| Date | November 2016 | Marks available | 5 | Reference code | 16N.2.hl.TZ0.11 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Hence and State | Question number | 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 \({\text{P}}(X = 4) = 0.0027\).
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.
Markscheme
\({\text{P}}(X = 3) = {(0.1)^3}\) 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]
