Sec 2 G1 Maths Probability worksheet with answers

A free Sec 2 G1 Maths worksheet on probability: 8 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 2 G1 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

Cards numbered

1

10

are shuffled and one is drawn at random. Find the probability that the number is a factor of 24.

Answer[2]

Cards numbered

1

25

are shuffled and one is drawn at random. Find the probability that the number is a multiple of 4.

Answer[2]

A bag contains

5

red,

6

blue and

9

green balls. A ball is taken at random. Find the probability that it is red or blue.

Answer[2]

(a)A bag contains

6

red,

2

blue and

4

green balls. A ball is taken at random. Find the probability that it is green or red.

Answer (a)[2]

(b)A bag contains

7

red,

5

blue,

3

green and

5

yellow balls. A ball is taken at random. Find the probability that it is not green.

Answer (b)[2]

(a)A bag contains

7

red,

5

blue,

3

green and

5

yellow balls. A ball is taken at random. Find the probability that it is red.

Answer (a)[2]

(b)A bag contains

3

red,

9

blue and

6

green balls. A ball is taken at random. Find the probability that it is green or red.

Answer (b)[2]

(a)A bag contains

8

red,

3

blue and

5

green balls. A ball is taken at random. Find the probability that it is not red.

Answer (a)[2]

(b)A bag contains

5

red,

3

blue and

4

green balls. A ball is taken at random. Find the probability that it is blue or green.

Answer (b)[2]

Section B

(a)A bag contains

6

red,

2

blue and

4

green balls. A ball is taken at random. Find the probability that it is green.

Answer (a)[2]

(b)Cards numbered

1

15

are shuffled and one is drawn at random. Find the probability that the number is a multiple of 4.

Answer (b)[2]

(c)A bag contains

8

red,

3

blue and

5

green balls. A ball is taken at random. Find the probability that it is not red.

Answer (c)[2]

(a)A bag contains

5

red,

3

blue and

4

green balls. A ball is taken at random. Find the probability that it is blue or green.

Answer (a)[2]

(b)(i)A bag contains

7

red,

5

blue,

3

green and

5

yellow balls. A ball is taken at random. Find the probability that it is not green.

Answer (b)(i)[2]

(b)(ii)A bag contains

3

red,

5

blue and

7

green balls. A ball is taken at random. Find the probability that it is not red.

Answer (b)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

P = \dfrac{3}{5}

[2]

1, 2, 3, 4, 6, 8

That is

6

numbers out of

10

\mathstrut P = \dfrac{6}{10}

P = \dfrac{3}{5}

Question 2

Answer:

P = \dfrac{6}{25}

[2]

4, 8, 12, 16, 20, 24

That is

6

numbers out of

25

P = \dfrac{6}{25}

Question 3

Answer:

P = \dfrac{11}{20}

[2]

\mathstrut \text{Favourable} = 5 + 6

\text{Favourable} = 11

P = \dfrac{11}{20}

Question 4

(a)Answer:

P = \dfrac{5}{6}

[2]

\mathstrut \text{Favourable} = 4 + 6

\text{Favourable} = 10

\mathstrut P = \dfrac{10}{12}

P = \dfrac{5}{6}

(b)Answer:

P = \dfrac{17}{20}

[2]

\mathstrut \text{Favourable} = 20 - 3

\text{Favourable} = 17

P = \dfrac{17}{20}

Question 5

(a)Answer:

\mathstrut P(\text{red}) = \dfrac{7}{20}

[2]

\mathstrut \text{Total} = 7 + 5 + 3 + 5

\text{Total} = 20

\mathstrut P(\text{red}) = \dfrac{7}{20}

(b)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 6 + 3

\text{Favourable} = 9

\mathstrut P = \dfrac{9}{18}

P = \dfrac{1}{2}

Question 6

(a)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 16 - 8

\text{Favourable} = 8

\mathstrut P = \dfrac{8}{16}

P = \dfrac{1}{2}

(b)Answer:

P = \dfrac{7}{12}

[2]

\mathstrut \text{Favourable} = 3 + 4

\text{Favourable} = 7

P = \dfrac{7}{12}

Question 7

(a)Answer:

P(\text{green}) = \dfrac{1}{3}

[2]

\mathstrut \text{Total} = 6 + 2 + 4

\text{Total} = 12

\mathstrut P(\text{green}) = \dfrac{4}{12}

P(\text{green}) = \dfrac{1}{3}

(b)Answer:

P = \dfrac{1}{5}

[2]

4, 8, 12

That is

3

numbers out of

15

\mathstrut P = \dfrac{3}{15}

P = \dfrac{1}{5}

(c)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 16 - 8

\text{Favourable} = 8

\mathstrut P = \dfrac{8}{16}

P = \dfrac{1}{2}

Question 8

(a)Answer:

P = \dfrac{7}{12}

[2]

\mathstrut \text{Favourable} = 3 + 4

\text{Favourable} = 7

P = \dfrac{7}{12}

(b)(i)Answer:

P = \dfrac{17}{20}

[2]

\mathstrut \text{Favourable} = 20 - 3

\text{Favourable} = 17

P = \dfrac{17}{20}

(b)(ii)Answer:

P = \dfrac{4}{5}

[2]

\mathstrut \text{Favourable} = 15 - 3

\text{Favourable} = 12

\mathstrut P = \dfrac{12}{15}

P = \dfrac{4}{5}

Sec 2 G1 Maths Probability worksheet with answers

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

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)A bag contains

6

red,

2

blue and

4

green balls. A ball is taken at random. Find the probability that it is green.

Answer (a)[2]

(b)Cards numbered

1

15

are shuffled and one is drawn at random. Find the probability that the number is a multiple of 4.

Answer (b)[2]

(c)A bag contains

8

red,

3

blue and

5

green balls. A ball is taken at random. Find the probability that it is not red.

Answer (c)[2]

Answers and worked solutions

Question 1

Answer:

P = \dfrac{3}{5}

[2]

1, 2, 3, 4, 6, 8

That is

6

numbers out of

10

\mathstrut P = \dfrac{6}{10}

P = \dfrac{3}{5}

Question 2

Answer:

P = \dfrac{6}{25}

[2]

4, 8, 12, 16, 20, 24

That is

6

numbers out of

25

P = \dfrac{6}{25}

Question 3

Answer:

P = \dfrac{11}{20}

[2]

\mathstrut \text{Favourable} = 5 + 6

\text{Favourable} = 11

P = \dfrac{11}{20}

Question 4

(a)Answer:

P = \dfrac{5}{6}

[2]

\mathstrut \text{Favourable} = 4 + 6

\text{Favourable} = 10

\mathstrut P = \dfrac{10}{12}

P = \dfrac{5}{6}

(b)Answer:

P = \dfrac{17}{20}

[2]

\mathstrut \text{Favourable} = 20 - 3

\text{Favourable} = 17

P = \dfrac{17}{20}

Question 5

(a)Answer:

\mathstrut P(\text{red}) = \dfrac{7}{20}

[2]

\mathstrut \text{Total} = 7 + 5 + 3 + 5

\text{Total} = 20

\mathstrut P(\text{red}) = \dfrac{7}{20}

(b)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 6 + 3

\text{Favourable} = 9

\mathstrut P = \dfrac{9}{18}

P = \dfrac{1}{2}

Question 6

(a)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 16 - 8

\text{Favourable} = 8

\mathstrut P = \dfrac{8}{16}

P = \dfrac{1}{2}

(b)Answer:

P = \dfrac{7}{12}

[2]

\mathstrut \text{Favourable} = 3 + 4

\text{Favourable} = 7

P = \dfrac{7}{12}

Question 7

(a)Answer:

P(\text{green}) = \dfrac{1}{3}

[2]

\mathstrut \text{Total} = 6 + 2 + 4

\text{Total} = 12

\mathstrut P(\text{green}) = \dfrac{4}{12}

P(\text{green}) = \dfrac{1}{3}

(b)Answer:

P = \dfrac{1}{5}

[2]

4, 8, 12

That is

3

numbers out of

15

\mathstrut P = \dfrac{3}{15}

P = \dfrac{1}{5}

(c)Answer:

P = \dfrac{1}{2}

[2]

\mathstrut \text{Favourable} = 16 - 8

\text{Favourable} = 8

\mathstrut P = \dfrac{8}{16}

P = \dfrac{1}{2}

Question 8

(a)Answer:

P = \dfrac{7}{12}

[2]

\mathstrut \text{Favourable} = 3 + 4

\text{Favourable} = 7

P = \dfrac{7}{12}

(b)(i)Answer:

P = \dfrac{17}{20}

[2]

\mathstrut \text{Favourable} = 20 - 3

\text{Favourable} = 17

P = \dfrac{17}{20}

(b)(ii)Answer:

P = \dfrac{4}{5}

[2]

\mathstrut \text{Favourable} = 15 - 3

\text{Favourable} = 12

\mathstrut P = \dfrac{12}{15}

P = \dfrac{4}{5}