Sec 4 · G3 (Express)

Probability

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1. Probability of an event

Rule	Formula
probability	$P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
range	$0 \leq P(E) \leq 1$
not happening	$P(E') = 1 - P(E)$ , where $E'$ means $E$ does not happen

2 red, 7 blue, 3 green: P(blue)

\displaystyle P(\text{blue}) = \dfrac{7}{12}

total: $2 + 7 + 3$ , which is $12$

P(not blue)

\displaystyle 1 - P(\text{blue})

e.g.

\displaystyle 1 - \dfrac{7}{12}

\displaystyle = \dfrac{5}{12}

6 red balls and P(red) known: the total n

\displaystyle \dfrac{6}{n} = \dfrac{2}{5}

e.g.

\displaystyle \dfrac{6}{n} = \dfrac{2}{5}

\displaystyle n = 15

blue balls: $15 - 6$ , which is $9$

📖 Worked example

Cards numbered $1$ to $15$ are shuffled and one is drawn at random. Find the probability that the number is a prime number.

Step 1 of 2

List the numbers that are a prime number

2, 3, 5, 7, 11, 13

That is

6

numbers out of

15

📖 Worked example

A bag contains $3$ red balls and $5$ blue balls. $2x$ red balls and $x$ blue balls are put into the bag. A ball is then taken at random. The probability that it is red is $\dfrac{13}{23}$ . Find the value of $x$ .

Step 1 of 3

The numbers of balls after the change, in terms of x

Red balls:

3 + 2x

; all the balls:

8 + 3x

📖 Worked example

The two-way table shows the students in a class. A student is chosen at random. Find the probability that the student is a boy who wears glasses.

	Wear glasses	Do not wear glasses	Total
Boys	12	8	20
Girls	9	11	20
Total	21	19	40

Step 1 of 3

Count the boys who wears glasses

From the table:

12

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2. Possibility diagrams

$+$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$\mathbf{9}$
$4$	$5$	$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$
$5$	$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$	$\mathbf{11}$
$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$	$\mathbf{11}$	$\mathbf{12}$

Two dice: P(sum > 8), the 10 bold outcomes out of 36

\displaystyle P = \dfrac{10}{36}

e.g.

\displaystyle \dfrac{10}{36}

\displaystyle = \dfrac{5}{18}

📖 Worked example

Spinner A has the numbers $1, 2, 3$ and spinner B has $0, 1, 2, 3, 4$ . Both are spun. Using a possibility diagram, find the probability that the sum of the two numbers is greater than $5$ .

+	0	1	2	3	4
1
2
3

Spinner A down the side, spinner B across the top

Step 1 of 3

Fill in the possibility diagram

There are

3 \times 5 = 15

equally likely outcomes.

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3. Tree diagrams

Rule	Example: 2 red, 5 blue, without replacement
along a branch: multiply	$P(RR) = \dfrac{2}{7} \times \dfrac{1}{6}$ , which is $\dfrac{1}{21}$
between branches: add	$P(\text{one of each}) = P(RB) + P(BR)$
each pair of branches adds up to $1$	$\dfrac{2}{7} + \dfrac{5}{7} = 1$

With replacement: the second draw is the same

\displaystyle P(RR)

e.g.

\displaystyle \dfrac{2}{7} \times \dfrac{2}{7}

\displaystyle = \dfrac{4}{49}

Without replacement: one fewer ball

\displaystyle P(\text{one of each})

e.g.

\displaystyle \dfrac{2}{7} \times \dfrac{5}{6} + \dfrac{5}{7} \times \dfrac{2}{6}

\displaystyle = \dfrac{10}{21}

📖 Worked example

A bag contains $2$ red and $5$ blue balls. Two balls are taken at random, one after the other, without replacement. The tree diagram shows the outcomes. Find the probability that both balls are blue.

Step 1 of 2

Read the probability on each branch of the tree

First:

P(R) = \dfrac{2}{7}

P(B) = \dfrac{5}{7}

.After a red:

P(R) = \dfrac{1}{6}

P(B) = \dfrac{5}{6}

.After a blue:

P(R) = \dfrac{1}{3}

P(B) = \dfrac{2}{3}

📖 Worked example

A bag contains $2$ red and $5$ blue balls. Two balls are taken at random, one after the other, without replacement. Draw a tree diagram, then find the probability that one ball of each colour is taken.

Step 1 of 3

Complete the tree: the first ball, then the second ball after each colour

An unknown number n of red balls (with 3 blue): P(both red) known

\displaystyle \dfrac{n}{n + 3} \times \dfrac{n - 1}{n + 2} = \dfrac{2}{7}

e.g.

\displaystyle 7n(n - 1) = 2(n + 3)(n + 2)

\displaystyle 5n^2 - 17n - 12 = 0

\displaystyle (n - 4)(5n + 3) = 0

$n = 4$ ; reject $n = -\dfrac{3}{5}$ (a number of balls)

📖 Worked example

A bag contains $n$ red balls and $5$ blue balls. Two balls are taken at random, one after the other, without replacement. The probability that both balls are red is $\dfrac{3}{28}$ . Find the total number of balls in the bag.

Step 1 of 4

Multiply along the red-red branch: n red out of n + k, then one red fewer and one ball fewer

\dfrac{n}{n + 5} \times \dfrac{n - 1}{n + 4} = \dfrac{3}{28}

📖 Worked example

A bag contains $n$ red balls and $2$ blue balls. Three balls are taken at random, one after the other, without replacement. The probability that all three balls are red is $\dfrac{1}{5}$ . Show that $2n^2 - 9n + 4 = 0$ , and hence find the number of red balls in the bag.

Step 1 of 4

Multiply along the red-red-red branch: one red fewer and one ball fewer each time

\dfrac{n}{n + 2} \times \dfrac{n - 1}{n + 1} \times \dfrac{n - 2}{n} = \dfrac{1}{5}

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4. Mutually exclusive and independent events

Events	Meaning	Rule
mutually exclusive	cannot both happen	$P(A \text{ or } B) = P(A) + P(B)$
independent	one does not change the other	$P(A \text{ and } B) = P(A) \times P(B)$
at least one		$1 - P(\text{neither})$

Mutually exclusive: neither A nor B

\displaystyle 1 - (P(A) + P(B))

e.g.

\displaystyle 1 - \left(\dfrac{2}{5} + \dfrac{1}{5}\right)

\displaystyle = \dfrac{2}{5}

Independent, P(A) = 0.7 and P(B) = 0.4: A but not B

\displaystyle P(A) \times (1 - P(B))

e.g.

\displaystyle 0.7 \times (1 - 0.4)

\displaystyle = 0.42

Independent: at least one

\displaystyle 1 - P(\text{neither})

e.g.

\displaystyle 1 - (1 - 0.7)(1 - 0.4)

\displaystyle = 0.82

📖 Worked example

$A$ and $B$ are mutually exclusive events with $P(A) = \dfrac{2}{5}$ and $P(B) = \dfrac{1}{5}$ . Find the probability that neither $A$ nor $B$ occurs.

Step 1 of 2

P(A or B) for mutually exclusive events

\mathstrut P(A \text{ or } B) = \dfrac{2}{5} + \dfrac{1}{5}

P(A \text{ or } B) = \dfrac{3}{5}

📖 Worked example

The probability that it rains tomorrow is $0.7$ and the probability that the bus is late is $0.4$ . The two events are independent. Find the probability that at least one of the two events occurs.

Step 1 of 3

The probability that each event does not occur

P(\text{first does not occur}) = 1 - 0.7

\phantom{P(\text{first does not occur})} = 0.3

P(\text{second does not occur}) = 1 - 0.4

\phantom{P(\text{second does not occur})} = 0.6

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Sec 4 · G3 (Express)

Probability

Practise probability with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Probability of an event

Rule	Formula
probability	$P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
range	$0 \leq P(E) \leq 1$
not happening	$P(E') = 1 - P(E)$ , where $E'$ means $E$ does not happen

2 red, 7 blue, 3 green: P(blue)

\displaystyle P(\text{blue}) = \dfrac{7}{12}

total: $2 + 7 + 3$ , which is $12$

P(not blue)

\displaystyle 1 - P(\text{blue})

e.g.

\displaystyle 1 - \dfrac{7}{12}

\displaystyle = \dfrac{5}{12}

6 red balls and P(red) known: the total n

\displaystyle \dfrac{6}{n} = \dfrac{2}{5}

e.g.

\displaystyle \dfrac{6}{n} = \dfrac{2}{5}

\displaystyle n = 15

blue balls: $15 - 6$ , which is $9$

📖 Worked example

Cards numbered $1$ to $15$ are shuffled and one is drawn at random. Find the probability that the number is a prime number.

Step 1 of 2

List the numbers that are a prime number

2, 3, 5, 7, 11, 13

That is

6

numbers out of

15

📖 Worked example

Step 1 of 3

The numbers of balls after the change, in terms of x

Red balls:

3 + 2x

; all the balls:

8 + 3x

📖 Worked example

The two-way table shows the students in a class. A student is chosen at random. Find the probability that the student is a boy who wears glasses.

	Wear glasses	Do not wear glasses	Total
Boys	12	8	20
Girls	9	11	20
Total	21	19	40

Step 1 of 3

Count the boys who wears glasses

From the table:

12

Practise this →

2. Possibility diagrams

$+$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$\mathbf{9}$
$4$	$5$	$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$
$5$	$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$	$\mathbf{11}$
$6$	$7$	$8$	$\mathbf{9}$	$\mathbf{10}$	$\mathbf{11}$	$\mathbf{12}$

Two dice: P(sum > 8), the 10 bold outcomes out of 36

\displaystyle P = \dfrac{10}{36}

e.g.

\displaystyle \dfrac{10}{36}

\displaystyle = \dfrac{5}{18}

📖 Worked example

Spinner A has the numbers $1, 2, 3$ and spinner B has $0, 1, 2, 3, 4$ . Both are spun. Using a possibility diagram, find the probability that the sum of the two numbers is greater than $5$ .

+	0	1	2	3	4
1
2
3

Spinner A down the side, spinner B across the top

Step 1 of 3

Fill in the possibility diagram

There are

3 \times 5 = 15

equally likely outcomes.

Practise this →

3. Tree diagrams

Rule	Example: 2 red, 5 blue, without replacement
along a branch: multiply	$P(RR) = \dfrac{2}{7} \times \dfrac{1}{6}$ , which is $\dfrac{1}{21}$
between branches: add	$P(\text{one of each}) = P(RB) + P(BR)$
each pair of branches adds up to $1$	$\dfrac{2}{7} + \dfrac{5}{7} = 1$

With replacement: the second draw is the same

\displaystyle P(RR)

e.g.

\displaystyle \dfrac{2}{7} \times \dfrac{2}{7}

\displaystyle = \dfrac{4}{49}

Without replacement: one fewer ball

\displaystyle P(\text{one of each})

e.g.

\displaystyle \dfrac{2}{7} \times \dfrac{5}{6} + \dfrac{5}{7} \times \dfrac{2}{6}

\displaystyle = \dfrac{10}{21}

📖 Worked example

A bag contains $2$ red and $5$ blue balls. Two balls are taken at random, one after the other, without replacement. The tree diagram shows the outcomes. Find the probability that both balls are blue.

Step 1 of 2

Read the probability on each branch of the tree

First:

P(R) = \dfrac{2}{7}

P(B) = \dfrac{5}{7}

.After a red:

P(R) = \dfrac{1}{6}

P(B) = \dfrac{5}{6}

.After a blue:

P(R) = \dfrac{1}{3}

P(B) = \dfrac{2}{3}

📖 Worked example

Step 1 of 3

Complete the tree: the first ball, then the second ball after each colour

An unknown number n of red balls (with 3 blue): P(both red) known

\displaystyle \dfrac{n}{n + 3} \times \dfrac{n - 1}{n + 2} = \dfrac{2}{7}

e.g.

\displaystyle 7n(n - 1) = 2(n + 3)(n + 2)

\displaystyle 5n^2 - 17n - 12 = 0

\displaystyle (n - 4)(5n + 3) = 0

$n = 4$ ; reject $n = -\dfrac{3}{5}$ (a number of balls)

📖 Worked example

Step 1 of 4

Multiply along the red-red branch: n red out of n + k, then one red fewer and one ball fewer

\dfrac{n}{n + 5} \times \dfrac{n - 1}{n + 4} = \dfrac{3}{28}

📖 Worked example

Step 1 of 4

Multiply along the red-red-red branch: one red fewer and one ball fewer each time

\dfrac{n}{n + 2} \times \dfrac{n - 1}{n + 1} \times \dfrac{n - 2}{n} = \dfrac{1}{5}

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4. Mutually exclusive and independent events

Events	Meaning	Rule
mutually exclusive	cannot both happen	$P(A \text{ or } B) = P(A) + P(B)$
independent	one does not change the other	$P(A \text{ and } B) = P(A) \times P(B)$
at least one		$1 - P(\text{neither})$

Mutually exclusive: neither A nor B

\displaystyle 1 - (P(A) + P(B))

e.g.

\displaystyle 1 - \left(\dfrac{2}{5} + \dfrac{1}{5}\right)

\displaystyle = \dfrac{2}{5}

Independent, P(A) = 0.7 and P(B) = 0.4: A but not B

\displaystyle P(A) \times (1 - P(B))

e.g.

\displaystyle 0.7 \times (1 - 0.4)

\displaystyle = 0.42

Independent: at least one

\displaystyle 1 - P(\text{neither})

e.g.

\displaystyle 1 - (1 - 0.7)(1 - 0.4)

\displaystyle = 0.82

📖 Worked example

$A$ and $B$ are mutually exclusive events with $P(A) = \dfrac{2}{5}$ and $P(B) = \dfrac{1}{5}$ . Find the probability that neither $A$ nor $B$ occurs.

Step 1 of 2

P(A or B) for mutually exclusive events

\mathstrut P(A \text{ or } B) = \dfrac{2}{5} + \dfrac{1}{5}

P(A \text{ or } B) = \dfrac{3}{5}

📖 Worked example

The probability that it rains tomorrow is $0.7$ and the probability that the bus is late is $0.4$ . The two events are independent. Find the probability that at least one of the two events occurs.

Step 1 of 3

The probability that each event does not occur

P(\text{first does not occur}) = 1 - 0.7

\phantom{P(\text{first does not occur})} = 0.3

P(\text{second does not occur}) = 1 - 0.4

\phantom{P(\text{second does not occur})} = 0.6

Practise this →

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