| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Nested binomial expected count |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question requiring standard calculations: deriving variance from given parameters, computing cumulative probability P(X≤2), and applying a nested binomial for part (iii). All steps are routine applications of formulas with no conceptual challenges beyond recognizing the two-stage structure. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(1.2 = 15p\); \(p = 0.08\); Var \(= npq = 15 \times 0.08 \times 0.92 = 1.104\) (AG) | M1 A1 | Attempt to find \(p\) using \(1.2 = 15p\); Correct answer |
| 2 marks | ||
| (ii) \(P(0, 1, 2) = (0.92)^{15} + ^{15}C_1(0.08)(0.92)^{14} + ^{15}C_2(0.08)^2(0.92)^{13}\) | M1 M1 A1 | Binomial expression \(^{15}C_r p^r(1-p)^{15-r}\), \(0 < p < 1\); Correct unsimplified expression for P(0, 1, 2); Correct answer |
| \(= 0.887\) | 3 marks | |
| (iii) \(P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15} = 0.7137\ldots\) | M1 | Attempt at P(0) or 1 – P(0) |
| \(P(\text{at least 1 faulty screw in 7 packets}) = ^7C_1(0.713\ldots)^1(0.2863\ldots) = 0.216\) | M1 A1 | Rounding to 0.71; Binomial expression \(^7C_p^r(1-p) < p < 1\) |
| 4 marks |
**(i)** $1.2 = 15p$; $p = 0.08$; Var $= npq = 15 \times 0.08 \times 0.92 = 1.104$ (AG) | M1 A1 | Attempt to find $p$ using $1.2 = 15p$; Correct answer |
| | | 2 marks |
**(ii)** $P(0, 1, 2) = (0.92)^{15} + ^{15}C_1(0.08)(0.92)^{14} + ^{15}C_2(0.08)^2(0.92)^{13}$ | M1 M1 A1 | Binomial expression $^{15}C_r p^r(1-p)^{15-r}$, $0 < p < 1$; Correct unsimplified expression for P(0, 1, 2); Correct answer |
| $= 0.887$ | | 3 marks |
**(iii)** $P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15} = 0.7137\ldots$ | M1 | Attempt at P(0) or 1 – P(0) |
| $P(\text{at least 1 faulty screw in 7 packets}) = ^7C_1(0.713\ldots)^1(0.2863\ldots) = 0.216$ | M1 A1 | Rounding to 0.71; Binomial expression $^7C_p^r(1-p) < p < 1$ |
| | | 4 marks |
---5 Screws are sold in packets of 15. Faulty screws occur randomly. A large number of packets are tested for faulty screws and the mean number of faulty screws per packet is found to be 1.2 .\\
(i) Show that the variance of the number of faulty screws in a packet is 1.104 .\\
(ii) Find the probability that a packet contains at most 2 faulty screws.
Damien buys 8 packets of screws at random.\\
(iii) Find the probability that there are exactly 7 packets in which there is at least 1 faulty screw.
\hfill \mbox{\textit{CAIE S1 2014 Q5 [9]}}