• This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. P(X = 3) = ${(0.1)^3}$ A1 $= 0.001$ AG P(X = 4) = P(VV̅V) + P(V̅VVV) + P(̅VVVV) $= 3 \times {(0.1)^3} \times 0.9$ (or equivalent) A1 $= 0.0027$ AG [3 marks]

METHOD 1 attempting to form equations in $a$ and $b$ M1 $\frac{{9 + 3a + b}}{{2000}} = \frac{1}{{1000}}{\text{ }}(3a + b = - 7)$ A1 $\frac{{16 + 4a + b}}{{2000}} \times \frac{9}{{10}} = \frac{{27}}{{10\,000}}{\text{ }}(4a + b = - 10)$ A1 attempting to solve simultaneously (M1) $a = - 3,{\text{ }}b = 2$ A1 METHOD 2 $P(X = n) = \left({\begin{matrix}{n - 1}\\2\end{matrix}}\right) \times {0.1^3} \times {0.9^{n - 3}}$ M1 $= \frac{{(n - 1)(n - 2)}}{{2000}} \times {0.9^{n - 3}}$ (M1)A1 $= \frac{{{n^2} - 3n + 2}}{{2000}} \times {0.9^{n - 3}}$ A1 $a = - 3,b = 2$ A1

**Note: **Condone the absence of $0.9^{n - 3}$ in the determination of the values of $a$ and $b$.

[5 marks]

METHOD 1 EITHER ${P}(X = n) = \frac{{n^2 - 3n + 2}}{{2000}} \times {0.9^{n - 3}}$ (M1) OR ${P}(X = n) = \left( {\begin{matrix} {n - 1} \\ 2 \end{matrix}} \right) \times {0.1^3} \times {0.9^{n - 3}}$ (M1) THEN ${ = \frac{{(n - 1)(n - 2)}}{{2000}} \times {0.9^{n - 3}}}$ A1 ${P}(X = n - 1) = \frac{{(n - 2)(n - 3)}}{{2000}} \times {0.9^{n - 4}}$ A1 ${\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n - 1)}} = \frac{{(n - 1)(n - 2)}}{{(n - 2)(n - 3)}} \times 0.9}$ A1 ${ = \frac{{0.9(n - 1)}}{{n - 3}}}$ AG METHOD 2 ${\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n - 1)}} = \frac{{\frac{{n^2 - 3n + 2}}{{2000}} \times {{0.9}^{n - 3}}}}{{\frac{{{{(n - 1)}^2} - 3(n - 1) + 2}}{{2000}} \times {{0.9}^{n - 4}}}}}$ (M1) ${ = \frac{{0.9({n^2} - 3n + 2)}}{{({n^2} - 5n + 6)}}}$ A1A1

**Note: **Award ***A1 ***for a correct numerator and ***A1 ***for a correct denominator.

${ = \frac{{0.9(n - 1)(n - 2)}}{{(n - 2)(n - 3)}}}$ A1 ${ = \frac{{0.9(n - 1)}}{{n - 3}}}$ AG [4 marks]