2 The random variable \(X\) is equally likely to take any of the \(n\) integer values from \(m + 1\) to \(m + n\) inclusive. It is given that \(\mathrm { E } ( 3 X ) = 30\) and \(\operatorname { Var } ( 3 X ) = 36\). Determine the value of \(m\) and the value of \(n\). 326 cards are each labelled with a different letter of the alphabet, A to Z. The letters A, E, I, O and U are vowels.

1. Five cards are selected at random without replacement. Determine the probability that the letters on at least three of the cards are vowels.
2. All 26 cards are arranged in a line, in random order.
   1. Show that the probability that all the vowels are next to one another is \(\frac { 1 } { 2990 }\).
   2. Determine the probability that three of the vowels are next to each other, and the other two vowels are next to each other, but the five vowels are not all next to each other. A biased spinner has five sides, numbered 1 to 5 . Elmer spins the spinner repeatedly and counts the number of spins, \(X\), up to and including the first time that the number 2 appears. He carries out this experiment 100 times and records the frequency \(f\) with which each value of \(X\) is obtained. His results are shown in Table 1, together with the values of \(x f\). \begin{table}[h] \end{table}

      Question			Answer	Mark	AO	Guidance
      \multirow[t]{8}{*}{1}	\multirow{8}{*}{(a)}		\(\mathrm { H } _ { 0 } : m _ { A } = m _ { B } , \mathrm { H } _ { 1 } : m _ { A } < m _ { B }\) where \(m _ { A }\) and \(m _ { B }\)	B1	1.1	OR: Median journey times equal, oe. Allow if \(m\) s used but not defined	Allow "mean" or "average" only if "population" stated
      				М1	1.1	Find either \(\mathrm { P } ( \geq 219 )\) (218.5) or \(\mathrm { P } ( \leq 141 )\) (141.5)	Use of 0.9559 is M0 here. For CV method see below
      			Consider correct tail, either 219 or 141 ( \(R _ { m } = 219 , m ( m + n + 1 ) - R _ { m } = 141\) ) \(p = \Phi \left( \frac { 141.5 - 180 } { \sqrt { 510 } } \right) = 0.0441 \ldots\) BC	М1	1.1	0.0421, 0.0401, 0.470 (no/wrong cc, \(\sqrt { }\) ): M1	\(0.9559 > 0.9 :\) A1A1 (M1A1) \(0.9559 > 0.1 :\) A1A0 M0A0
      			0.0441 < 0.1	A1ft	1.1	Explicit comparison. FT on
      			Alternative:
      			CV \(180 - z \times \sqrt { 5 } 10\) 141 (141.5) used \(z = 1.282 \quad ( \mathrm { CV } = 151.05,151.058 .\). \(141.5 < 151.05 ( 85 ) \quad\) or \(218.5 > 208.95\)	М1 M1 A1 A1		Allow \(\sqrt { }\) errors Stated or implied CV and cc correct e.g. 141 < 150.55	\(180 + 1.282 \sqrt { } 510\) etc is M0 unless 219 (218.5) used, in which case give M2(A1A1) E.g. \(219 > 209.45\)
      			Reject \(\mathrm { H } _ { 0 }\). Significant evidence that route B takes longer	M1ft A1ft [8]	1.1 2.2b	Correct first conclusion Contextualised, not too definite	Needs like-with-like, e.g. 0.9559 with 0.9
      SC Sum of A's ranks \(= 435 - 219 = 216\) used: B1B0 M0M1A0A1 M1A1 max 5/8
      			Exx: \(\alpha\) : \(\quad \mathrm { H } _ { 0 }\) : Journey times are the same, \(\mathrm { H } _ { 1 }\) : journey times for \(B\) are higher: \(\beta\) : \(\quad \mathrm { H } _ { 0 }\) : No evidence that median journey times are different, etc:			B0 B0
      1	(b)		Must be a random sample (of all journeys) Or distributions must be same shape (necessary assumption for Wilcoxon ranksum test!)	B1 [1]	3.5b	Or equivalent. Allow "(journeys) independent"	Not "representative".
      2			\(\begin{aligned}	3 \mathrm { E } ( X ) = 30 \text { or } \mathrm { E } ( X ) = 10
      	9 \times \operatorname { Var } ( X ) = 36 \text { or } \operatorname { Var } ( X ) = 4 \end{aligned}\)	B1 B1	2.2a 2.2a	Used, stated or implied One of these, used, stated or implied

      Question			Answer	Mark	AO	Guidance
      \multirow{5}{*}{}	\multirow{5}{*}{}	\multirow{5}{*}{}	\(\frac { 1 } { 1 } \left( n ^ { 2 } - 1 \right) = 4\)	M1	2.2a	\(n = 7\) only, no need for "reject -7"	\multirow[b]{2}{*}{Allow if \(\mathrm { E } ( 3 X + m )\) used rather than \(\mathrm { E } [ 3 ( X + m ) ]\)}
      			\(\mathrm { E } ( X - m ) = \frac { 1 } { 2 } ( n + 1 )\)	M1	3.1b	Use expectation of uniform, e.g. \(2 m + n + 1 = 20\).
      			Alternative: \(\operatorname { Var } ( Y + m ) = \frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)\)	М1 A1 М1		\(n = 7\) only Use expectation of uniform, e.g. \(2 m + n + 1 = 20\).	No need for "reject -7"
      			\(10 - m = 4\)	М1	2.1	Validly derive single equation for \(m\)
      			\(m = 6\)	A1 [7]	2.2a	\(m = 6\) only	NB: \(\operatorname { Var } = ( n - 1 ) ^ { 2 } / 12\) is from continuous uniform!
      \multirow{4}{*}{3}	\multirow{4}{*}{(a)}	\multirow{4}{*}{}	\({ } ^ { 5 } C _ { 3 } \times { } ^ { 21 } C _ { 2 } + { } ^ { 5 } C _ { 4 } \times { } ^ { 21 } C _ { 1 } + 1 \quad [ = 2100 + 105 + 1 ]\)	M1dep	3.1b	Any correct pair of \({ } ^ { n } C _ { r }\) s multiplied	Or \(1 - \mathrm { P } ( 0,1,2 ) = 1 - .9665\)
      				A1	1.1	All terms correct
      			\(\div { } ^ { 26 } C _ { 5 } [ = 65780 ]\)	*M1	1.1
      			\(\frac { 1103 } { 32890 }\) or \(0.0335 \ldots\)	A1 [4]	3.2a	Awrt 0.0335 or any exact fraction	e.g. \(\frac { 2206 } { 65780 }\) or \(\frac { 264720 } { 7893600 }\)
      			Alternative: \(\frac { 5 } { 26 } \times \frac { 4 } { 25 } \times \frac { 3 } { 24 } \times \frac { 2 } { 23 } \times \frac { 1 } { 22 }\)	B1		Must have 5 oe, e.g. \({ } ^ { 5 } C _ { 1 }\)

      3

      (b)

      (i)

      \(\frac { 22 ! \times 5 ! } { 26 ! } \left( = \frac { 1 \times 2 \times 3 \times 4 \times 5 } { 23 \times 24 \times 25 \times 26 } = \frac { 120 } { 358800 } \right)\)

      M1 A1

      1.1 2.1

      Oe. Allow M1 for 21! instead of 22! Fully correct

      \(\frac { 1 \times 2 \times 3 \times 4 \times 5 } { 22 \times 23 \times 24 \times 25 \times 26 } :\) M1

      Question			Answer	Mark	AO	Guidance
      			\(= \frac { 1 } { 2990 } \quad\) AG	A1 [3]	2.2a	Correctly obtain AG using exact method	Allow even if no working after \(22 ! \times 5 ! \div 26\) !
      \multirow{8}{*}{3}	\multirow{8}{*}{(b)}	\multirow{8}{*}{(ii)}	22 fences: 22 for [VVV] \(\times 21\) for [VV]	M1	3.1b	Correct strategy, allow \({ } ^ { 22 } C _ { 2 }\) for \({ } ^ { 22 } P _ { 2 }\)
      			Consonants arranged in 21! ways	M1	1.1	At least one of these, no subtraction	\(21 ! \times 3 ! \times 2 ! \div 26 !\) M0M1 \(21 ! \times 3 ! \times 2 ! \times 22 \times 21\) : M2A0
      			Vowels arranged in 5! ways ( \(= { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 }\) )	A1	2.1	Both correct	NB: \({ } ^ { 5 } C _ { 3 } \times 3 ! \times 2 ! = 5 !\)
      			\(\begin{aligned}	\text { Product } \div 26 ! = \frac { 21 } { 2990 }
      	\left( = 2.832 \times 10 ^ { 24 } \div 4.0329 \times 10 ^ { 26 } \right) \end{aligned}\)	A1 [4]	3.2a	Allow from calculator but must be exact fraction
      			Alternative: Treat 21 consonants, [VVV] and [VV] as 23	М1	3.1b	Correct strategy, allow \(23 ! \times 2 ! \times 3 !\)	(Must subtract \(2 \times 1 / 2990\) as 23! method counts
      				A1	2.1	Correct \(\left( 5 ! = { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 } = \right. \left. { } ^ { 5 } C _ { 3 } \times 2 ! \times 3 ! \right)\)
      						M1 also for subtracting \(1 \times\)	[VVVVV] twice, once as [VVV][VV] and once as [VV][VVV]) (11/1495 is M1A1M1A0)
      			Answer is \(\frac { 21 } { 2990 }\)	A1	1.1	1/2990 Final answer, exact fraction

      \begin{center} \begin{tabular}{ | l | l | l | l | l | l | }