Show that ${\text{P}}(X = 3) = 0.001$ and ${\text{P}}(X = 4) = 0.0027$.

Find the values of the constants $a$ and $b$.

Deduce that $\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n - 1)}} = \frac{{0.9(n - 1)}}{{n - 3}}$ for $n > 3$.

(i)     Hence show that $X$ has two modes ${m_1}$ and ${m_2}$.

(ii)     State the values of ${m_1}$ and ${m_2}$.

Determine the minimum value of $x$ such that the probability Kati receives at least one free gift is greater than 0.5.

${\text{P}}(X = 3) = {(0.1)^3}$    A1

$= 0.001$    AG

${\text{P}}(X = 4) = {\text{P}}(VV\bar VV) + {\text{P}}(V\bar VVV) + {\text{P}}(\bar VVVV)$    (M1)

$= 3 \times {(0.1)^3} \times 0.9$ (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

${\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n - 3}}$    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

${\text{P}}(X = n) = \frac{{{n^2} - 3n + 2}}{{2000}} \times {0.9^{n - 3}}$    (M1)

OR

${\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n - 3}}$    (M1)

THEN

$= \frac{{(n - 1)(n - 2)}}{{2000}} \times {0.9^{n - 3}}$    A1

${\text{P}}(X = n - 1) = \frac{{(n - 2)(n - 3)}}{{2000}} \times {0.9^{n - 4}}$    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]

(i)     attempting to solve $\frac{{0.9(n - 1)}}{{n - 3}} = 1$ for $n$     M1

$n = 21$    A1

$\frac{{0.9(n - 1)}}{{n - 3}} < 1 \Rightarrow n > 21$    R1

$\frac{{0.9(n - 1)}}{{n - 3}} > 1 \Rightarrow n < 21$    R1

$X$ has two modes     AG

Note: Award R1R1 for a clearly labelled graphical representation of the two inequalities (using $\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n - 1)}}$).

(ii)     the modes are 20 and 21     A1

[5 marks]

METHOD 1

$Y \sim {\text{B}}(x,{\text{ }}0.1)$    (A1)

attempting to solve ${\text{P}}(Y \geqslant 3) > 0.5$ (or equivalent eg $1 - {\text{P}}(Y \leqslant 2) > 0.5$) for $x$     (M1)

Note: Award (M1) for attempting to solve an equality (obtaining $x = 26.4$).

$x = 27$    A1

METHOD 2

$\sum\limits_{n = 0}^x {{\text{P}}(X = n) > 0.5}$    (A1)

attempting to solve for $x$     (M1)

$x = 27$    A1

[3 marks]