Challenging +1.8 This problem requires students to set up and solve an inequality involving factorials and combinatorial reasoning about permutations. While the conceptual setup is accessible (counting favorable outcomes vs total outcomes), it demands careful enumeration of cases (exactly correct OR one adjacent transposition), algebraic manipulation of the resulting expression (n+1)/n! < 0.01, and systematic testing of values. The multi-step reasoning and need to handle factorial inequalities places it well above average difficulty.
8 Alex claims that he can read people's minds. A volunteer, Jane, arranges the integers 1 to \(n\) in an order of Jane's own choice and Alex tells Jane what order he believes was chosen.
They agree that Alex's claim will be accepted if he gets the order completely correct or if he gets the order correct apart from two numbers which are the wrong way round.
They use a value of \(n\) such that, if Alex chooses the order of the integers at random, the probability that Alex's claim will be accepted is less than \(1 \%\).
Determine the smallest possible value of \(n\).
\section*{END OF QUESTION PAPER}
Probability of this or better \(= \dfrac{1 + {}^nC_2}{n!}\)
M1
\({}^nC_2\) or \({}^nP_2 = n(n-1)\) or \({}^nC_1 = n\) seen
\(+1\)
A1
2 omitted, or \((1+{}^nC_1)/n!\): M1A1A1M1 A0A0A0 \(= 4/7\)
\(\div\ n!\)
A1
\(\dfrac{1+{}^nC_2}{n!} < 0.01\)
M1
\(< 0.01\) and one relevant number substituted *or* attempt to simplify factorials
\(n = 6,\ p = 0.0222\ldots > 0.01\)
A1
One correct \(p\), compared with 0.01
\(n = 7,\ p = 0.004365\ldots < 0.01\)
A1
Both of these
The smallest value of \(n\) is 7
A1 [7]
Correct conclusion stated, allow if only one probability seen, www
# Question 8:
Probability of this or better $= \dfrac{1 + {}^nC_2}{n!}$ | **M1** | ${}^nC_2$ or ${}^nP_2 = n(n-1)$ or ${}^nC_1 = n$ seen
$+1$ | **A1** | 2 omitted, or $(1+{}^nC_1)/n!$: M1A1A1M1 A0A0A0 $= 4/7$
$\div\ n!$ | **A1**
$\dfrac{1+{}^nC_2}{n!} < 0.01$ | **M1** | $< 0.01$ and one relevant number substituted *or* attempt to simplify factorials | Needs attempt at non-1 term
$n = 6,\ p = 0.0222\ldots > 0.01$ | **A1** | One correct $p$, compared with 0.01 | OR $100 + 50n(n-1) < n!$
$n = 7,\ p = 0.004365\ldots < 0.01$ | **A1** | Both of these | $n(n-1)[(n-2)! - 50] > 100$
The smallest value of $n$ is 7 | **A1 [7]** | Correct conclusion stated, allow if only one probability seen, www | SR: Spearman: $(n=6,\ 0.9905 > 0.9429)$: SC B2
8 Alex claims that he can read people's minds. A volunteer, Jane, arranges the integers 1 to $n$ in an order of Jane's own choice and Alex tells Jane what order he believes was chosen.
They agree that Alex's claim will be accepted if he gets the order completely correct or if he gets the order correct apart from two numbers which are the wrong way round.
They use a value of $n$ such that, if Alex chooses the order of the integers at random, the probability that Alex's claim will be accepted is less than $1 \%$.
Determine the smallest possible value of $n$.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR Further Statistics AS 2019 Q8 [7]}}