IB Questionbank

User interface language: English | Español

Date	November 2016	Marks available	3	Reference code	16N.2.AHL.TZ0.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Determine	Question number	H_11	Adapted from	N/A

Question

A Chocolate Shop advertises free gifts to customers that collect three vouchers. The vouchers are placed at random into 10% of all chocolate bars sold at this shop. Kati buys some of these bars and she opens them one at a time to see if they contain a voucher. Let ${\text{P}}(X = n)$ be the probability that Kati obtains her third voucher on the $n{\text{th}}$ bar opened.

(It is assumed that the probability that a chocolate bar contains a voucher stays at 10% throughout the question.)

It is given that ${\text{P}}(X = n) = \frac{{{n^2} + an + b}}{{2000}} \times {0.9^{n - 3}}$ for $n \geqslant 3,{\text{ }}n \in \mathbb{N}$ .

Kati’s mother goes to the shop and buys $x$ chocolate bars. She takes the bars home for Kati to open.

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

[3]

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

[5]

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

[4]

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

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

[5]

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

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{P}}(X = 3) = {(0.1)^3}$ 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)

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