificate. (b) Estimate the number of students who receive the certificate, out of $800$ students who sat for the test.

Show worked solution

(a) The top $0.13\%$ is a right tail of $0.0013$ , which matches $P(Z>3)=1-0.9987=0.0013$ . Standardise the unknown minimum score $x$ :

\frac{x-65}{5}=3 \;\Rightarrow\; x=65+3(5)=80

(b) Out of a large group, the expected number meeting a condition is the probability multiplied by the group size:

800\times 0.0013=1.04

Answer

The minimum score is $80$ , and about $1$ student (from $1.04$ , rounded to the nearest whole student) receives the certificate. A tail probability this small on a modest group size should give an expected count near $1$ , which is exactly what the calculation shows.

Key method points

These three examples show how the chapter's tools work in reverse and in combination. Keep the following points in mind as you practise.

Divide variance by mean to isolate $q$ : $\dfrac{npq}{np}=q$ ; then $p=1-q$ and $n=\dfrac{\text{mean}}{p}$ .
Two normal probabilities give two equations in $\mu$ and $\sigma$ ; eliminate $\mu$ by subtraction.
Convert every probability to a left-area before reading a $Z$ -score, and use symmetry for negatives.
For 'at least one', set $1-q^{n}>$ target, giving $q^{n}<$ a bound.
Taking logs of a value below $1$ gives a negative number, so the inequality flips when you divide.
Always confirm the boundary case, here $n=4$ and $n=5$ , so the 'smallest' is genuinely correct.

How a teacher helps

Hard Probability Distribution questions are less about new formulas and more about setting up the right equations, which is where a teacher earns their keep. One-to-one, we slow the opening down, what does 'mean and variance' let you recover, which probability becomes which $Z$ -score, when does an inequality flip, so the plan is sound before the algebra starts.

Then we check the boundary case together, the step students most often skip. Our teachers are experienced.

Lessons are taught in English, while SPM papers are set in Malay and English, so every symbol stays familiar in both.

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Frequently asked questions

How do I get

n

and

p

from the mean and variance?

Divide: $\dfrac{\text{variance}}{\text{mean}}=\dfrac{npq}{np}=q$ . Then $p=1-q$ , and $n=\dfrac{\text{mean}}{p}$ .

Two short steps recover both parameters.

Why subtract the two equations in the normal problem?

Both equations contain $-\mu$ . Subtracting one from the other cancels $\mu$ and leaves a single equation in $\sigma$ , which you solve first, then back-substitute for $\mu$ .

Why does the inequality flip when I take logarithms?

$\lg 0.6$ is negative. Dividing or multiplying an inequality by a negative number reverses its direction, so $n\,\lg 0.6<\lg 0.1$ becomes $n>\dfrac{\lg 0.1}{\lg 0.6}$ .

Once I get

n>4.51

, why is the answer

5

and not

4

$n$ counts trials, so it must be a whole number. The smallest whole number greater than $4.51$ is $5$ ; checking $n=4$ gives only $0.8704$ , which fails the requirement.

Source:SRC-DSKP-EN

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

Probability Distribution, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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