An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
Paper 42 (Extended)
Answer all questions. Calculators are permitted. Diagrams are not to scale unless specified.
11 题目 · 121 分
题目 1 · Structured Problem Solving
11 分
Let \(f(x) = x^3 - 3x^2 - 9x + 5\) and \(g(x) = 2x - 3\).
(a) Sketch the graph of \(y = f(x)\) for \(-3 \le x \le 5\). [2]
(b) Find the coordinates of: (i) the local maximum, [2] (ii) the local minimum. [2]
(c) On the same diagram, sketch the line \(y = g(x)\). [2]
(d) Find the \(x\)-coordinates of the points of intersection of \(y = f(x)\) and \(y = g(x)\). [3]
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解题
(a) The sketch should show a cubic curve with a local maximum in the second quadrant and a local minimum in the fourth quadrant, passing through the y-intercept at \((0,5)\).
(b) Using a graphic display calculator: (i) The local maximum is at \(x = -1\). \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\). Coordinates: \((-1, 10)\). (ii) The local minimum is at \(x = 3\). \(f(3) = 3^3 - 3(3)^2 - 9(3) + 5 = -22\). Coordinates: \((3, -22)\).
(c) The line \(y = 2x - 3\) is a straight line with gradient 2 and y-intercept at \((0, -3)\).
(d) Set \(f(x) = g(x)\): \(x^3 - 3x^2 - 9x + 5 = 2x - 3 \implies x^3 - 3x^2 - 11x + 8 = 0\). Using the intersection or solver function on the GDC, the \(x\)-coordinates are: \(x \approx -2.12\), \(x \approx 0.73\), \(x \approx 4.39\) (correct to 3 significant figures).
评分标准
(a) **B1** for correct cubic shape. **B1** for correct intercepts or key features shown.
(b)(i) **B2** for \((-1, 10)\). (If not fully correct, **B1** for \(x = -1\) or \(y = 10\)).
(b)(ii) **B2** for \((3, -22)\). (If not fully correct, **B1** for \(x = 3\) or \(y = -22\)).
(c) **B2** for a straight line with positive gradient and negative y-intercept. (If not fully correct, **B1** for a line through \((0, -3)\) or with positive gradient).
(d) **M1** for setting \(x^3 - 3x^2 - 9x + 5 = 2x - 3\). **A2** for all three values: \(-2.12\), \(0.73\), \(4.39\). (If only one or two correct, **A1\).
题目 2 · Structured Problem Solving
11 分
A drawer contains 12 socks. Some are blue and the rest are red. Let the number of blue socks be \(b\) and the number of red socks be \(r\). Two socks are taken from the drawer at random, one after the other, without replacement.
(a) (i) Show that the probability that both socks are blue is \(\frac{b(b-1)}{132}\). [2] (ii) Given that the probability that both socks are blue is \(\frac{5}{33}\), show that \(b^2 - b - 20 = 0\). [3] (iii) Find the number of blue socks in the drawer. [2]
(b) A second drawer contains 15 socks: 6 blue, 4 red, and 5 green. Two socks are taken from this drawer at random without replacement. Find the probability that: (i) both socks are the same colour, [2] (ii) at least one sock is green. [2]
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解题
(a) (i) The probability of selecting the first blue sock is \(\frac{b}{12}\). The probability of selecting the second blue sock without replacement is \(\frac{b-1}{11}\). The probability that both are blue is \(\frac{b}{12} \times \frac{b-1}{11} = \frac{b(b-1)}{132}\).
(ii) Set the probability equal to \(\frac{5}{33}\): \(\frac{b(b-1)}{132} = \frac{5}{33} \implies b(b-1) = \frac{5 \times 132}{33} = 20\). Thus, \(b^2 - b = 20 \implies b^2 - b - 20 = 0\).
(iii) Solve the quadratic equation: \((b - 5)(b + 4) = 0 \implies b = 5\) or \(b = -4\). Since the number of socks must be positive, \(b = 5\).
(ii) P(no green sock) = \(\frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}\). P(at least one green sock) = \(1 - \frac{3}{7} = \frac{4}{7}\).
评分标准
(a)(i) **M1** for \(\frac{b}{12} \times \frac{b-1}{11}\). **A1** for fully convincing show.
(a)(ii) **M1** for \(\frac{b(b-1)}{132} = \frac{5}{33}\). **M1** for \(b(b-1) = 20\). **A1** for \(b^2 - b - 20 = 0\).
(a)(iii) **M1** for factorising \((b-5)(b+4) = 0\) or using formula. **A1** for \(b = 5\).
(b)(i) **M1** for \(\frac{6}{15}\times\frac{5}{14} + \frac{4}{15}\times\frac{3}{14} + \frac{5}{15}\times\frac{4}{14}\). **A1** for \(\frac{31}{105}\) or equivalent decimal \(0.295\).
(b)(ii) **M1** for \(1 - \frac{10}{15}\times\frac{9}{14}\). **A1** for \(\frac{4}{7}\) or equivalent decimal \(0.571\).
题目 3 · Structured Problem Solving
11 分
A right pyramid \(VABCD\) has a square base \(ABCD\) of side 10 cm. The apex \(V\) is vertically above the centre of the base \(M\). The slant height (edge length) \(VA = 13\) cm.
(a) Show that the height of the pyramid, \(VM\), is \(\sqrt{119}\) cm. [3]
(b) Calculate the angle between the sloping face \(VBC\) and the base \(ABCD\). [4]
(c) Calculate the angle between the sloping edge \(VA\) and the base \(ABCD\). [4]
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解题
(a) Find the diagonal \(AC\) of the base: \(AC^2 = 10^2 + 10^2 = 200 \implies AC = 10\sqrt{2}\) cm. Since \(M\) is the midpoint of \(AC\), \(AM = 5\sqrt{2}\) cm. In the right-angled triangle \(VMA\): \(VM^2 = VA^2 - AM^2 = 13^2 - (5\sqrt{2})^2 = 169 - 50 = 119\). Therefore, \(VM = \sqrt{119}\) cm.
(b) Let \(N\) be the midpoint of \(BC\). Then \(MN = 5\) cm (half of the side length of the base). The angle between the sloping face \(VBC\) and the base \(ABCD\) is the angle \(VNM\) in the right-angled triangle \(VMN\). \(\tan(\angle VNM) = \frac{VM}{MN} = \frac{\sqrt{119}}{5}\). \(\angle VNM = \tan^{-1}\left(\frac{\sqrt{119}}{5}\right) \approx 65.4^\circ\).
(c) The angle between the sloping edge \(VA\) and the base \(ABCD\) is the angle \(VAM\). In the right-angled triangle \(VMA\): \(\sin(\angle VAM) = \frac{VM}{VA} = \frac{\sqrt{119}}{13}\). \(\angle VAM = \sin^{-1}\left(\frac{\sqrt{119}}{13}\right) \approx 57.0^\circ\).
评分标准
(a) **M1** for \(AC^2 = 10^2 + 10^2\) or \(AM = \sqrt{5^2 + 5^2}\). **M1** for \(VM^2 = 13^2 - AM^2\). **A1** for reaching \(\sqrt{119}\) with no errors.
(b) **M1** for identifying the angle \(VNM\). **B1** for \(MN = 5\) cm. **M1** for \(\tan \theta = \frac{\sqrt{119}}{5}\). **A1** for \(65.4^\circ\) or \(65.37^\circ\) to \(65.38^\circ\).
(c) **M1** for identifying the angle \(VAM\). **M1** for \(\sin \phi = \frac{\sqrt{119}}{13}\) or \(\cos \phi = \frac{5\sqrt{2}}{13}\). **A1** for \(5\sqrt{2}\) or \(7.07\). **A1** for \(57.0^\circ\) or \(57.04^\circ\) to \(57.05^\circ\).
题目 4 · Structured Problem Solving
11 分
Triangle \(T\) has vertices at \((1, 1)\), \((3, 1)\), and \((1, 4)\).
(a) State the coordinates of the vertices of the image when triangle \(T\) is: (i) Reflected in the line \(y = -x\). Label the image \(A\). [2] (ii) Translated by the vector \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\). Label the image \(B\). [2] (iii) Rotated by \(90^\circ\) anticlockwise about the origin. Label the image \(C\). [2]
(b) Describe fully the single transformation that maps: (i) Triangle \(T\) onto Triangle \(D\), which has vertices at \((2, 2)\), \((6, 2)\), and \((2, 8)\). [2] (ii) Triangle \(T\) onto Triangle \(E\), which has vertices at \((1, -1)\), \((3, -1)\), and \((1, -4)\). [3]
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解题
(a) (i) A reflection in the line \(y = -x\) maps \((x, y) \to (-y, -x)\). Vertices: \((-1, -1)\), \((-1, -3)\), \((-4, -1)\). (ii) Translation by \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\) maps \((x, y) \to (x-3, y+2)\). Vertices: \((-2, 3)\), \((0, 3)\), \((-2, 6)\). (iii) Rotation of \(90^\circ\) anticlockwise about the origin maps \((x, y) \to (-y, x)\). Vertices: \((-1, 1)\), \((-1, 3)\), \((-4, 1)\).
(b) (i) Enlargement with scale factor 2 and centre at the origin \((0, 0)\), since all coordinates are doubled. (ii) Reflection in the x-axis (or the line \(y = 0\)), since the y-coordinates are negated while the x-coordinates remain unchanged.
评分标准
(a)(i) **B2** for all three vertices correct. (If 1 or 2 correct, **B1**).
(a)(ii) **B2** for all three vertices correct. (If 1 or 2 correct, **B1**).
(a)(iii) **B2** for all three vertices correct. (If 1 or 2 correct, **B1**).
(b)(i) **B1** for "Enlargement, scale factor 2". **B1** for "centre \((0, 0)\)" or "origin".
(b)(ii) **B2** for "Reflection in the x-axis" or "Reflection in the line \(y = 0\)". (If not fully correct, **B1** for "Reflection" but wrong or missing line).
题目 5 · Structured Problem Solving
11 分
A triangular field \(ABC\) has sides \(AB = 75\) m, \(BC = 110\) m, and angle \(ABC = 64^\circ\).
(a) Calculate the distance \(AC\). [3]
(b) Calculate the angle \(ACB\). [3]
(c) Calculate the area of the field. [2]
(d) A vertical tower of height \(h\) m stands at \(A\). The angle of elevation of the top of the tower from \(B\) is \(18^\circ\). Calculate the value of \(h\). [3]
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解题
(a) Use the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(B)\) \(AC^2 = 75^2 + 110^2 - 2(75)(110)\cos(64^circ) \approx 10491.9\) \(AC \approx 102.43 \approx 102\) m.
(b) Use the Sine Rule to find angle \(ACB\) (let it be \(C\)): \(\frac{\sin(C)}{AB} = \frac{\sin(B)}{AC}\) \(\sin(C) = \frac{75 \times \sin(64^\circ)}{102.43} \approx 0.6581\) \(C = \sin^{-1}(0.6581) \approx 41.2^\circ\). (If using rounded value of \(102\) m for \(AC\), the answer is \(41.4^\circ\). Answers between \(41.1^\circ\) and \(41.4^\circ\) are accepted).
(c) \(\text{Area} = \frac{1}{2} a c \sin(B) = \frac{1}{2} \times 110 \times 75 \times \sin(64^\circ) \approx 3710\) m\(^2\).
(d) The triangle \(AB\text{Top}\) is right-angled at \(A\): \(\tan(18^\circ) = \frac{h}{AB} = \frac{h}{75}\) \(h = 75 \tan(18^\circ) \approx 24.4\) m.
评分标准
(a) **M1** for \(75^2 + 110^2 - 2(75)(110)\cos 64^\circ\). **A1** for \(10490\) to \(10500\). **A1** for \(102\) or \(102.4\).
(b) **M1** for \(\frac{\sin C}{75} = \frac{\sin 64^\circ}{\text{their } AC}\). **M1** for \(\sin C = \frac{75\sin 64^\circ}{\text{their } AC}\). **A1** for \(41.1^\circ\) to \(41.4^\circ\).
(c) **M1** for \(\frac{1}{2} \times 110 \times 75 \sin 64^\circ\). **A1** for \(3710\) or \(3708\).
(d) **M1** for identifying the right-angled triangle. **M1** for \(\tan 18^\circ = \frac{h}{75}\). **A1** for \(24.4\) or \(24.37\).
题目 6 · Structured Problem Solving
11 分
(a) Liam invests $5000 in a bank account that pays compound interest at a rate of 3.2% per year. (i) Calculate the value of the investment at the end of 5 years. Give your answer correct to the nearest dollar. [3] (ii) Find the number of complete years it takes for the investment to double in value. [3]
(b) Sophia buys a car for $24 000. It depreciates at a rate of \(r\%\) per year. At the end of 4 years, the value of the car is $15 746.40. (i) Show that \(r = 10\). [3] (ii) Find the value of the car at the end of 8 years. [2]
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解题
(a) (i) Value = \(5000 \times (1 + 0.032)^5 = 5000 \times (1.032)^5 \approx 5852.86 \approx \$5853\). (ii) \(5000 \times (1.032)^n \ge 10000 \implies (1.032)^n \ge 2\). Taking logarithms: \(n \ge \frac{\log(2)}{\log(1.032)} \approx 21.999\) years. Thus, the number of complete years is 22.
(b) (i) \(15746.40 = 24000 \times (1 - \frac{r}{100})^4\) \((1 - \frac{r}{100})^4 = 0.6561\) \(1 - \frac{r}{100} = \sqrt[4]{0.6561} = 0.9 \implies r = 10\). (ii) Value at 8 years = \(24000 \times (0.9)^8 = \$10331.21\).
评分标准
(a)(i) **M2** for \(5000 \times 1.032^5\). (If not M2, **M1** for \(5000 \times 1.032^k\) with \(k \ne 5\)). **A1** for \(5853\).
(a)(ii) **M1** for \(1.032^n = 2\). **M1** for \(n = \frac{\log 2}{\log 1.032}\). **A1** for \(22\).
(b)(i) **M1** for \(24000 \times (1 - \frac{r}{100})^4 = 15746.40\). **M1** for \((1 - \frac{r}{100}) = \sqrt[4]{0.6561}\). **A1** for convincing conclusion \(r = 10\).
(b)(ii) **M1** for \(24000 \times 0.9^8\) or \(15746.40 \times 0.9^4\). **A1** for \(10331.21\) (or \(10330\)).
题目 7 · Structured Problem Solving
11 分
Let \(f(x) = 3x - 5\) and \(g(x) = \frac{x+2}{2x-1}\), \(x \ne 0.5\).
(a) Find: (i) \(f(2)\), [1] (ii) \(g(f(x))\) in its simplest form. [3]
(a)(ii) **M1** for substituting \(3x-5\) into \(g(x)\). **A1** for numerator \(3x-3\) correct. **A1** for denominator \(6x-11\) correct.
(b) **M1** for \(x = 3y-5\) or \(y+5 = 3x\). **A1** for \(\frac{x+5}{3\}. (c) **M1** for \)y(2x-1) = x+2\). **M1** for collecting \(x\) terms: \(x(2y-1) = y+2\). **A1** for \(\frac{x+2}{2x-1}\).
(d) **M1** for \(x+2 = 2(2x-1)\). **A1** for \(\frac{4}{3}\) or \(1.33\).
题目 8 · Structured Problem Solving
11 分
A shape is made from a sector of a circle \(OAB\) of radius \(r\) cm and angle \(AOB = 120^\circ\), and an equilateral triangle \(OBC\) of side length \(r\) cm attached along the radius \(OB\).
(a) Write down an expression in terms of \(\pi\) and \(r\) for: (i) the area of the sector \(OAB\), [2] (ii) the arc length \(AB\). [2]
(b) Show that the total area of the shape is \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\). [3]
(c) Given that the total area of the shape is 150 cm\(^2\), calculate: (i) the value of \(r\), [3] (ii) the perimeter of the shape. [2]
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解题
(a) (i) Area of sector = \(\frac{120}{360} \times \pi r^2 = \frac{\pi r^2}{3}\). (ii) Arc length = \(\frac{120}{360} \times 2 \pi r = \frac{2\pi r}{3}\).
(b) Area of equilateral triangle \(OBC\) is \(\frac{1}{2} \times r \times r \times \sin(60^\circ) = \frac{\sqrt{3}}{4}r^2\). Total Area = Sector Area + Triangle Area \(\text{Total Area} = \frac{\pi r^2}{3} + \frac{\sqrt{3}}{4}r^2 = r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\).
(c) (i) \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) = 150 \implies 1.48021 r^2 = 150 \implies r^2 = 101.337 \implies r \approx 10.1\) cm. (ii) The perimeter consists of the arc \(AB\), and the three straight edges \(OA\), \(BC\), and \(CO\) of length \(r\). \(\text{Perimeter} = \frac{2\pi r}{3} + 3r \approx 10.0666 \times (2.0944 + 3) \approx 51.3\) cm.
评分标准
(a)(i) **M1** for \(\frac{120}{360} \times \pi r^2\). **A1** for \(\frac{\pi r^2}{3}\).
(a)(ii) **M1** for \(\frac{120}{360} \times 2 \pi r\). **A1** for \(\frac{2\pi r}{3}\).
(b) **M1** for Area of triangle = \(\frac{\sqrt{3}}{4}r^2\). **M1** for summing the two areas. **A1** for factoring out \(r^2\) to get \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\).
(c)(i) **M1** for \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) = 150\). **M1** for \(1.48... r^2 = 150\). **A1** for \(10.1\) or \(10.07\).
(c)(ii) **M1** for Perimeter = \(\text{their } AB + 3r\). **A1** for \(51.3\) or \(51.28\).
题目 9 · Structured Problem Solving
11 分
A pyramid \(VABCD\) has a rectangular base \(ABCD\). \(AB = 8\text{ cm}\) and \(BC = 6\text{ cm}\). The vertex \(V\) is vertically above the center \(O\) of the base. The height of the pyramid, \(VO\), is \(12\text{ cm}\).
(a) Show that \(AC = 10\text{ cm}\).
(b) Calculate the length of a sloping edge, \(VA\).
(c) Calculate the angle between the sloping edge \(VA\) and the base \(ABCD\).
(d) Calculate the angle between the triangular face \(VBC\) and the base \(ABCD\).
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解题
(a) Since \(ABCD\) is a rectangle, angle \(ABC = 90^\circ\). Applying Pythagoras' theorem in triangle \(ABC\): \(AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100\) \(AC = \sqrt{100} = 10\text{ cm}\).
(b) Since \(O\) is the center of the rectangular base, \(AO = \frac{1}{2} AC = 5\text{ cm}\). In the right-angled triangle \(VOA\): \(VA^2 = VO^2 + AO^2 = 12^2 + 5^2 = 144 + 25 = 169\) \(VA = \sqrt{169} = 13\text{ cm}\).
(c) The angle between the sloping edge \(VA\) and the base \(ABCD\) is the angle \(\angle VAO\). Let this angle be \(\theta\). In triangle \(VOA\): \(\tan(\theta) = \frac{VO}{AO} = \frac{12}{5} = 2.4\) \[\theta = \arctan(2.4) \approx 67.38^\circ \approx 67.4^\circ\ (to 1 d.p.)\]
(d) Let \(M\) be the midpoint of \(BC\). The line \(VM\) is the perpendicular height of the triangular face \(VBC\). Since \(ABCD\) is a rectangle with center \(O\), \(OM\) is parallel to \(AB\) and \(OM = \frac{1}{2} AB = 4\text{ cm}\). The angle between the face \(VBC\) and the base \(ABCD\) is \(\angle VMO\). Let this angle be \(\phi\). In the right-angled triangle \(VOM\): \(\tan(\phi) = \frac{VO}{OM} = \frac{12}{4} = 3\) \[\phi = \arctan(3) \approx 71.57^\circ \approx 71.6^\circ\ (to 1 d.p.)\]
评分标准
(a) [2 marks] M1 for \(8^2 + 6^2\) seen A1 for \(\sqrt{100} = 10\) clearly shown
(b) [2 marks] M1 for \(12^2 + 5^2\) (where \(AO = 5\) is either stated or used) A1 for \(13\text{ cm}\)
(c) [3 marks] M1 for identifying angle \(VAO\) as the required angle M1 for correct trig ratio, e.g. \(\tan(\theta) = \frac{12}{5}\) or \(\sin(\theta) = \frac{12}{13}\) A1 for \(67.4^\circ\) (accept \(67.38^\circ\) to \(67.4^\circ\))
(d) [4 marks] M1 for identifying angle \(VMO\) as the required angle (where \(M\) is the midpoint of \(BC\)) B1 for stating or using \(OM = 4\) M1 for correct trig ratio, e.g. \(\tan(\phi) = \frac{12}{4}\) A1 for \(71.6^\circ\) (accept \(71.56^\circ\) to \(71.6^\circ\))
题目 10 · Structured Problem Solving
11 分
Three functions are defined as: \(f(x) = 3x - 5\) \(g(x) = \frac{12}{x+2}\) for \(x \neq -2\) \(h(x) = x^2 - 2x\)
(a) Find the value of: (i) \(f(-2)\) (ii) \(g(4)\) (iii) \(h(f(3))\)
(a) [4 marks] (i) B1 for \(-11\) (ii) B1 for \(2\) (iii) M1 for finding \(f(3) = 4\) seen or implied, A1 for \(8\)
(b) [3 marks] M1 for multiplying by \(x+2\) to get \(y(x+2) = 12\) or \(x(y+2) = 12\) M1 for isolating \(x\) (or \(y\)) correctly A1 for \(g^{-1}(x) = \frac{12-2x}{x}\) or equivalent, e.g. \(\frac{12}{x}-2\)
(c) [4 marks] M1 for setting up equation \(3x - 5 = x^2 - 2x\) M1 for rearranging into quadratic standard form \(x^2 - 5x + 5 = 0\) A1 for \(1.38\) (accept \(1.381\) to \(1.382\)) A1 for \(3.62\) (accept \(3.617\) to \(3.618\))
题目 11 · Structured Problem Solving
11 分
A bag contains \(n\) marbles, of which 6 are green and the rest are yellow. Two marbles are drawn at random from the bag without replacement.
(a) The probability that both marbles are green is \(\frac{1}{3}\). (i) Show that \(n^2 - n - 90 = 0\). (ii) Solve this equation to find the value of \(n\).
(b) Using the value of \(n\) found in part (a), find the probability that: (i) the two marbles drawn have different colors, (ii) at least one green marble is drawn.
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解题
(a) (i) The probability of drawing the first green marble is \(\frac{6}{n}\). Since the first green marble is not replaced, the probability of drawing a second green marble is \(\frac{5}{n-1}\). Therefore, the probability of both being green is: \(P(G, G) = \frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\) Set this equal to \(\frac{1}{3}\): \(\frac{30}{n(n-1)} = \frac{1}{3}\) \(n(n-1) = 90\) \(n^2 - n = 90\) \(n^2 - n - 90 = 0\).
(ii) Factorise the quadratic equation: \((n - 10)(n + 9) = 0\) So, \(n = 10\) or \(n = -9\). Since the total number of marbles must be positive, \(n = 10\).
(b) With \(n = 10\), the bag contains 10 marbles in total: 6 green and \(10 - 6 = 4\) yellow. (i) The two drawn marbles have different colors if the outcome is Green then Yellow (GY) or Yellow then Green (YG). \(P(G, Y) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\) \(P(Y, G) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\) \(P(\text{different}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\) (or \(\approx 0.533\)).
(ii) The probability of drawing at least one green marble is: \(P(\text{at least one green}) = 1 - P(Y, Y)\) \(P(Y, Y) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\) \(P(\text{at least one green}) = 1 - \frac{2}{15} = \frac{13}{15}\) (or \(\approx 0.867\)).
评分标准
(a) (i) [4 marks] M1 for probability of first green being \(\frac{6}{n}\) M1 for probability of second green being \(\frac{5}{n-1}\) M1 for setting product equal to \(\frac{1}{3}\) A1 for correct expansion and algebraic steps leading to \(n^2 - n - 90 = 0\)
(ii) [2 marks] M1 for factorisation \((n-10)(n+9) = 0\) or correct use of quadratic formula A1 for \(n = 10\) (rejecting \(-9\))
(b) [5 marks] (i) [3 marks] M1 for correct calculation of one branch, e.g. \(\frac{6}{10} \times \frac{4}{9}\) M1 for adding the two possible different outcomes: \(\frac{24}{90} + \frac{24}{90}\) A1 for \(\frac{8}{15}\) or \(\frac{48}{90}\) or \(0.533\) (accept \(0.5333...\))
(ii) [2 marks] M1 for using \(1 - P(Y, Y)\) or \(P(G, G) + P(G, Y) + P(Y, G)\) A1 for \(\frac{13}{15}\) or \(\frac{78}{90}\) or \(0.867\) (accept \(0.8666...\))