OCR S1 2009 June — Question 7 8 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2009
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPermutations & Arrangements
TypeCorrect ordering probability
DifficultyModerate -0.8 This is a straightforward combinations and probability question requiring basic counting principles. Part (i) is a direct application of C(8,3), part (ii) uses complementary counting or conditional probability with C(7,2)/C(8,3), and part (iii) applies permutations with 3!/(8×7×6). All techniques are standard S1 material with no conceptual challenges or novel problem-solving required.
Spec5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems

Three letters are selected at random from the 8 letters of the word COMPUTER, without regard to order.
  1. Find the number of possible selections of 3 letters. [2]
  2. Find the probability that the letter P is included in the selection. [3]
Three letters are now selected at random, one at a time, from the 8 letters of the word COMPUTER, and are placed in order in a line.
  1. Find the probability that the 3 letters form the word TOP. [3]

Part i
Answer: \(^nC_3 = 56\)
Marks: M1, A1 2
Part ii
Answer: \(^nC_2\) or or \(^nP_3/^nP_3\) \(= \frac{1}{8}\) not from incorrect
Marks: M1, A1 3
Guidance: \(^nC_1 + ^nC_1^nC_1\) or 21 or 8×7×6 or \(\frac{1}{8} × \frac{1}{7} × \frac{1}{6}\) or \(\frac{1}{8} × 3\) only or \(\frac{1}{8} + \frac{1}{8} × \frac{1}{7} + \frac{1}{8} × \frac{1}{7} × \frac{1}{6}\) \(= \frac{1}{8}\)
\(^nC_1 + ^nC_1^nC_1\) or 21 or 8×7×6 or \(\frac{1}{8} × \frac{1}{7} × \frac{1}{6}\) (M1). indep, dep ans < 1 (A1 3)
Part iii
Answer: \(^8P_3\) or 8×7×6 or \(^nC_1 × ^nC_1^nC_1\) or 336
Marks: M1, M1, A1 3
Guidance: \(\frac{1}{8} × ^8P_3\) only = \(\frac{1}{336}\) or 0.00298 (3 sf) (M1 A1 3). \(\frac{1}{8} × \frac{1}{7} × \frac{1}{6}\) only M2. If × or ÷: M1 (\(\frac{1}{8}\)) M1
### Part i
**Answer:** $^nC_3 = 56$
**Marks:** M1, A1 2

### Part ii
**Answer:** $^nC_2$ or or $^nP_3/^nP_3$ $= \frac{1}{8}$ not from incorrect
**Marks:** M1, A1 3
**Guidance:** $^nC_1 + ^nC_1^nC_1$ or 21 or 8×7×6 or $\frac{1}{8} × \frac{1}{7} × \frac{1}{6}$ or $\frac{1}{8} × 3$ only or $\frac{1}{8} + \frac{1}{8} × \frac{1}{7} + \frac{1}{8} × \frac{1}{7} × \frac{1}{6}$ $= \frac{1}{8}$

$^nC_1 + ^nC_1^nC_1$ or 21 or 8×7×6 or $\frac{1}{8} × \frac{1}{7} × \frac{1}{6}$ (M1). indep, dep ans < 1 (A1 3)

### Part iii
**Answer:** $^8P_3$ or 8×7×6 or $^nC_1 × ^nC_1^nC_1$ or 336
**Marks:** M1, M1, A1 3
**Guidance:** $\frac{1}{8} × ^8P_3$ only = $\frac{1}{336}$ or 0.00298 (3 sf) (M1 A1 3). $\frac{1}{8} × \frac{1}{7} × \frac{1}{6}$ only M2. If × or ÷: M1 ($\frac{1}{8}$) M1

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Three letters are selected at random from the 8 letters of the word COMPUTER, without regard to order.

\begin{enumerate}[label=(\roman*)]
\item Find the number of possible selections of 3 letters. [2]
\item Find the probability that the letter P is included in the selection. [3]
\end{enumerate}

Three letters are now selected at random, one at a time, from the 8 letters of the word COMPUTER, and are placed in order in a line.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the probability that the 3 letters form the word TOP. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR S1 2009 Q7 [8]}}