Sec 2 Maths: Probability of Single Events, practice questions & worked solutions

Q: How do I find a probability when items are added to a bag?

Add x to both the number of favourable items and the total, set the fraction equal to the given probability, then cross-multiply and solve. For example, (6+x)/(24+x) = 1/3 gives x = 3.

Q: What is the complement rule?

The probability that an event does not happen is 1 minus the probability that it does: P(not A) = 1 - P(A). It is often quicker, for example P(not red) = 1 - 10/24 = 7/12.

Probability questions from 2025 Singapore Secondary 2 End-of-Year examination papers. They run from finding the probability of drawing a coloured marble to forming an equation when items are added or removed, and every question has a full step-by-step worked solution.

About this topic & key rules

In Secondary 2 Mathematics, probability is introduced through single events with equally likely outcomes, such as drawing one ball from a bag. The questions below practise the basic formula, the complement rule $P(\text{not }A)=1-P(A)$ , and a common exam type: some items are added to or removed from the bag, the new probability is given, and you form and solve an equation to find how many.

In those questions, the key step is to write down the new number of favourable outcomes and the new total before setting up the fraction. The worked solutions below do this on one line, then cross-multiply. For a structured programme covering this topic, see our Sec 2 Maths tuition.

Key rules

Probability of an event: $P(A)=\dfrac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}$ .
Range: $0\leq P(A)\leq 1$ ; an impossible event has probability $0$ and a certain event has probability $1$ .
Complement: $P(\text{not }A)=1-P(A)$ .
Adding or removing items: both the favourable count and the total change, for example $\dfrac{6+x}{24+x}$ after adding $x$ green marbles.
Simplify: give probabilities as fractions in their lowest terms unless told otherwise.

Questions & worked solutions

Q1Greenridge Secondary2025EOY P1 Q11●○○ Foundational4 marks

Probability of a single event, its complement, and adding items

(a)

A bag contains 10 red marbles, 7 blue marbles, 6 green marbles and 1 white marble. A marble is drawn at random. Find the probability that the marble is

(i)

blue,

(ii)

not red.

(b)

$x$ more green marbles were added in the bag. Given that the probability of drawing a green marble is $\dfrac13$ after the addition, find the value of $x$ .

Show worked solution▾

Total number of marbles $=10+7+6+1=24$ .

(a)(i)

$P(\text{blue})=\frac{7}{24}$

(a)(ii)

$\begin{aligned}P(\text{not red})&=1-\frac{10}{24}\\ &=\frac{14}{24}\\ &=\frac{7}{12}\end{aligned}$

(b)

After adding $x$ green marbles there are $6+x$ green marbles out of $24+x$ .

$\begin{aligned}\frac{6+x}{24+x}&=\frac13\\ 3(6+x)&=24+x\\ 18+3x&=24+x\\ 2x&=6\\ x&=3\end{aligned}$

Check: $\dfrac{9}{27}=\dfrac13$ .

Q2Regent Secondary2025EOY P1 Q9●○○ Foundational3 marks

Adding items to reach a given probability

A bag contains 5 blue balls, 8 green balls and 3 yellow balls.

(a)

A ball is drawn at random. Find the probability of getting a green ball.

(b)

$x$ yellow balls are added to the bag. The probability of getting a yellow ball becomes $\dfrac12$ . Find the value of $x$ .

Show worked solution▾

Total number of balls $=5+8+3=16$ .

(a)

$\begin{aligned}P(\text{green})&=\frac{8}{16}\\ &=\frac12\end{aligned}$

(b)

After adding $x$ yellow balls there are $3+x$ yellow balls out of $16+x$ .

$\begin{aligned}\frac{3+x}{16+x}&=\frac12\\ 2(3+x)&=16+x\\ 6+2x&=16+x\\ x&=10\end{aligned}$

Check: $\dfrac{13}{26}=\dfrac12$ .

Q3Ngee Ann Secondary2025EOY Q3●●○ Standard4 marks

Probability of either of two outcomes, then adding items

A bag contains 18 coloured balls, of which 6 are red, 4 are yellow and the rest are purple. A ball is chosen at random from the bag.

(a)

Find the probability that either a red or yellow ball is chosen.

(b)

$x$ red balls are added into the bag such that the probability of choosing a red ball becomes $\dfrac{2}{5}$ .

Find the value of $x$ .

Show worked solution▾

(a)

There are $6+4=10$ balls that are red or yellow.

$\begin{aligned}P(\text{red or yellow})&=\frac{10}{18}\\ &=\frac{5}{9}\end{aligned}$

(b)

After adding $x$ red balls there are $6+x$ red balls out of $18+x$ .

$\begin{aligned}\frac{6+x}{18+x}&=\frac{2}{5}\\ 5(6+x)&=2(18+x)\\ 30+5x&=36+2x\\ 3x&=6\\ x&=2\end{aligned}$

Check: $\dfrac{8}{20}=\dfrac25$ .

Q4Queensway Secondary2025EOY P2 Q2●●○ Standard2 marks

Removing items to reach a given probability

There are 24 marbles in a jar. 15 of the marbles are blue, and the rest are green. After $x$ green marbles are removed from the jar, the probability of picking a green marble is $\dfrac14$ .

By forming an equation in terms of $x$ , find the value of $x$ .

Show worked solution▾

There are $24-15=9$ green marbles. After removing $x$ of them there are $9-x$ green marbles out of $24-x$ .

$\begin{aligned}\frac{9-x}{24-x}&=\frac14\\ 4(9-x)&=24-x\\ 36-4x&=24-x\\ 12&=3x\\ x&=4\end{aligned}$

Check: $\dfrac{5}{20}=\dfrac14$ .

Q5Chung Cheng High (Yishun)2025EOY Q6●●○ Standard2 marks

Two groups added, forming a probability equation

There are 25 adults and 7 children in the theatre room. After another $2y$ adults and $y$ children entered the theatre room, the probability of selecting a child at random becomes $\dfrac{2}{7}$ . Find the value of $y$ .

Show worked solution▾

After the new arrivals there are $7+y$ children and $25+2y$ adults, so the total is $25+7+2y+y=32+3y$ people.

$\begin{aligned}\frac{7+y}{32+3y}&=\frac{2}{7}\\ 7(7+y)&=2(32+3y)\\ 49+7y&=64+6y\\ y&=15\end{aligned}$

Check: $22$ children out of $77$ people, and $\dfrac{22}{77}=\dfrac27$ .

Q6Woodgrove Secondary2025EOY P1 Q9●●○ Standard4 marks

Probability from fractions of two groups

Among a group of 45 students, 30 are girls. $\dfrac{3}{5}$ of the girls and $\dfrac{4}{5}$ of the boys passed a recent science test. A student is selected randomly. Find, as a fraction in its lowest terms, the probability that the student

(a)

fail the test,

(b)

is a boy who passed the test,

(c)

is a girl who failed the test.

Show worked solution▾

There are $45-30=15$ boys. Work out each group first:

$\begin{aligned}\text{Girls who passed}&=\tfrac35\times 30=18,\quad\text{girls who failed}=30-18=12\\ \text{Boys who passed}&=\tfrac45\times 15=12,\quad\text{boys who failed}=15-12=3\end{aligned}$

(a)

$\begin{aligned}P(\text{fail})&=\frac{12+3}{45}\\ &=\frac{15}{45}\\ &=\frac13\end{aligned}$

(b)

$\begin{aligned}P(\text{boy who passed})&=\frac{12}{45}\\ &=\frac{4}{15}\end{aligned}$

(c)

$\begin{aligned}P(\text{girl who failed})&=\frac{12}{45}\\ &=\frac{4}{15}\end{aligned}$

Not sure how to count the outcomes before you write a probability? Listing outcomes carefully is exactly what we practise in our Secondary 2 Maths programme, every method behind these questions taught step by step.

Frequently asked questions

How do I find a probability when items are added to a bag?▾

Add

x

to both the number of favourable items and the total, set the fraction equal to the given probability, then cross-multiply and solve. For example,

\dfrac{6+x}{24+x}=\dfrac13

gives

x=3

, as in Q1 above.

What is the complement rule?▾

The probability that an event does not happen is

1

minus the probability that it does:

P(\text{not }A)=1-P(A)

. It is often quicker, for example

P(\text{not red})=1-\dfrac{10}{24}=\dfrac{7}{12}

Should probabilities be given as fractions or decimals?▾

Follow the question. If no form is stated, a fraction in its lowest terms is the usual choice at Secondary 2, since it is exact.

Sec 2 Maths: Probability of Single Events, practice questions & worked solutions

Key rules

Questions & worked solutions

Frequently asked questions

Related Sec 2 topics