Sec 4 E-Math Probability worksheet with answers

A free Sec 4 E-Math 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 E-Math worksheet: Probability

Total: 30 marksSuggested time: 45 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

+	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

11

Answer[2]

(a)A bag contains

6

red and

4

blue balls. Two balls are taken at random, the first being put back before the second is taken. Draw a tree diagram, then find the probability that both balls are red.

Answer (a)[1]

(b)A bag contains

7

red and

3

blue balls. Two balls are taken at random, the first being put back before the second is taken. Draw a tree diagram, then find the probability that both balls are blue.

Answer (b)[1]

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 green.

Answer[2]

(a)The probability that a randomly chosen car is white is

0.35

and the probability that it is black is

0.45

. Find the probability that it is white or black.

Answer (a)[2]

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

1

15

. State whether the events "an odd number" and "a multiple of 3" are mutually exclusive.

Answer (b)[1]

Section B

A bag contains

7

red and

3

blue balls. Two balls are taken at random, one after the other, without replacement.

(a)Find the probability that both balls are red.

Answer (a)[2]

(b)Find the probability that one ball of each colour is taken.

Answer (b)[2]

(c)Find the probability that at least one ball is red.

Answer (c)[2]

The table shows how the

28

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	4	$x$

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

\dfrac{3}{7}

. Find the value of

x

Answer (a)[2]

(b)One of the

28

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

Answer (b)[1]

(c)Two of the

28

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

Answer (c)[2]

(d)Three of the

28

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

6

blue,

1

black and

4

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 red.

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 black.

Answer (d)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

P = \dfrac{1}{18}

[2]

There are

6 \times 6 = 36

equally likely outcomes.

(5, 6),\ (6, 5)

That is

2

outcomes.

\mathstrut P = \dfrac{2}{36}

P = \dfrac{1}{18}

Question 2

(a)Answer:

P(RR) = \dfrac{9}{25}

[1]

\mathstrut P(RR) = \dfrac{3}{5} \times \dfrac{3}{5}

P(RR) = \dfrac{9}{25}

(b)Answer:

P(BB) = \dfrac{9}{100}

[1]

\mathstrut P(BB) = \dfrac{3}{10} \times \dfrac{3}{10}

P(BB) = \dfrac{9}{100}

Question 3

Answer:

\mathstrut P(\text{green}) = \dfrac{3}{20}

[2]

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

\text{Total} = 20

\mathstrut P(\text{green}) = \dfrac{3}{20}

Question 4

(a)Answer:

P = 0.8

[2]

The two can't both happen: they are mutually exclusive, so add.

P(A \text{ or } B) = P(A) + P(B)

\mathstrut P = 0.35 + 0.45

P = 0.8

(b)Answer: NO[1]

3, 9, 15

are both an odd number and a multiple of 3.No.

Question 5

(a)Answer:

\dfrac{7}{15}

[2]

P(RR) = \dfrac{7}{10} \times \dfrac{6}{9}

\phantom{P(RR)} = \dfrac{7}{15}

(b)Answer:

\dfrac{7}{15}

[2]

P(RB) + P(BR) = \dfrac{7}{10} \times \dfrac{3}{9} + \dfrac{3}{10} \times \dfrac{7}{9}

\phantom{P(RB) + P(BR)} = \dfrac{7}{15}

(c)Answer:

\dfrac{14}{15}

[2]

P(\text{at least one red}) = 1 - P(BB)

\phantom{P(\text{at least one red})} = 1 - \dfrac{3}{10} \times \dfrac{2}{9}

\phantom{P(\text{at least one red})} = \dfrac{14}{15}

Question 6

(a)Answer:

x = 3

[2]

\dfrac{x}{4 + x} = \dfrac{3}{7}

x = 3

(b)Answer:

\dfrac{4}{7}

[1]

P = \dfrac{12 + 4}{28}

\phantom{P} = \dfrac{4}{7}

(c)Answer:

\dfrac{1}{18}

[2]

P = \dfrac{7}{28} \times \dfrac{6}{27}

\phantom{P} = \dfrac{1}{18}

(d)Answer:

\dfrac{97}{117}

[2]

12

people chose badminton; the other

16

chose tennis.

P(\text{none}) = \dfrac{16}{28} \times \dfrac{15}{27} \times \dfrac{14}{26}

\phantom{P(\text{none})} = \dfrac{20}{117}

P(\text{at least one}) = 1 - \dfrac{20}{117}

\phantom{P(\text{at least one})} = \dfrac{97}{117}

Question 7

(a)Answer:

\dfrac{6}{55}

[2]

P(\text{both red}) = \dfrac{4}{11} \times \dfrac{3}{10}

\phantom{P(\text{both red})} = \dfrac{6}{55}

(b)Answer:

\dfrac{34}{55}

[2]

P(\text{same}) = \dfrac{6}{11} \times \dfrac{5}{10} + \dfrac{4}{11} \times \dfrac{3}{10}

\phantom{P(\text{same})} = \dfrac{21}{55}

P(\text{different}) = 1 - \dfrac{21}{55}

\phantom{P(\text{different})} = \dfrac{34}{55}

(c)Answer:

\dfrac{8}{55}

[2]

There are only

1

black pens, so three black is impossible.

P(\text{all the same}) = \dfrac{6}{11} \times \dfrac{5}{10} \times \dfrac{4}{9} + \dfrac{4}{11} \times \dfrac{3}{10} \times \dfrac{2}{9}

\phantom{P(\text{all the same})} = \dfrac{8}{55}

(d)Answer:

\dfrac{3}{11}

[2]

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

P = 3 \times \dfrac{1}{11} \times \dfrac{10}{10} \times \dfrac{9}{9}

\phantom{P} = \dfrac{3}{11}

Practice questions by skill

Sec 4 E-Math 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

A bag contains

7

red and

3

blue balls. Two balls are taken at random, one after the other, without replacement.

(a)Find the probability that both balls are red.

Answer (a)[2]

(b)Find the probability that one ball of each colour is taken.

Answer (b)[2]

(c)Find the probability that at least one ball is red.

Answer (c)[2]

Tennis

Badminton

Adults

Children

x

Answers and worked solutions

Question 1

Answer:

P = \dfrac{1}{18}

[2]

There are

6 \times 6 = 36

equally likely outcomes.

(5, 6),\ (6, 5)

That is

2

outcomes.

\mathstrut P = \dfrac{2}{36}

P = \dfrac{1}{18}

Question 2

(a)Answer:

P(RR) = \dfrac{9}{25}

[1]

\mathstrut P(RR) = \dfrac{3}{5} \times \dfrac{3}{5}

P(RR) = \dfrac{9}{25}

(b)Answer:

P(BB) = \dfrac{9}{100}

[1]

\mathstrut P(BB) = \dfrac{3}{10} \times \dfrac{3}{10}

P(BB) = \dfrac{9}{100}

Question 3

Answer:

\mathstrut P(\text{green}) = \dfrac{3}{20}

[2]

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

\text{Total} = 20

\mathstrut P(\text{green}) = \dfrac{3}{20}

Question 4

(a)Answer:

P = 0.8

[2]

The two can't both happen: they are mutually exclusive, so add.

P(A \text{ or } B) = P(A) + P(B)

\mathstrut P = 0.35 + 0.45

P = 0.8

(b)Answer: NO[1]

3, 9, 15

are both an odd number and a multiple of 3.No.

Question 5

(a)Answer:

\dfrac{7}{15}

[2]

P(RR) = \dfrac{7}{10} \times \dfrac{6}{9}

\phantom{P(RR)} = \dfrac{7}{15}

(b)Answer:

\dfrac{7}{15}

[2]

P(RB) + P(BR) = \dfrac{7}{10} \times \dfrac{3}{9} + \dfrac{3}{10} \times \dfrac{7}{9}

\phantom{P(RB) + P(BR)} = \dfrac{7}{15}

(c)Answer:

\dfrac{14}{15}

[2]

P(\text{at least one red}) = 1 - P(BB)

\phantom{P(\text{at least one red})} = 1 - \dfrac{3}{10} \times \dfrac{2}{9}

\phantom{P(\text{at least one red})} = \dfrac{14}{15}

Question 6

(a)Answer:

x = 3

[2]

\dfrac{x}{4 + x} = \dfrac{3}{7}

x = 3

(b)Answer:

\dfrac{4}{7}

[1]

P = \dfrac{12 + 4}{28}

\phantom{P} = \dfrac{4}{7}

(c)Answer:

\dfrac{1}{18}

[2]

P = \dfrac{7}{28} \times \dfrac{6}{27}

\phantom{P} = \dfrac{1}{18}

(d)Answer:

\dfrac{97}{117}

[2]

12

people chose badminton; the other

16

chose tennis.

P(\text{none}) = \dfrac{16}{28} \times \dfrac{15}{27} \times \dfrac{14}{26}

\phantom{P(\text{none})} = \dfrac{20}{117}

P(\text{at least one}) = 1 - \dfrac{20}{117}

\phantom{P(\text{at least one})} = \dfrac{97}{117}

Question 7

(a)Answer:

\dfrac{6}{55}

[2]

P(\text{both red}) = \dfrac{4}{11} \times \dfrac{3}{10}

\phantom{P(\text{both red})} = \dfrac{6}{55}

(b)Answer:

\dfrac{34}{55}

[2]

P(\text{same}) = \dfrac{6}{11} \times \dfrac{5}{10} + \dfrac{4}{11} \times \dfrac{3}{10}

\phantom{P(\text{same})} = \dfrac{21}{55}

P(\text{different}) = 1 - \dfrac{21}{55}

\phantom{P(\text{different})} = \dfrac{34}{55}

(c)Answer:

\dfrac{8}{55}

[2]

There are only

1

black pens, so three black is impossible.

P(\text{all the same}) = \dfrac{6}{11} \times \dfrac{5}{10} \times \dfrac{4}{9} + \dfrac{4}{11} \times \dfrac{3}{10} \times \dfrac{2}{9}

\phantom{P(\text{all the same})} = \dfrac{8}{55}

(d)Answer:

\dfrac{3}{11}

[2]

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

P = 3 \times \dfrac{1}{11} \times \dfrac{10}{10} \times \dfrac{9}{9}

\phantom{P} = \dfrac{3}{11}