A fair coin is thrown twice. The tree diagram shows all the possible outcomes.
(a)Write down the probability that the first throw is a Head.(1)
(b)Find the probability that both throws are Heads.(2)
(Total for Question 1 is 3 marks)
2
Answer the following short questions about branch probabilities on a tree diagram.
(i)Which pair of probabilities could correctly label the two branches leading from the same point on a tree diagram?(1)
A) 0.3 and 0.6
B) 0.45 and 0.55
C) 0.2 and 0.9
D) 0.5 and 0.6
(ii)A tree diagram has two branches from a point, one labelled 0.35. What is the correct value of the other branch?(1)
A) 0.35
B) 0.65
C) 1.35
D) 0.7
(Total for Question 2 is 2 marks)
3
A bag contains 5 counters: 3 red and 2 blue. Sara takes a counter at random from the bag, notes its colour, and replaces it in the bag. She then takes a second counter at random from the bag. The tree diagram shows the first pick, and part of the second pick.
(a)Write down the missing probability for Blue on the first pick.(1)
(b)Complete the tree diagram by writing the two missing probabilities in the dashed boxes for the second pick, following the Red outcome.(2)
(c)Find the probability that Sara takes two red counters.(2)
(d)Find the probability that Sara takes one counter of each colour (in either order).(3)
(Total for Question 3 is 8 marks)
4
For each statement about tree diagrams, write down whether it is True or False.
(i)The probabilities on a set of branches from the same point on a tree diagram must add up to 1.(1)
(ii)To find the probability of two events both happening (an 'and' outcome), you add the branch probabilities together.(1)
(iii)To find the probability of either of two different end-of-tree outcomes (an 'or' outcome), you add the relevant end-of-tree probabilities.(1)
(Total for Question 4 is 3 marks)
5
150 people at a gym were asked whether they attend Yoga classes, split by whether they are Morning members or Evening members. The frequency tree shows some of the results.
Of the 150 people, 80 are Morning members and 70 are Evening members. Of the Morning members, 20 attend Yoga classes. In total (Morning and Evening members combined), 55 people attend Yoga classes.
(a)Work out the number of Morning members who do not attend Yoga classes.(2)
(b)Given that a total of 55 people at the gym attend Yoga classes, work out the number of Evening members who attend Yoga classes.(2)
(c)One person is chosen at random from all 150 people. Find the probability that this person is an Evening member who does not attend Yoga classes.(2)
(Total for Question 5 is 6 marks)
6
A spinner is divided into three unequal sections: red, blue and green. The probability of landing on red is 1/2, on blue is 1/3, and on green is 1/6. The spinner is spun once, then a fair coin is thrown. The tree diagram shows the spinner outcomes and part of the coin outcomes.
(a)Write down the probability that the spinner lands on green.(1)
(b)Complete the tree diagram by writing the two missing probabilities for the coin throw following the Green outcome.(2)
(c)Find the probability that the spinner lands on blue and the coin shows tails.(2)
(d)Show that the probability the coin shows heads is 1/2, using all three spinner outcomes, and explain why this value makes sense.(3)
(Total for Question 6 is 8 marks)
7
The probability that it rains on any given day in a particular week is 0.2, independently of any other day. Sam wants to show the possible outcomes for two consecutive days on a tree diagram.
(a)Draw a fully labelled tree diagram to show the possible outcomes (Rain, No rain) for Day 1 and Day 2, with the correct probability on each branch.(3)
(b)Find the probability that it rains on both days.(2)
(c)Find the probability that it rains on at least one of the two days.(3)
(Total for Question 7 is 8 marks)
8
A bag contains 10 marbles: 6 red and 4 blue. Two marbles are drawn at random from the bag, one after another, without replacement. The tree diagram shows the first draw, and part of the second draw.
(a)Complete the tree diagram by writing the four missing second-draw probabilities.(2)
(b)Find the probability that both marbles are red.(2)
(c)Find the probability that exactly one of the two marbles is blue.(3)
(d)Find the probability that at least one of the two marbles is red.(3)
(Total for Question 8 is 10 marks)
9
A student is investigating whether, when two counters are drawn without replacement from a bag containing red and blue counters, the two counters are more likely to be the same colour or different colours. She plans to model this using a tree diagram, then collect data by repeating a physical trial.
(a)State one reason why carrying out a large number of trials would make the experimental relative frequency a more reliable estimate of the theoretical probability.(1)
(b)The bag contains 8 counters: 5 red and 3 blue. Using a tree diagram, find the theoretical probability that two counters drawn at random without replacement are of different colours.(3)
(c)The student repeats the experiment 56 times and records different colours on 24 occasions. Find the experimental relative frequency, and comment on how it compares with the theoretical probability found in part (b).(2)
(Total for Question 9 is 6 marks)
10
A basketball player has a probability of 0.7 of scoring with any free throw, independently of any other attempt. She takes three free throws. The tree diagram shows the structure of the three attempts (every branching point uses the same probabilities: 0.7 for Score, 0.3 for Miss).
(a)Find the probability that she scores with all three free throws.(2)
(b)Find the probability that she scores with exactly two of the three free throws.(3)
(c)Find the probability that she scores with at least one of the three free throws.(2)
(Total for Question 10 is 7 marks)
11
To play a game at a school fair, a player pays £2. The player spins two independent fair spinners. Each spinner has a probability of 0.3 of landing on a star. If both spinners land on a star, the player wins a £10 prize. Otherwise, the player wins nothing.
(a)Complete the tree diagram by writing the missing probability of No star on Spinner 1, and the two missing probabilities for Spinner 2 shown in the dashed boxes.(2)
(b)Find the probability that the player wins the £10 prize (both spinners land on a star).(2)
(c)The game is played 200 times. Work out an estimate for the number of times the £10 prize is won.(2)
(d)Determine whether the school fair is likely to make a profit from this game over 200 plays, showing your working.(3)
(Total for Question 11 is 9 marks)
12
A bag contains 6 counters: 3 red, 2 yellow and 1 green. Two counters are drawn at random from the bag, one after another, without replacement. The tree diagram shows the first draw, and part of the second draw following a Red first counter.
(a)Complete the tree diagram by writing the three missing second-draw probabilities following a Red first counter.(2)
(b)Find the probability that both counters drawn are red.(2)
(c)Find the probability that one counter is red and the other is yellow (in either order).(3)
(d)Find the probability that at least one of the two counters drawn is green.(3)
(Total for Question 12 is 10 marks)
13
A bag contains counters that are either red or yellow. Ben takes a counter at random from the bag, notes its colour, and replaces it. He then takes a second counter at random. The probability that a counter taken from the bag is red is p. The probability that Ben takes two red counters is 4/25.
(a)Show that p satisfies the equation 25p2 = 4, and find the value of p.(3)
(b)Using your value of p, find the probability that Ben takes exactly one red counter from his two picks.(3)
(Total for Question 13 is 6 marks)
14
A box contains 10 tokens: 7 silver and 3 gold. Three tokens are drawn at random from the box, one after another, without replacement.
(a)Find the probability that all three tokens drawn are silver.(2)
(b)Find the probability that exactly two of the three tokens drawn are silver (and the other is gold).(4)
(c)Find the probability that at least one gold token is drawn.(2)
(Total for Question 14 is 8 marks)
15
A factory has two machines, A and B, that produce identical components. Machine A produces 60% of all the components, and Machine B produces the remaining 40%. 5% of the components from Machine A are defective, and 8% of the components from Machine B are defective. A component is selected at random from the factory's output. The tree diagram shows these probabilities.
(a)Find the probability that a component is from Machine A and is defective.(2)
(b)Find the probability that a randomly selected component is defective.(3)
(c)Given that a component is defective, find the probability that it came from Machine B.(3)
(Total for Question 15 is 8 marks)
Mark scheme · S28 Tree Diagrams
Question 1
(a) B1 0.5 oe cao
(a) Answer: 0.5
(b) M1 0.5 x 0.5, oe
(b) A1 0.25 oe cao
(b) Answer: 0.25
Question 2
(i) B1 B cao
(i) Answer: B) 0.45 and 0.55
(ii) B1 B cao
(ii) Answer: B) 0.65
Question 3
(a) B1 2/5 oe cao
(a) Answer: 2/5
(b) B1 Red = 3/5
(b) B1 Blue = 2/5
(b) Answer: Red = 3/5, Blue = 2/5
(c) M1 3/5 x 3/5, oe
(c) A1 9/25 oe cao
(c) Answer: 9/25
(d) M1 3/5 x 2/5, oe
(d) M1 2/5 x 3/5, oe
(d) A1 12/25 oe cao
(d) Answer: 12/25
Question 4
(i) B1 True cao
(i) Answer: True
(ii) B1 False cao
(ii) Answer: False
(iii) B1 True cao
(iii) Answer: True
Question 5
(a) M1 80 - 20, oe
(a) A1 60 cao
(a) Answer: 60
(b) M1 55 - 20, oe
(b) A1 35 cao
(b) Answer: 35
(c) M1 70 - 35 (= 35 evening, no yoga), then 35/150, oe, ft from part (b)
(c) A1 7/30 oe cao
(c) Answer: 7/30
Question 6
(a) B1 1/6 oe cao
(a) Answer: 1/6
(b) B1 Heads = 1/2
(b) B1 Tails = 1/2
(b) Answer: Heads = 1/2, Tails = 1/2
(c) M1 1/3 x 1/2, oe
(c) A1 1/6 oe cao
(c) Answer: 1/6
(d) M1 (1/2 x 1/2) + (1/3 x 1/2) + (1/6 x 1/2), oe
(d) B1 valid explanation, e.g. the coin throw is independent of the spinner, so P(Heads) must equal 1/2 whatever the spinner shows
(d) Answer: 1/2, because the coin is independent of the spinner so P(Heads) is always 1/2.
Question 7
(a) B1 correct branch structure: two branches at each of the two stages
(a) B1 all branches correctly labelled Rain and No rain
(a) B1 correct probabilities on every branch: 0.2 for Rain and 0.8 for No rain, at both stages
(a) Answer: Tree diagram with Rain (0.2) and No rain (0.8) branches at both Day 1 and Day 2.
(b) M1 0.2 x 0.2, oe
(b) A1 0.04 oe cao
(b) Answer: 0.04
(c) M1 0.8 x 0.8 (= 0.64), the probability of no rain on either day
(c) M1 1 - 0.64 (dependent on the first method mark)
(c) A1 0.36 cao
(c) Answer: 0.36
Question 8
(a) B1 after Red: Red = 5/9 and Blue = 4/9
(a) B1 after Blue: Red = 6/9 and Blue = 3/9
(a) Answer: After first Red: 5/9 Red, 4/9 Blue. After first Blue: 6/9 Red, 3/9 Blue.
(b) M1 6/10 x 5/9, oe
(b) A1 1/3 oe cao
(b) Answer: 1/3
(c) M1 6/10 x 4/9, oe
(c) M1 4/10 x 6/9, oe
(c) A1 8/15 oe cao
(c) Answer: 8/15
(d) M1 4/10 x 3/9 (= 12/90), the probability of no red marbles
(d) M1 1 - 12/90 (dependent on the first method mark)
(d) A1 13/15 oe cao
(d) Answer: 13/15
Question 9
(a) B1 valid reason, e.g. relative frequency from a small number of trials is subject to random variation; more trials reduce this variation so the relative frequency settles closer to the true probability
(a) Answer: With more trials, random variation has less effect, so the relative frequency becomes a more reliable estimate of the true (theoretical) probability.
(b) M1 5/8 x 3/7, oe
(b) M1 3/8 x 5/7, oe
(b) A1 15/28 oe cao
(b) Answer: 15/28
(c) B1 relative frequency = 24/56 = 3/7 (approximately 0.429)
(c) B1 valid comparison, e.g. this is somewhat lower than the theoretical probability of 15/28 (approximately 0.536), which is plausible sampling variation for a moderate number of trials
(c) Answer: Relative frequency = 24/56 = 3/7, approximately 0.429, which is lower than the theoretical probability of 15/28, approximately 0.536; this difference is consistent with normal sampling variation.
Question 10
(a) M1 0.7 x 0.7 x 0.7, oe
(a) A1 0.343 cao
(a) Answer: 0.343
(b) M1 0.7 x 0.7 x 0.3 (= 0.147), one arrangement of two scores and one miss
(b) M1 identifies there are 3 such arrangements (Score-Score-Miss, Score-Miss-Score, Miss-Score-Score)
(b) A1 0.441 cao
(b) Answer: 0.441
(c) M1 1 - 0.3 x 0.3 x 0.3, oe
(c) A1 0.973 cao
(c) Answer: 0.973
Question 11
(a) B1 No star (Spinner 1) = 0.7
(a) B1 Star = 0.3 and No star = 0.7 for Spinner 2
(a) Answer: No star (Spinner 1) = 0.7; Spinner 2: Star = 0.3, No star = 0.7
(b) M1 0.3 x 0.3, oe
(b) A1 0.09 oe cao
(b) Answer: 0.09
(c) M1 200 x 0.09, ft from part (b)
(c) A1 18 cao
(c) Answer: 18
(d) M1 total revenue = 200 x £2 = £400
(d) M1 total expected payout = 18 x £10 = £180, ft from part (c)
(d) A1 correct conclusion: profit of £220 (400 - 180), so yes, the fair is likely to make a profit
(d) Answer: Yes, an estimated profit of £220 (£400 revenue minus an estimated £180 paid out in prizes).
Question 12
(a) B1 Red = 2/5 and Yellow = 2/5
(a) B1 Green = 1/5
(a) Answer: Red = 2/5, Yellow = 2/5, Green = 1/5
(b) M1 3/6 x 2/5, oe
(b) A1 1/5 oe cao
(b) Answer: 1/5
(c) M1 3/6 x 2/5, oe (red then yellow)
(c) M1 2/6 x 3/5, oe (yellow then red)
(c) A1 2/5 oe cao
(c) Answer: 2/5
(d) M1 5/6 x 4/5 (= 20/30), the probability that neither counter is green
(d) M1 1 - 20/30 (dependent on the first method mark)
(d) A1 1/3 oe cao
(d) Answer: 1/3
Question 13
(a) M1 p x p = 4/25, oe (probability of red on both picks)
(a) M1 rearranges to 25p2 = 4
(a) A1 p = 2/5, rejecting the negative root since p is a probability
(a) Answer: p = 2/5
(b) M1 1 - p = 3/5
(b) M1 2 x (2/5 x 3/5), oe
(b) A1 12/25 oe cao
(b) Answer: 12/25
Question 14
(a) M1 7/10 x 6/9 x 5/8, oe
(a) A1 7/24 oe cao
(a) Answer: 7/24
(b) M1 one arrangement, e.g. 7/10 x 6/9 x 3/8, oe (= 126/720)
(b) M1 identifies there are 3 arrangements (SSG, SGS, GSS)
(b) M1 recognises each arrangement has the same probability (126/720 = 7/40) by symmetry, and multiplies by 3
(b) A1 21/40 oe cao
(b) Answer: 21/40
(c) M1 1 - 7/24, ft from part (a)
(c) A1 17/24 oe cao
(c) Answer: 17/24
Question 15
(a) M1 0.6 x 0.05, oe
(a) A1 0.03 cao
(a) Answer: 0.03
(b) M1 0.6 x 0.05 (+) 0.4 x 0.08, oe
(b) M1 0.03 + 0.032 (dependent, correctly adding both products)
(b) A1 0.062 cao
(b) Answer: 0.062
(c) M1 0.4 x 0.08 (= 0.032) identified as the numerator
(c) M1 divides by P(defective) = 0.062 from part (b), ft