| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | State general binomial conditions |
| Difficulty | Easy -1.2 Part (i) is pure recall of standard binomial conditions (fixed n, constant p, independence, binary outcomes). Parts (ii) and (iii) involve straightforward application of binomial probability formulas with no conceptual challenges—just substitution into P(X=r) and basic probability arithmetic. This is a routine textbook exercise testing recall and basic calculation rather than problem-solving. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Boxes are independent | B1 | Both must be in context |
| Probability same for each box | B1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B(8, 0.1)\) | M1 | \(B(8, 0.1)\) stated or \(0.1\), \(0.9\) seen and sum of powers \(= 8\) |
| \(0.4305\) | A1 | \(0.43[05]\) correct |
| \(1 - P(\leq 1)\) | M1 | \(1 - 0.8131\) or \(1 - (0.9^8 + 8 \times 0.9^7 \times 0.1)\) correct |
| \(0.1869\) | A1 | 4 marks, answer a.r.t. \(0.187\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times 0.4305 \times 0.1869\) | M1 | \((a) \times (b)\) |
| M1 | \(2 \times (a) \times (b)\) | |
| \(0.16092\) | A1 | 3 marks, answer a.r.t. \(0.161\) |
# Question 7:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Boxes are independent | B1 | Both must be in context |
| Probability same for each box | B1 | 2 marks total |
## Part (ii)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(8, 0.1)$ | M1 | $B(8, 0.1)$ stated or $0.1$, $0.9$ seen and sum of powers $= 8$ |
| $0.4305$ | A1 | $0.43[05]$ correct |
| $1 - P(\leq 1)$ | M1 | $1 - 0.8131$ or $1 - (0.9^8 + 8 \times 0.9^7 \times 0.1)$ correct |
| $0.1869$ | A1 | 4 marks, answer a.r.t. $0.187$ |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times 0.4305 \times 0.1869$ | M1 | $(a) \times (b)$ |
| | M1 | $2 \times (a) \times (b)$ |
| $0.16092$ | A1 | 3 marks, answer a.r.t. $0.161$ |
---7 It is known that, on average, one match box in 10 contains fewer than 42 matches. Eight boxes are selected, and the number of boxes that contain fewer than 42 matches is denoted by $Y$.\\
(i) State two conditions needed to model $Y$ by a binomial distribution.
Assume now that a binomial model is valid.\\
(ii) Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( Y = 0 )$,
\item $\mathrm { P } ( Y \geqslant 2 )$.\\
(iii) On Wednesday 8 boxes are selected, and on Thursday another 8 boxes are selected. Find the probability that on one of these days the number of boxes containing fewer than 42 matches is 0 , and that on the other day the number is 2 or more.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2005 Q7 [9]}}