Challenging +1.2 This is a standard negative binomial distribution question requiring formula application, algebraic manipulation of probability ratios, and finding a mode. While it involves Further Maths content (negative binomial), the techniques are routine: direct probability calculation, verifying a given formula, simplifying ratios, and solving an inequality. The multi-part structure and 10 marks indicate moderate length, but no novel insight is required—just systematic application of learned methods.
An archer shoots at a target. It may be assumed that each shot is independent of all other shots and that, on average, she hits the bull's-eye with 3 shots in 20. Find the probability that she requires at least 6 shots to hit the bull's-eye. [3]
When she hits the bull's-eye for the third time her total number of shots is \(Y\). Show that
$$\mathrm{P}(Y = r) = \frac{1}{2}(r - 1)(r - 2)\left(\frac{3}{20}\right)^3\left(\frac{17}{20}\right)^{r-3}.$$ [3]
Simplify \(\frac{\mathrm{P}(Y = r + 1)}{\mathrm{P}(Y = r)}\), and hence find the set of values of \(r\) for which \(\mathrm{P}(Y = r + 1) < \mathrm{P}(Y = r)\). Deduce the most probable value of \(Y\). [4]
EITHER: State physical significance: Equivalent to √z for zero speed B1
OR: State expected relationship: Constant force so v 2 ∝ z (B1)
Valid comment, √ on 0⋅021: 0⋅021 or 0⋅021 2 ≈ 0 as expected
Answer
Marks
or ≠ 0 suggests error in data B1
(2)
(1)
(4)
(5)
Answer
Marks
(2)
[14]
Question 10:
10 | Use geometric distribution [with p = 3/20 = 0⋅15] M1
5 2 3 4
Find prob. of missing in 5 shots: q or 1 – p (1 + q + q + q + q ) M1
(q 6 or 0⋅377 earns M1 A0) = 0⋅85 5 = 0⋅444 A1
Separate probabilities of hits: P(Y = r) = P(2 hits in r–1 shots) ×
th
P(hit on r shot) M1
Substitute using geometric distribution: = r–1 C2 × 0⋅15 2 × 0⋅85 r–3 × 0⋅15 M1
Replace r–1 C2 : A.G. = ½(r – 1)(r – 2) 0⋅15 3 × 0⋅85 r–3 A1
Simplify P(Y = r+1) / P(Y = r) : 0⋅85 r / (r – 2) or 17 r / 20 (r – 2) B1
Find required values of r: 17 r < 20 (r – 2) M1
r > 40/3 or 13⋅3 or 13 A1
Deduce most probable value of Y : 14 (√ on values of r) A1√ | (3)
(3)
(4) | [10]
Page 8 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2009 | 9231 | 02
11
EITHER | Find extension e0 at equilibrium position: 4mge0/l = mg, e0 = ¼l B1
2 2
Apply Newton’s law at general point: m d x/dt = mg – 4mg(e0 + x) / l M1 A1
Combine: d 2 x/dt 2 = – 4gx / l A.G. A1
2 ω2 2 2 2
Find SHM amplitude A (or A ) from initial speed: gl = A , A = ¼ l or A = ½ l M1 A1
2 ω2 2 2
Find speed v when string’s length is l: v = (A – e0 )
= (4g/l)(¼ l 2 – l 2 /16), v = √(¾gl) M1 A1
Find time from x = A to x = – e0 by e.g.: t1 = ω –1 cos –1 (– e0/A)
or ½T – ω –1 cos –1 (e0/A)
or ω –1 sin –1 (– e0/A) – ¼T M1
Substitute for ω, e0, A, T: t1 = ½√(l/g) cos –1 (– ½)
or ½π√(l/g) – ½√(l/g) cos –1 (½)
or ½√(l/g) sin –1 (– ½) – ¼π√(l/g) A1
Simplify: t1 = (π/3) √(l/g) A1
Find further time to rest (A.E.F.): t2 = v/g = √(¾l/g) or ½√(3l/g) M1 A1
Combine times: t1 + t2 = (⅓π + √3/2) √(l/g) A.G. A1 | (4)
(4)
(6) | [14]
Page 9 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2009 | 9231 | 02
11
OR | State what is minimised (A.E.F.): Sum of squares of residuals B1
Show residuals on copy of diagram: Verticals between points and line B1
State answer with valid reason (A.E.F.) e.g.: No, since points not collinear B1
(i) Calculate correlation coefficient:
r = (1022⋅15 – 340 × 22⋅41 / 8) / √ {(15 500 – 340 2 / 8) (67⋅65 – 22⋅41 2 / 8)} M1 A1
= 69⋅725/√(1050 × 4⋅874) = 0⋅975 *A1
State valid comment on diagram (A.E.F.): Points lie close to a straight line B1√
(B0 for ‘points are nearly correlated’)
(ii) State valid reason (A.E.F.; dep *A1): 0⋅975 closer to 1 than 0⋅965 B1
Find coefficient b in regression line for y: b = (1022⋅15 – 340 × 22⋅41 / 8) /
2
(15 500 – 340 / 8)
= 69⋅725/1050 = 0⋅0664 M1 A1
Find equation of regression line: √z = b(v – 42⋅5) + 2⋅80125
= 0⋅0664 v – 0⋅0207[5]
or 0⋅066 v – 0⋅021 M1 A1
(iii) All answers A.E.F.:
EITHER: State physical significance: Equivalent to √z for zero speed B1
OR: State expected relationship: Constant force so v 2 ∝ z (B1)
Valid comment, √ on 0⋅021: 0⋅021 or 0⋅021 2 ≈ 0 as expected
or ≠ 0 suggests error in data B1 | (2)
(1)
(4)
(5)
(2) | [14]
An archer shoots at a target. It may be assumed that each shot is independent of all other shots and that, on average, she hits the bull's-eye with 3 shots in 20. Find the probability that she requires at least 6 shots to hit the bull's-eye. [3]
When she hits the bull's-eye for the third time her total number of shots is $Y$. Show that
$$\mathrm{P}(Y = r) = \frac{1}{2}(r - 1)(r - 2)\left(\frac{3}{20}\right)^3\left(\frac{17}{20}\right)^{r-3}.$$ [3]
Simplify $\frac{\mathrm{P}(Y = r + 1)}{\mathrm{P}(Y = r)}$, and hence find the set of values of $r$ for which $\mathrm{P}(Y = r + 1) < \mathrm{P}(Y = r)$. Deduce the most probable value of $Y$. [4]
\hfill \mbox{\textit{CAIE FP2 2009 Q10 [10]}}