Exam questions · Maths
Probability
- 30 exam questions
- 99 marks
- 45 quick checks
Probability Basics and Relative Frequency
Just this lesson-
1 Write down [2 marks]
A bag contains 6 red counters, 3 blue counters and 1 green counter. Kay takes a counter at random. Write down the probability that the counter is (a) blue, (b) not red.
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Model answer
(a) \(\dfrac{3}{10}\). (b) There are 4 counters that are not red, so \(\dfrac{4}{10} = \dfrac{2}{5}\).
Mark scheme
- (a) \(\dfrac{3}{10}\) or 0.3 — B1
- (b) \(\dfrac{4}{10}\) or \(\dfrac{2}{5}\) or 0.4 — B1
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2 Work out [4 marks]
Here are the probabilities of the scores on a biased dice. The probability of a 1 is 0.1, of a 2 is 0.15, of a 3 is 0.2, of a 4 is \(x\), of a 5 is 0.25 and of a 6 is 0.1. (a) Work out the value of \(x\). (2 marks) (b) Mel throws the dice 200 times. Work out an estimate for the number of times she gets a 5. (2 marks)
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Model answer
(a) The other probabilities add up to \(0.1 + 0.15 + 0.2 + 0.25 + 0.1 = 0.8\), so \(x = 1 - 0.8 = 0.2\). (b) \(0.25 \times 200 = 50\).
Mark scheme
- (a) \(0.1 + 0.15 + 0.2 + 0.25 + 0.1 = 0.8\) or \(1 - 0.8\) — M1
- (a) \(0.2\) — A1
- (b) \(0.25 \times 200\) — M1
- (b) 50 — A1
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3 Explain [3 marks]
Ali throws a dice 150 times. It lands on 6 a total of 45 times. (a) Work out the relative frequency of getting a 6. (1 mark) (b) Ali says, “The dice is fair.” Is Ali correct? Give a reason for your answer. (2 marks)
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Model answer
(a) \(\dfrac{45}{150} = 0.3\). (b) For a fair dice the probability of a 6 is \(\dfrac{1}{6}\), which is about 0.17. The relative frequency 0.3 is much larger, so the dice is probably not fair.
Mark scheme
- (a) \(0.3\) or \(\dfrac{3}{10}\) — B1
- (b) Compares 0.3 with \(\dfrac{1}{6}\) or about 0.17 — M1
- (b) Not fair, because the relative frequency is much bigger than for a fair dice — C1
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4 Work out [2 marks]
The probability that a train is late is 0.08. Mia takes the train on 250 days. Work out an estimate for the number of days that the train is late.
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Model answer
\(0.08 \times 250 = 20\).
Mark scheme
- \(0.08 \times 250\) or \(\dfrac{8}{100} \times 250\) — M1
- 20 — A1
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5 Work out [3 marks]
Jo plays a game. The probability that she wins is 0.35. The probability that she draws is 0.25. (a) Work out the probability that she loses. (1 mark) Jo plays the game 40 times. (b) Work out an estimate for the number of times she wins. (2 marks)
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Model answer
(a) \(1 - 0.35 - 0.25 = 0.4\). (b) \(0.35 \times 40 = 14\).
Mark scheme
- (a) \(0.4\) — B1
- (b) \(0.35 \times 40\) — M1
- (b) 14 — A1
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6 Work out [3 marks]
In a school, the probability that a student chosen at random is a girl is 0.6. The probability that a student plays chess is 0.3. The probability that a student is a girl and plays chess is 0.2. Work out the probability that a student chosen at random is a girl or plays chess.
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Model answer
\(P(\text{girl or chess}) = P(\text{girl}) + P(\text{chess}) - P(\text{girl and chess}) = 0.6 + 0.3 - 0.2 = 0.7\).
Mark scheme
- \(0.6 + 0.3\) — M1
- \(0.6 + 0.3 - 0.2\) — M1
- \(0.7\) — A1
Quick check
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1
What is the probability of an impossible event?
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B: \(0\)
An impossible event has probability 0.
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2
A bag has 3 red, 5 blue and 2 green counters. One is taken at random. What is the probability it is blue?
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A: \(\dfrac{1}{2}\)
There are 10 counters and 5 are blue, so \(\dfrac{5}{10} = \dfrac{1}{2}\).
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3
A bag has 3 red, 5 blue and 2 green counters. What is the probability that a counter taken at random is not green?
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D: \(\dfrac{4}{5}\)
\(1 - \dfrac{2}{10} = \dfrac{8}{10} = \dfrac{4}{5}\).
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4
A dice is thrown 120 times and a 6 comes up 30 times. What is the relative frequency of a 6?
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C: \(\dfrac{1}{4}\)
\(\dfrac{30}{120} = \dfrac{1}{4}\).
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5
A fair dice is thrown 600 times. How many sixes are expected?
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B: \(100\)
\(\dfrac{1}{6} \times 600 = 100\).
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6
Two fair dice are thrown. What is the probability that the total score is 7?
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A: \(\dfrac{1}{6}\)
There are 6 ways to make 7 out of 36, so \(\dfrac{6}{36} = \dfrac{1}{6}\).
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7
A bag has 3 red, 5 blue and 2 green counters. What is the probability of taking a red or a green counter?
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D: \(\dfrac{1}{2}\)
The events are mutually exclusive, so \(\dfrac{3}{10} + \dfrac{2}{10} = \dfrac{1}{2}\).
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8
A spinner is spun 200 times and lands on red 74 times. How many reds would be expected if the probability of red were \(\dfrac{1}{4}\)?
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C: \(50\)
\(\dfrac{1}{4} \times 200 = 50\). The result of 74 suggests the spinner may be biased.
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9
A card is taken from a pack of 52. What is the probability that it is a heart or a king?
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B: \(\dfrac{4}{13}\)
\(\dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}\).
Tree Diagrams
Just this lesson-
1 Work out [2 marks]
Tom flips a fair coin twice. Work out the probability that he gets two heads.
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Model answer
\(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
Mark scheme
- \(\dfrac{1}{2} \times \dfrac{1}{2}\) — M1
- \(\dfrac{1}{4}\) — A1
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2 Work out [4 marks]
Mina throws a biased coin twice. The probability that it lands on heads is 0.6. The incomplete probability tree diagram shows the first throw and the second throw. (a) Complete the probability tree diagram. (2 marks) (b) Work out the probability that Mina gets one head and one tail. (2 marks)
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Model answer
(a) The probability of tails is \(1 - 0.6 = 0.4\). The first-throw tails branch is 0.4, and the second-throw branches are 0.6 for heads and 0.4 for tails on all four. (b) \(0.6 \times 0.4 + 0.4 \times 0.6 = 0.24 + 0.24 = 0.48\).
Mark scheme
- (a) 0.4 on the first tails branch — B1
- (a) 0.6 and 0.4 on each pair of second-throw branches — B1
- (b) \(0.6 \times 0.4\) and \(0.4 \times 0.6\) added — M1
- (b) \(0.48\) — A1
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3 Work out [3 marks]
A bag contains 5 red pens and 3 blue pens. Eve takes two pens at random without replacement. Work out the probability that both pens are red.
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Model answer
\(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\).
Mark scheme
- \(\dfrac{5}{8}\) and \(\dfrac{4}{7}\) seen — M1
- \(\dfrac{5}{8} \times \dfrac{4}{7}\) — M1
- \(\dfrac{5}{14}\) or \(\dfrac{20}{56}\) — A1
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4 Work out [3 marks]
Sam has a written test and a practical test. The probability that he passes the written test is 0.7. The probability that he passes the practical test is 0.8. The tests are independent. Work out the probability that Sam passes at least one of the tests.
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Model answer
The probability that he fails both is \(0.3 \times 0.2 = 0.06\), so the probability he passes at least one is \(1 - 0.06 = 0.94\).
Mark scheme
- \(0.3\) and \(0.2\) seen — M1
- \(0.3 \times 0.2 = 0.06\) — M1
- \(0.94\) — A1
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5 Work out [3 marks]
A fair dice is rolled twice. Work out the probability of getting at least one 6.
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Model answer
No sixes in two rolls: \(\dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{36}\). So the probability of at least one 6 is \(1 - \dfrac{25}{36} = \dfrac{11}{36}\).
Mark scheme
- \(\dfrac{5}{6} \times \dfrac{5}{6}\) — M1
- \(1 - \dfrac{25}{36}\) — M1
- \(\dfrac{11}{36}\) — A1
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6 Show that [5 marks]
A bag contains 3 red counters and \(n\) blue counters. Two counters are taken at random without replacement. The probability that both counters are red is \(\dfrac{1}{15}\). (a) Show that \(n^2 + 5n - 84 = 0\). (3 marks) (b) Hence find the value of \(n\). (2 marks)
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Model answer
(a) There are \(n + 3\) counters, so \(\dfrac{3}{n + 3} \times \dfrac{2}{n + 2} = \dfrac{1}{15}\). Then \((n + 3)(n + 2) = 90\), so \(n^2 + 5n + 6 = 90\) and \(n^2 + 5n - 84 = 0\). (b) \((n + 12)(n - 7) = 0\), so \(n = 7\), because \(n\) cannot be negative.
Mark scheme
- (a) \(\dfrac{3}{n + 3} \times \dfrac{2}{n + 2}\) — M1
- (a) \(6 \times 15 = (n + 3)(n + 2)\) or equivalent — M1
- (a) \(n^2 + 5n - 84 = 0\) shown — C1
- (b) \((n + 12)(n - 7)\) — M1
- (b) \(n = 7\) — A1
Quick check
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1
The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on both of two days?
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C: \(0.09\)
\(0.3 \times 0.3 = 0.09\).
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2
The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on at least one of two days?
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B: \(0.51\)
\(1 - 0.7 \times 0.7 = 1 - 0.49 = 0.51\).
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3
Two fair coins are tossed. What is the probability of two heads?
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A: \(\dfrac{1}{4}\)
\(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
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4
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that both are red?
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D: \(\dfrac{3}{10}\)
\(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}\).
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5
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are different colours?
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C: \(\dfrac{3}{5}\)
\(\dfrac{3}{5} \times \dfrac{2}{4} + \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{12}{20} = \dfrac{3}{5}\).
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6
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are the same colour?
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B: \(\dfrac{2}{5}\)
\(\dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}\).
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7
On a tree diagram, one branch has probability 0.6. What is the other branch from the same point?
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A: \(0.4\)
The branches from one point add up to 1.
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8
A bag has 4 red and 6 blue counters. Two are taken without replacement. What is the probability that both are blue?
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D: \(\dfrac{1}{3}\)
\(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).
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9
A bag has 4 red counters and some blue counters, \(n\) in all. Two are taken without replacement and \(P(\text{two reds}) = \dfrac{2}{15}\). How many counters are in the bag?
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C: \(10\)
\(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\), so \(n = 10\).
Venn Diagrams and Set Notation
Just this lesson-
1 Write down [2 marks]
\(\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). \(A\) is the set of even numbers and \(B = \{2, 3, 5, 7\}\). (a) Write down \(A \cap B\). (1 mark) (b) Write down \(A \cup B\). (1 mark)
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Model answer
(a) \(A \cap B = \{2\}\). (b) \(A \cup B = \{2, 3, 4, 5, 6, 7, 8, 10\}\).
Mark scheme
- (a) \(\{2\}\) — B1
- (b) \(\{2, 3, 4, 5, 6, 7, 8, 10\}\) — B1
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2 Work out [4 marks]
There are 40 students in a class. 22 play football and 19 play tennis. The incomplete Venn diagram shows 6 who play both sports and 5 who play neither. (a) Complete the Venn diagram. (2 marks) (b) A student is chosen at random. Work out the probability that the student plays exactly one of the two sports. (2 marks)
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Model answer
(a) Football only is \(22 - 6 = 16\) and tennis only is \(19 - 6 = 13\). Check: \(16 + 6 + 13 + 5 = 40\). (b) \(\dfrac{16 + 13}{40} = \dfrac{29}{40}\).
Mark scheme
- (a) 16 — B1
- (a) 13 — B1
- (b) \(16 + 13 = 29\) seen — M1
- (b) \(\dfrac{29}{40}\) — A1
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3 Work out [3 marks]
\(\xi = \{1, 2, 3, \ldots, 12\}\). \(A\) is the set of multiples of 3 and \(B\) is the set of factors of 12. (a) List the members of \(A \cap B\). (1 mark) (b) Work out \(n((A \cup B)')\). (2 marks)
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Model answer
(a) \(A = \{3, 6, 9, 12\}\) and \(B = \{1, 2, 3, 4, 6, 12\}\), so \(A \cap B = \{3, 6, 12\}\). (b) \(A \cup B = \{1, 2, 3, 4, 6, 9, 12\}\), which has 7 members, so \(n((A \cup B)') = 12 - 7 = 5\).
Mark scheme
- (a) \(\{3, 6, 12\}\) — B1
- (b) Lists or counts 7 members of \(A \cup B\) — M1
- (b) 5 — A1
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4 Work out [3 marks]
In a group of 50 people, 30 like tea and 25 like coffee. 8 people like neither drink. Work out the number of people who like both tea and coffee.
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Model answer
The number who like at least one drink is \(50 - 8 = 42\). Then \(30 + 25 - 42 = 13\) like both.
Mark scheme
- \(50 - 8 = 42\) — M1
- \(30 + 25 - 42\) — M1
- 13 — A1
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5 Work out [4 marks]
In a Venn diagram of 52 people, the number in set \(A\) only is \(2x\), the number in both sets is \(x\), the number in set \(B\) only is \(x + 4\), and 8 people are in neither set. (a) Work out the value of \(x\). (2 marks) (b) A person is chosen at random. Work out the probability that the person is in set \(A\). Give your answer in its simplest form. (2 marks)
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Model answer
(a) \(2x + x + x + 4 + 8 = 52\), so \(4x + 12 = 52\) and \(x = 10\). (b) Set \(A\) has \(2x + x = 30\) people, so the probability is \(\dfrac{30}{52} = \dfrac{15}{26}\).
Mark scheme
- (a) \(4x + 12 = 52\) — M1
- (a) \(x = 10\) — A1
- (b) \(\dfrac{30}{52}\) — M1
- (b) \(\dfrac{15}{26}\) — A1
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6 Work out [3 marks]
There are 30 students in a class. 18 study French and 12 study Spanish. 5 study both. A student is chosen at random. Work out the probability that the student studies neither French nor Spanish.
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Model answer
At least one language: \(18 + 12 - 5 = 25\). Neither: \(30 - 25 = 5\). The probability is \(\dfrac{5}{30} = \dfrac{1}{6}\).
Mark scheme
- \(18 + 12 - 5 = 25\) — M1
- 5 students study neither — A1
- \(\dfrac{1}{6}\) — A1
Quick check
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1
What does \(A \cap B\) mean?
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D: The items in both \(A\) and \(B\)
\(\cap\) is the intersection, the overlap.
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2
What does \(A \cup B\) mean?
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C: The items in \(A\) or \(B\) or both
\(\cup\) is the union, everything in either circle.
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3
What does \(A'\) mean?
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B: The items not in \(A\)
\(A'\) is the complement of \(A\).
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4
\(A = \{2, 4, 6, 8, 10\}\) and \(B = \{3, 6, 9\}\). What is \(A \cap B\)?
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A: \(\{6\}\)
Only 6 is in both sets.
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5
18 students play football and 6 of them also play tennis. How many play football only?
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D: \(12\)
\(18 - 6 = 12\).
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6
30 students: 18 play football, 14 play tennis and 6 play both. How many play neither?
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C: \(4\)
\(12 + 6 + 8 = 26\) play at least one, so \(30 - 26 = 4\).
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7
In the same survey, what is the probability that a student chosen at random plays football only?
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B: \(\dfrac{2}{5}\)
\(\dfrac{12}{30} = \dfrac{2}{5}\). The fraction \(\dfrac{18}{30}\) would include those who play both.
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8
In a class of 40, 22 play football, 19 play tennis and 5 play neither. How many play both?
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A: \(6\)
\(40 - 5 = 35\) play at least one, and \(22 + 19 - 35 = 6\).
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9
\(n(A) = 15\), \(n(B) = 12\) and \(n(A \cap B) = 5\). What is \(n(A \cup B)\)?
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D: \(22\)
\(15 + 12 - 5 = 22\).
Frequency Trees and Two-Way Tables
Just this lesson-
1 Work out [3 marks]
There are 60 people at a gym. 36 of them are women. 9 of the women and 6 of the men use the pool. (a) Work out the number of men. (1 mark) (b) Work out the number of people who do not use the pool. (2 marks)
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Model answer
(a) \(60 - 36 = 24\). (b) \(9 + 6 = 15\) people use the pool, so \(60 - 15 = 45\) do not.
Mark scheme
- (a) 24 — B1
- (b) \(9 + 6 = 15\) — M1
- (b) 45 — A1
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2 Work out [4 marks]
The two-way table shows information about how 100 students travel to school. Some of the numbers are missing. (a) Complete the two-way table. (3 marks) (b) A student is chosen at random. Work out the probability that the student is in Year 11 and walks. (1 mark)
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Model answer
(a) Year 10 cycle: \(50 - 24 - 18 = 8\). Year 11 bus: \(44 - 24 = 20\). Year 11 total: 50. Total who walk: \(18 + 25 = 43\). (b) \(\dfrac{25}{100} = \dfrac{1}{4}\).
Mark scheme
- (a) At least two correct values — B1
- (a) At least three correct values — B1
- (a) All four correct: 8, 20, 50 and 43 — B1
- (b) \(\dfrac{25}{100}\) or \(\dfrac{1}{4}\) — B1
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3 Work out [3 marks]
A survey asks 200 people if they own a dog. 120 of the people are female, and the rest are male. 30% of the females and 25% of the males own a dog. Work out the total number of people who own a dog.
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Model answer
Females: \(0.3 \times 120 = 36\). There are 80 males, and \(0.25 \times 80 = 20\). The total is \(36 + 20 = 56\).
Mark scheme
- \(0.3 \times 120 = 36\) — M1
- \(0.25 \times 80 = 20\) — M1
- 56 — A1
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4 Work out [3 marks]
In a school, 60% of the students are girls. 20% of the girls and 40% of the boys play hockey. A student is chosen at random. Work out the probability that the student plays hockey.
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Model answer
\(0.6 \times 0.2 + 0.4 \times 0.4 = 0.12 + 0.16 = 0.28\).
Mark scheme
- \(0.6 \times 0.2\) or \(0.4 \times 0.4\) — M1
- \(0.6 \times 0.2 + 0.4 \times 0.4\) — M1
- \(0.28\) — A1
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5 Work out [4 marks]
There are 150 students in Year 11. 90 of them are boys. 35 of the boys and 25 of the girls are left-handed. (a) Work out the number of girls. (1 mark) (b) Work out the number of students who are right-handed. (2 marks) (c) A student is chosen at random. Work out the probability that the student is a left-handed girl. (1 mark)
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Model answer
(a) \(150 - 90 = 60\). (b) Right-handed boys \(90 - 35 = 55\), right-handed girls \(60 - 25 = 35\), so \(55 + 35 = 90\). (c) \(\dfrac{25}{150} = \dfrac{1}{6}\).
Mark scheme
- (a) 60 — B1
- (b) \(55\) and \(35\) seen, or \(150 - 35 - 25\) — M1
- (b) 90 — A1
- (c) \(\dfrac{1}{6}\) or \(\dfrac{25}{150}\) — B1
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6 Work out [4 marks]
There are 80 pupils on a trip. The ratio of boys to girls is \(3 : 5\). 40% of the boys and 20% of the girls bring a packed lunch. Work out the fraction of all the pupils who bring a packed lunch. Give your answer in its simplest form.
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Model answer
Boys: \(80 \div 8 \times 3 = 30\). Girls: 50. Packed lunches: \(0.4 \times 30 = 12\) and \(0.2 \times 50 = 10\), which is 22. The fraction is \(\dfrac{22}{80} = \dfrac{11}{40}\).
Mark scheme
- 30 boys and 50 girls — M1
- \(0.4 \times 30 = 12\) and \(0.2 \times 50 = 10\) — M1
- 22 — A1
- \(\dfrac{11}{40}\) — A1
Quick check
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1
A frequency tree starts with 100 pupils, and 60 are girls. How many are boys?
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A: \(40\)
\(100 - 60 = 40\).
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2
40% of 60 girls walk to school. How many girls walk?
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D: \(24\)
\(0.4 \times 60 = 24\).
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3
In a two-way table the Year 10 row total is 50. 24 take the bus and 18 walk. How many cycle?
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C: \(8\)
\(50 - 24 - 18 = 8\).
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4
In a survey of 100 students, 24 are in Year 10 and take the bus. What is the probability that a student is in Year 10 and takes the bus?
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B: \(\dfrac{6}{25}\)
\(\dfrac{24}{100} = \dfrac{6}{25}\).
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5
13 out of 100 students cycle to school. How many of 500 students would you expect to cycle?
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A: \(65\)
\(\dfrac{13}{100} \times 500 = 65\).
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6
What is 25% of 120?
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D: \(30\)
\(0.25 \times 120 = 30\).
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7
200 people go to a gym. 60% are women. 25% of the women and 40% of the men go in the morning. How many people go in the morning?
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C: \(62\)
120 women and 80 men. \(30 + 32 = 62\).
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8
50 people were asked about pets. 28 have a cat, 20 have a dog and 6 have both. How many have neither?
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B: \(8\)
At least one: \(22 + 6 + 14 = 42\). \(50 - 42 = 8\).
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9
In a frequency tree, the branches after “60 girls” show 24 who walk and some who take the bus. How many take the bus?
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A: \(36\)
The branches add up to the number they came from: \(60 - 24 = 36\).
Conditional Probability
Just this lesson-
1 Work out [2 marks]
There are 3 red sweets and 5 blue sweets in a bag. Tom takes a sweet at random and eats it. He then takes a second sweet. Given that the first sweet was red, work out the probability that the second sweet is blue.
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Model answer
After the red sweet is eaten there are 7 sweets left, 5 of them blue, so the probability is \(\dfrac{5}{7}\).
Mark scheme
- 7 sweets remaining, or the denominator 7 — M1
- \(\dfrac{5}{7}\) — A1
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2 Work out [5 marks]
The two-way table shows how 100 people travel to work. (a) One of the people is chosen at random. Work out the probability that the person drives. (1 mark) (b) One of the females is chosen at random. Work out the probability that she cycles. (2 marks) (c) One of the people who take the bus is chosen at random. Work out the probability that the person is male. (2 marks)
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Model answer
(a) \(\dfrac{50}{100} = \dfrac{1}{2}\). (b) There are 40 females and 16 cycle, so \(\dfrac{16}{40} = \dfrac{2}{5}\). (c) There are 22 people who take the bus and 18 of them are male, so \(\dfrac{18}{22} = \dfrac{9}{11}\).
Mark scheme
- (a) \(\dfrac{50}{100}\) or \(\dfrac{1}{2}\) — B1
- (b) \(\dfrac{16}{40}\) — M1
- (b) \(\dfrac{2}{5}\) — A1
- (c) \(\dfrac{18}{22}\) — M1
- (c) \(\dfrac{9}{11}\) — A1
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3 Work out [3 marks]
\(A\) and \(B\) are two events. \(P(A) = 0.4\), \(P(B) = 0.5\) and \(P(A \text{ and } B) = 0.2\). (a) Work out \(P(A \text{ given } B)\). (2 marks) (b) Are \(A\) and \(B\) independent? Give a reason for your answer. (1 mark)
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Model answer
(a) \(P(A \text{ given } B) = \dfrac{0.2}{0.5} = 0.4\). (b) Yes, because \(P(A \text{ given } B) = 0.4 = P(A)\), so knowing \(B\) happened does not change the probability of \(A\).
Mark scheme
- (a) \(\dfrac{0.2}{0.5}\) or \(0.2 = P(A \text{ given } B) \times 0.5\) — M1
- (a) \(0.4\) — A1
- (b) Yes, with the reason that \(P(A \text{ given } B) = P(A)\) — C1
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4 Work out [4 marks]
There are 40 students in a class. 22 play football (F) and 19 play tennis (T). 6 play both and 5 play neither. (a) A student who plays tennis is chosen at random. Work out the probability that the student also plays football. (2 marks) (b) A student who does not play football is chosen at random. Work out the probability that the student plays tennis. (2 marks)
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Model answer
(a) 19 play tennis and 6 of them also play football, so \(\dfrac{6}{19}\). (b) \(40 - 22 = 18\) do not play football, and 13 of them play tennis, so \(\dfrac{13}{18}\).
Mark scheme
- (a) \(\dfrac{6}{19}\) with 19 as the denominator — M1
- (a) \(\dfrac{6}{19}\) — A1
- (b) \(\dfrac{13}{18}\) with 18 as the denominator — M1
- (b) \(\dfrac{13}{18}\) — A1
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5 Work out [4 marks]
A bag contains 5 red counters and 3 blue counters. Two counters are taken at random without replacement. Given that the two counters are different colours, work out the probability that the first counter is red.
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Model answer
\(P(\text{red then blue}) = \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}\) and \(P(\text{blue then red}) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}\). The probability that they are different is \(\dfrac{30}{56}\), so the probability that the first is red is \(\dfrac{15}{56} \div \dfrac{30}{56} = \dfrac{1}{2}\).
Mark scheme
- \(\dfrac{5}{8} \times \dfrac{3}{7}\) — M1
- \(\dfrac{3}{8} \times \dfrac{5}{7}\) and the two added — M1
- \(\dfrac{15}{56} \div \dfrac{30}{56}\) or \(\dfrac{15}{30}\) — M1
- \(\dfrac{1}{2}\) — A1
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6 Work out [4 marks]
Machine \(X\) makes 60% of the items in a factory and machine \(Y\) makes 40%. 5% of the items from \(X\) are faulty and 10% of the items from \(Y\) are faulty. An item is chosen at random and is found to be faulty. Work out the probability that it was made by machine \(X\).
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Model answer
\(P(X \text{ and faulty}) = 0.6 \times 0.05 = 0.03\) and \(P(Y \text{ and faulty}) = 0.4 \times 0.1 = 0.04\). So \(P(\text{faulty}) = 0.07\), and the probability is \(\dfrac{0.03}{0.07} = \dfrac{3}{7}\).
Mark scheme
- \(0.6 \times 0.05 = 0.03\) — M1
- \(0.4 \times 0.1 = 0.04\) — M1
- \(\dfrac{0.03}{0.07}\) — M1
- \(\dfrac{3}{7}\) — A1
Quick check
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1
What does \(P(A \mid B)\) mean?
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B: The probability of \(A\) given that \(B\) has happened
The vertical line means “given that”.
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2
14 students play tennis, and 6 of them also play football. What is the probability that a tennis player also plays football?
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A: \(\dfrac{3}{7}\)
The group is the 14 tennis players: \(\dfrac{6}{14} = \dfrac{3}{7}\).
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3
Of 50 people who drive to work, 20 are female. A driver is chosen at random. What is the probability that the person is male?
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D: \(\dfrac{3}{5}\)
\(\dfrac{30}{50} = \dfrac{3}{5}\). The group is the 50 drivers.
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4
Of 100 people, 36 male drivers and 24 female drivers drive to work. A driver is chosen at random. What is the probability that the driver is female?
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C: \(\dfrac{2}{5}\)
The group is \(36 + 24 = 60\) drivers, and \(\dfrac{24}{60} = \dfrac{2}{5}\).
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5
A bag has 3 red and 2 blue counters. One red counter is taken and not replaced. What is the probability that the next counter is red?
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B: \(\dfrac{1}{2}\)
2 red counters are left among 4, so \(\dfrac{2}{4} = \dfrac{1}{2}\).
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6
Is \(P(A \mid B)\) always equal to \(P(B \mid A)\)?
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A: No, because they use different groups
The group on the bottom is different in each.
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7
\(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ and } B) = 0.2\). Are \(A\) and \(B\) independent?
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D: Yes, because \(0.5 \times 0.4 = 0.2\)
Independent events have \(P(A \text{ and } B) = P(A) \times P(B)\).
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8
\(P(A \text{ and } B) = 0.1\) and \(P(B) = 0.4\). What is \(P(A \mid B)\)?
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C: \(0.25\)
\(\dfrac{0.1}{0.4} = 0.25\).
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9
\(P(A \mid B) = 0.5\) and \(P(B) = 0.6\). What is \(P(A \text{ and } B)\)?
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B: \(0.3\)
\(P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.6 = 0.3\).