A biased coin is flipped. The probability of getting Heads is 0.7.
(a)Write down the probability of getting Tails.(1)
(b)The coin is flipped twice. The probability tree diagram shows the possible outcomes. Two of the probabilities, labelled P and Q, are missing. Write down the values of P and Q.(2)
(c)Calculate the probability of getting Heads on both flips.(2)
(Total for Question 1 is 5 marks)
2
Two multiple-choice statements about using probability tree diagrams.
(a)To find the probability of a sequence of two outcomes (one after another) using a tree diagram, you should:(1)
A) Add the two branch probabilities
B) Multiply the two branch probabilities
C) Subtract the smaller probability from the larger
D) Divide the first probability by the second
(b)To find the probability of two different, mutually exclusive routes through a tree diagram (route 1 OR route 2), you should:(1)
A) Multiply the two route probabilities
B) Add the two route probabilities
C) Subtract one route probability from the other
D) Divide one route probability by the other
(Total for Question 2 is 2 marks)
3
State whether each statement about probability tree diagrams is True or False.
(a)On a probability tree diagram, you multiply the probabilities along a branch to find the probability of a sequence of outcomes.(1)
True
False
(b)If two events are independent, the probabilities on the second set of branches depend on the outcome of the first event.(1)
True
False
(c)The probabilities on any set of branches that start from the same point must add up to 1.(1)
True
False
(Total for Question 3 is 3 marks)
4
Tom plays a game twice. In each round, the probability that Tom wins is 0.4, independently of any other round.
(a)Complete the probability tree diagram for the two rounds by writing the missing probability on each of the four second-stage branches.(2)
(b)Work out the probability that Tom wins exactly one of the two rounds.(3)
(Total for Question 4 is 5 marks)
5
The probability that it rains on any given day in April this year is 0.2, independently of other days. Sanjay walks to school on two randomly chosen April days.
(a)Find the probability that it rains on both days.(2)
(b)Find the probability that it rains on at least one of the two days.(3)
(Total for Question 5 is 5 marks)
6
The probability that Josh's bus to work is late is 0.25 on any given day, independently of other days. Consider Monday and Tuesday next week.
(a)Complete the probability tree diagram for the two days.(2)
(b)Calculate the probability that the bus is late on exactly one of the two days.(3)
(c)Calculate the probability that the bus is late on at least one of the two days.(2)
(Total for Question 6 is 7 marks)
7
A bag contains 5 red counters and 3 blue counters only. Bilal takes a counter at random from the bag and does not replace it. He then takes a second counter at random from the bag.
(a)Complete the probability tree diagram, writing each missing probability as a fraction.(2)
(b)Calculate the probability that both counters are red.(2)
(c)Calculate the probability that the two counters are different colours.(3)
(Total for Question 7 is 7 marks)
8
A drawer contains 4 black socks and 2 white socks only. Priya takes a sock at random from the drawer and does not put it back. She then takes a second sock at random.
(a)Complete the probability tree diagram, writing each missing probability as a fraction.(2)
(b)Work out the probability that both socks are the same colour.(3)
(c)Work out the probability that at least one of the socks is black.(2)
(Total for Question 8 is 7 marks)
9
In a factory, machines A and B operate independently of each other on any given day. The probability that machine A breaks down on a given day is 0.1. The probability that machine B breaks down on a given day is 0.05.
(a)Complete the probability tree diagram for a single day, showing both machines.(2)
(b)Find the probability that neither machine breaks down on a given day.(2)
(c)Find the probability that exactly one of the two machines breaks down on a given day.(3)
(Total for Question 9 is 7 marks)
10
A fair spinner has three equal sections coloured red, green and yellow. Meera spins the spinner twice.
(a)Complete the probability tree diagram for the two spins, writing the probability of each outcome on the branches.(2)
(b)Find the probability that the spinner lands on the same colour both times.(3)
(c)Find the probability that the spinner lands on red at least once.(3)
(Total for Question 10 is 8 marks)
11
A youth club has 10 boys and 15 girls. Grace, the club leader, picks two members at random, one after another without replacement, to represent the club at a regional meeting.
(a)Find the probability that both members picked are girls.(3)
(b)Find the probability that one boy and one girl are picked, in either order.(3)
(Total for Question 11 is 6 marks)
12
A charity raffle sells 50 tickets, 4 of which win a prize. Chloe buys two of the raffle tickets.
(a)Find the probability that both of Chloe's tickets win a prize.(2)
(b)Find the probability that at least one of Chloe's tickets wins a prize.(3)
(Total for Question 12 is 5 marks)
13
A box for a game show contains 3 gold tickets and 7 silver tickets. A contestant draws two tickets from the box at random, one after another without replacement.
(a)Complete the probability tree diagram, writing each missing probability as a fraction.(2)
(b)Calculate the probability that the contestant draws exactly one gold ticket.(3)
(c)Given that the first ticket drawn was silver, write down the probability that the second ticket drawn is gold.(1)
(Total for Question 13 is 6 marks)
14
A netball team has a probability of 0.85 of winning any match, independently of other matches. Calculate the probability that the team wins at least one of their next three matches.
(Total for Question 14 is 3 marks)
15
A bag contains 3 red counters and n blue counters only, where n > 0. Two counters are taken from the bag at random, one after another, without replacement.
(a)Show that the probability that both counters are red is 6 / ((n+3)(n+2)).(3)
(b)Given that the probability that both counters are red is 1/5, show that n2 + 5n - 24 = 0.(2)
(c)Hence find the number of blue counters in the bag.(3)
(Total for Question 15 is 8 marks)
16
A factory makes phone screens. 3% of the screens produced are faulty. Every screen is tested. If a screen is faulty, the test correctly shows it as faulty 90% of the time. If a screen is not faulty, the test incorrectly shows it as faulty 5% of the time.
(a)Complete the probability tree diagram for a randomly chosen screen.(3)
(b)Calculate the probability that a randomly selected screen is shown as faulty by the test.(3)
(c)A screen is tested and shown as faulty. Find the probability that it is actually faulty.(3)
(Total for Question 16 is 9 marks)
Mark scheme · 5.22 Probability Trees
Question 1
(a) B1 0.3 cao
(a) Answer: 0.3
(b) B1 P = 0.3
(b) B1 Q = 0.7
(b) Answer: P = 0.3, Q = 0.7
(c) M1 0.7 x 0.7
(c) A1 0.49 cao
(c) Answer: 0.49
Question 2
(a) B1 B
(a) Answer: B
(b) B1 B
(b) Answer: B
Question 3
(a) B1 True
(a) Answer: True
(b) B1 False
(b) Answer: False
(c) B1 True
(c) Answer: True
Question 4
(a) B1 0.4 (Win) on both second-stage branch pairs
(a) B1 0.6 (Lose) on both second-stage branch pairs
(a) Answer: Win = 0.4 and Lose = 0.6 on each set of second-stage branches
(b) M1 0.4 x 0.6 (or 0.6 x 0.4) for one correct route
(b) M1 0.4 x 0.6 + 0.6 x 0.4 (both routes identified and added)
(b) A1 0.48 cao
(b) Answer: 0.48
Question 5
(a) M1 0.2 x 0.2
(a) A1 0.04 cao
(a) Answer: 0.04
(b) M1 0.8 x 0.8 ( = P(no rain either day) )
(b) M1 1 - 0.8 x 0.8
(b) A1 0.36 cao
(b) Answer: 0.36
Question 6
(a) B1 0.25 (Late) on both second-stage branch pairs
(a) B1 0.75 (Not late) on both second-stage branch pairs
(a) Answer: Late = 0.25, Not late = 0.75 on each set of second-stage branches
(b) M1 0.25 x 0.75 (one correct route)
(b) M1 0.25 x 0.75 + 0.75 x 0.25 (both routes)
(b) A1 0.375 cao
(b) Answer: 0.375
(c) M1 1 - 0.75 x 0.75
(c) A1 0.4375 cao
(c) Answer: 0.4375
Question 7
(a) B1 4/7 and 3/7 on the branches after Red
(a) B1 5/7 and 2/7 on the branches after Blue
(a) Answer: After Red: Red 4/7, Blue 3/7. After Blue: Red 5/7, Blue 2/7
(b) M1 5/8 x 4/7
(b) A1 5/14 oe (e.g. 20/56)
(b) Answer: 5/14
(c) M1 5/8 x 3/7 (or 3/8 x 5/7) for one correct route
(c) M1 5/8 x 3/7 + 3/8 x 5/7 (both routes added)
(c) A1 15/28 oe (e.g. 30/56)
(c) Answer: 15/28
Question 8
(a) B1 3/5 and 2/5 on the branches after Black
(a) B1 4/5 and 1/5 on the branches after White
(a) Answer: After Black: Black 3/5, White 2/5. After White: Black 4/5, White 1/5
(b) M1 4/6 x 3/5 (or 2/6 x 1/5) for one correct route
(b) M1 4/6 x 3/5 + 2/6 x 1/5 (both routes added)
(b) A1 7/15 oe (e.g. 14/30)
(b) Answer: 7/15
(c) M1 1 - (2/6 x 1/5) ( = 1 - P(both white) )
(c) A1 14/15 oe (e.g. 28/30)
(c) Answer: 14/15
Question 9
(a) B1 0.9 and 0.1 on the machine A branches
(a) B1 0.95 and 0.05 on the machine B branches
(a) Answer: A: Breaks down 0.1, Does not break down 0.9. B: Breaks down 0.05, Does not break down 0.95
(b) M1 0.9 x 0.95
(b) A1 0.855 cao
(b) Answer: 0.855
(c) M1 0.1 x 0.95 (or 0.9 x 0.05) for one correct route
(c) M1 0.1 x 0.95 + 0.9 x 0.05 (both routes added)
(c) A1 0.14 cao
(c) Answer: 0.14
Question 10
(a) B1 1/3 on all three first-stage branches
(a) B1 1/3 on all three branches from each of the three first-stage outcomes
(a) Answer: Every branch is labelled 1/3
(b) M1 1/3 x 1/3 ( = 1/9, one matching route)
(b) M1 3 x (1/3 x 1/3) (all three matching routes added)
(b) A1 1/3 oe (e.g. 3/9)
(b) Answer: 1/3
(c) M1 2/3 x 2/3 ( = P(no red either spin) )
(c) M1 1 - 2/3 x 2/3
(c) A1 5/9 oe
(c) Answer: 5/9
Question 11
(a) M1 15/25 x 14/24
(a) M1 210/600 oe simplification started
(a) A1 7/20 cao
(a) Answer: 7/20
(b) M1 10/25 x 15/24 (or 15/25 x 10/24) for one correct route
(b) M1 2 x (10/25 x 15/24) (both routes added)
(b) A1 1/2 cao
(b) Answer: 1/2
Question 12
(a) M1 4/50 x 3/49
(a) A1 6/1225 oe (e.g. 12/2450, awrt 0.0049)
(a) Answer: 6/1225
(b) M1 46/50 x 45/49 ( = P(neither ticket wins) )
(b) M1 1 - 46/50 x 45/49
(b) A1 38/245 oe (e.g. 380/2450, awrt 0.155)
(b) Answer: 38/245
Question 13
(a) B1 2/9 and 7/9 on the branches after Gold
(a) B1 3/9 and 6/9 on the branches after Silver
(a) Answer: After Gold: Gold 2/9, Silver 7/9. After Silver: Gold 3/9, Silver 6/9
(b) M1 3/10 x 7/9 (or 7/10 x 3/9) for one correct route
(b) M1 3/10 x 7/9 + 7/10 x 3/9 (both routes added)
(c) M1 n = 3 selected, rejecting n = -8 since n > 0
(c) A1 n = 3 cao
(c) Answer: 3 blue counters
Question 16
(a) B1 0.9 and 0.1 on the branches after Faulty
(a) B1 0.05 and 0.95 on the branches after Not faulty
(a) B1 0.03 and 0.97 correctly retained on the first-stage branches
(a) Answer: First stage: Faulty 0.03, Not faulty 0.97. After Faulty: Test faulty 0.9, Test not faulty 0.1. After Not faulty: Test faulty 0.05, Test not faulty 0.95
(b) M1 0.03 x 0.9 (= 0.027, one correct route)
(b) M1 0.03 x 0.9 + 0.97 x 0.05 (both routes added)
(b) A1 0.0755 cao
(b) Answer: 0.0755
(c) M1 identifies P(faulty and shown faulty) = 0.027
(c) M1 0.027 / 0.0755 (divides by P(shown faulty) from part (b))