Sec 4 G2 Maths Probability worksheet with answers

A free Sec 4 G2 Maths worksheet on probability: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 G2 Maths worksheet: Probability

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The probability that Ali passes his driving test is

0.45

and the probability that Ben passes his driving test is

0.2

. The two events are independent. Find the probability that Ali passes his driving test but Ben fails his driving test.

Answer[2]

+	1	2	3	4	5	6
1
2
3
4
5
6

First die down the side, second die across the top

Two fair dice are thrown. Using a possibility diagram, find the probability that the sum of the two numbers is

8

Answer[2]

A group of

40

students is made up of

15

boys and

25

girls.

\dfrac{4}{5}

of the boys and

\dfrac{3}{5}

of the girls passed a swimming test. One of the students is chosen at random. Find the probability that the student passed.

Answer[2]

(a)A card is drawn at random from cards numbered

1

15

. State whether the events "a prime number" and "greater than 10" are mutually exclusive.

Answer (a)[1]

(b)

A

and

B

are mutually exclusive events with

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

and

P(B) = \dfrac{1}{4}

. Find the probability that neither

A

nor

B

occurs.

Answer (b)[2]

Section B

The probability that Mei takes the bus to school (otherwise she walks) is

0.55

. The probability that Mei is late is

x

if she takes the bus and

0.2

if she walks. The tree diagram shows this information.

(a)The probability that Mei takes the bus and is late is

0.165

. Find the value of

x

Answer (a)[1]

(b)Find the probability that Mei is late.

Answer (b)[2]

(c)Using your answer to part (b), write down the probability that Mei is on time.

Answer (c)[1]

(d)Assume that the outcomes are independent. Find the probability that, on two days, chosen at random, Mei is late on exactly one of the two days. Give your answer correct to 3 significant figures.

Answer (d)[2]

The table shows how the

31

members of a sports club chose one sport each. Give each probability as a fraction in its simplest form.

	Tennis	Badminton
Adults	12	9
Children	7	$x$

(a)One of the children is chosen at random. The probability that this child chose badminton is

\dfrac{3}{10}

. Find the value of

x

Answer (a)[2]

(b)One of the

31

members of a sports club is chosen at random. Find the probability that this person chose tennis.

Answer (b)[1]

(c)Two of the

31

members of a sports club are chosen at random. Find the probability that both are children.

Answer (c)[2]

(d)Three of the

31

members of a sports club are chosen at random. Find the probability that at least one of them chose badminton.

Answer (d)[2]

A box contains

5

blue,

4

black and

1

red pens. Pens are taken out at random, one at a time, without replacement. Give each answer as a fraction in its simplest form.

(a)Two pens are taken. Find the probability that both are black.

Answer (a)[2]

(b)Two pens are taken. Find the probability that they are of different colours.

Answer (b)[2]

(c)Instead, three pens are taken. Find the probability that all three are the same colour.

Answer (c)[2]

(d)Three pens are taken. Find the probability that exactly one of them is blue.

Answer (d)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

P = 0.36

[2]

P(\text{second does not occur}) = 1 - 0.2

\phantom{P(\text{second does not occur})} = 0.8

\mathstrut P = 0.45 \times 0.8

P = 0.36

Question 2

Answer:

P = \dfrac{5}{36}

[2]

There are

6 \times 6 = 36

equally likely outcomes.

(2, 6),\ (3, 5),\ (4, 4),\ (5, 3),\ (6, 2)

That is

5

outcomes.

P = \dfrac{5}{36}

Question 3

Answer:

P = \dfrac{27}{40}

[2]

boys:

\dfrac{4}{5} \times 15 = 12

girls:

\dfrac{3}{5} \times 25 = 15

12 + 15 = 27

P = \dfrac{27}{40}

Question 4

(a)Answer: NO[1]

11, 13

are both a prime number and greater than 10.No.

(b)Answer:

P(\text{neither}) = \dfrac{7}{12}

[2]

\mathstrut P(A \text{ or } B) = \dfrac{1}{6} + \dfrac{1}{4}

P(A \text{ or } B) = \dfrac{5}{12}

\mathstrut P(\text{neither}) = 1 - \dfrac{5}{12}

P(\text{neither}) = \dfrac{7}{12}

Question 5

(a)Answer:

x = 0.3

[1]

0.55 \times x = 0.165

x = \dfrac{0.165}{0.55}

x = 0.3

(b)Answer:

0.255

[2]

P = 0.55 \times 0.3 + 0.45 \times 0.2

= 0.165 + 0.09

= 0.255

(c)Answer:

0.745

[1]

P = 1 - 0.255

\phantom{P} = 0.745

(d)Answer:

0.380

[2]

P = 0.255 \times 0.745 + 0.745 \times 0.255

= 0.37995

= 0.380

(3 s.f.)

Question 6

(a)Answer:

x = 3

[2]

\dfrac{x}{7 + x} = \dfrac{3}{10}

x = 3

(b)Answer:

\dfrac{19}{31}

[1]

P = \dfrac{12 + 7}{31}

\phantom{P} = \dfrac{19}{31}

(c)Answer:

\dfrac{3}{31}

[2]

P = \dfrac{10}{31} \times \dfrac{9}{30}

\phantom{P} = \dfrac{3}{31}

(d)Answer:

\dfrac{3526}{4495}

[2]

12

people chose badminton; the other

19

chose tennis.

P(\text{none}) = \dfrac{19}{31} \times \dfrac{18}{30} \times \dfrac{17}{29}

\phantom{P(\text{none})} = \dfrac{969}{4495}

P(\text{at least one}) = 1 - \dfrac{969}{4495}

\phantom{P(\text{at least one})} = \dfrac{3526}{4495}

Question 7

(a)Answer:

\dfrac{2}{15}

[2]

P(\text{both black}) = \dfrac{4}{10} \times \dfrac{3}{9}

\phantom{P(\text{both black})} = \dfrac{2}{15}

(b)Answer:

\dfrac{29}{45}

[2]

P(\text{same}) = \dfrac{5}{10} \times \dfrac{4}{9} + \dfrac{4}{10} \times \dfrac{3}{9}

\phantom{P(\text{same})} = \dfrac{16}{45}

P(\text{different}) = 1 - \dfrac{16}{45}

\phantom{P(\text{different})} = \dfrac{29}{45}

(c)Answer:

\dfrac{7}{60}

[2]

There are only

1

red pens, so three red is impossible.

P(\text{all the same}) = \dfrac{5}{10} \times \dfrac{4}{9} \times \dfrac{3}{8} + \dfrac{4}{10} \times \dfrac{3}{9} \times \dfrac{2}{8}

\phantom{P(\text{all the same})} = \dfrac{7}{60}

(d)Answer:

\dfrac{5}{12}

[2]

The blue one can be taken first, second or third: each order has the same probability.

P = 3 \times \dfrac{5}{10} \times \dfrac{5}{9} \times \dfrac{4}{8}

\phantom{P} = \dfrac{5}{12}

Sec 4 G2 Maths Probability worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Probability notes. Or practise it online with step-by-step help when you get one wrong.

Section B

The probability that Mei takes the bus to school (otherwise she walks) is

0.55

. The probability that Mei is late is

x

if she takes the bus and

0.2

if she walks. The tree diagram shows this information.

(a)The probability that Mei takes the bus and is late is

0.165

. Find the value of

x

Answer (a)[1]

(b)Find the probability that Mei is late.

Answer (b)[2]

(c)Using your answer to part (b), write down the probability that Mei is on time.

Answer (c)[1]

(d)Assume that the outcomes are independent. Find the probability that, on two days, chosen at random, Mei is late on exactly one of the two days. Give your answer correct to 3 significant figures.

Answer (d)[2]

Tennis

Badminton

Adults

Children

x

Answers and worked solutions

Question 1

Answer:

P = 0.36

[2]

P(\text{second does not occur}) = 1 - 0.2

\phantom{P(\text{second does not occur})} = 0.8

\mathstrut P = 0.45 \times 0.8

P = 0.36

Question 2

Answer:

P = \dfrac{5}{36}

[2]

There are

6 \times 6 = 36

equally likely outcomes.

(2, 6),\ (3, 5),\ (4, 4),\ (5, 3),\ (6, 2)

That is

5

outcomes.

P = \dfrac{5}{36}

Question 3

Answer:

P = \dfrac{27}{40}

[2]

boys:

\dfrac{4}{5} \times 15 = 12

girls:

\dfrac{3}{5} \times 25 = 15

12 + 15 = 27

P = \dfrac{27}{40}

Question 4

(a)Answer: NO[1]

11, 13

are both a prime number and greater than 10.No.

(b)Answer:

P(\text{neither}) = \dfrac{7}{12}

[2]

\mathstrut P(A \text{ or } B) = \dfrac{1}{6} + \dfrac{1}{4}

P(A \text{ or } B) = \dfrac{5}{12}

\mathstrut P(\text{neither}) = 1 - \dfrac{5}{12}

P(\text{neither}) = \dfrac{7}{12}

Question 5

(a)Answer:

x = 0.3

[1]

0.55 \times x = 0.165

x = \dfrac{0.165}{0.55}

x = 0.3

(b)Answer:

0.255

[2]

P = 0.55 \times 0.3 + 0.45 \times 0.2

= 0.165 + 0.09

= 0.255

(c)Answer:

0.745

[1]

P = 1 - 0.255

\phantom{P} = 0.745

(d)Answer:

0.380

[2]

P = 0.255 \times 0.745 + 0.745 \times 0.255

= 0.37995

= 0.380

(3 s.f.)

Question 6

(a)Answer:

x = 3

[2]

\dfrac{x}{7 + x} = \dfrac{3}{10}

x = 3

(b)Answer:

\dfrac{19}{31}

[1]

P = \dfrac{12 + 7}{31}

\phantom{P} = \dfrac{19}{31}

(c)Answer:

\dfrac{3}{31}

[2]

P = \dfrac{10}{31} \times \dfrac{9}{30}

\phantom{P} = \dfrac{3}{31}

(d)Answer:

\dfrac{3526}{4495}

[2]

12

people chose badminton; the other

19

chose tennis.

P(\text{none}) = \dfrac{19}{31} \times \dfrac{18}{30} \times \dfrac{17}{29}

\phantom{P(\text{none})} = \dfrac{969}{4495}

P(\text{at least one}) = 1 - \dfrac{969}{4495}

\phantom{P(\text{at least one})} = \dfrac{3526}{4495}

Question 7

(a)Answer:

\dfrac{2}{15}

[2]

P(\text{both black}) = \dfrac{4}{10} \times \dfrac{3}{9}

\phantom{P(\text{both black})} = \dfrac{2}{15}

(b)Answer:

\dfrac{29}{45}

[2]

P(\text{same}) = \dfrac{5}{10} \times \dfrac{4}{9} + \dfrac{4}{10} \times \dfrac{3}{9}

\phantom{P(\text{same})} = \dfrac{16}{45}

P(\text{different}) = 1 - \dfrac{16}{45}

\phantom{P(\text{different})} = \dfrac{29}{45}

(c)Answer:

\dfrac{7}{60}

[2]

There are only

1

red pens, so three red is impossible.

P(\text{all the same}) = \dfrac{5}{10} \times \dfrac{4}{9} \times \dfrac{3}{8} + \dfrac{4}{10} \times \dfrac{3}{9} \times \dfrac{2}{8}

\phantom{P(\text{all the same})} = \dfrac{7}{60}

(d)Answer:

\dfrac{5}{12}

[2]

The blue one can be taken first, second or third: each order has the same probability.

P = 3 \times \dfrac{5}{10} \times \dfrac{5}{9} \times \dfrac{4}{8}

\phantom{P} = \dfrac{5}{12}