Binomial

Question

The flight times, T minutes, between two cities can be modelled by a normal distribution with a mean of 75 minutes and a standard deviation of σ minutes.

(a) Given that 2 % of the flight times are longer than 82 minutes, find the value of σ . [3]
(b) Find the probability that a randomly selected flight will have a flight time of more than 80 minutes. [2]
(c) Given that a flight between the two cities takes longer than 80 minutes, find the probability that it takes less than 82 minutes. [4]
On a particular day, there are 64 flights scheduled between these two cities.
(d) Find the expected number of flights that will have a flight time of more than 80 minutes. [3]
(e) Find the probability that more than 6 of the flights on this particular day will have a flight time of more than 80 minutes. [3]

Answer/Explanation

Ans:

(a) $T\sim \text{N}\left(75,\sigma ^2\right)$.
Since $\text{P}\left(T\gt 82\right)=0.02$, we have $$\begin{eqnarray} \text{P}\left(T\lt 82\right)=0.98 \nonumber \\ \text{P}\left(\frac{T-75}{\sigma}\lt \frac{82-75}{\sigma}\right)=0.98 \nonumber \\ \text{P}\left(Z\lt \frac{82-75}{\sigma}\right)=0.98 \nonumber \\ \frac{82-75}{\sigma} = 2.05375 \nonumber \\ \sigma = 3.41. \end{eqnarray}$$ (b) From the graphing calculator, we have $\text{P}\left(T\gt 80\right)=0.0712$.
(c) $$\begin{eqnarray} \text{P}\left(T\lt 82|T\gt 80\right) &=& \frac{\text{P}\left(T\lt 82\cap T\gt 80\right)}{\text{P}\left(T\gt 80\right)} \nonumber \\ &=& \frac{\text{P}\left(80\lt T\lt 82\right)}{\text{P}\left(T\gt 80\right)} \nonumber \\ &=& 0.719. \end{eqnarray}$$ (d) Let $F$ be the number of flights with a flight time of more than $80$ minutes in $64$ flights, then $F\sim \text{B}\left(64,\text{P}\left(T\gt 80\right)\right)$.
$$\begin{eqnarray} \text{E}\left(F\right) &=& 64\text{P}\left(T\gt 80\right) \nonumber \\ &=& 4.56. \end{eqnarray}$$ (e) $$\begin{eqnarray} \text{P}\left(F\gt 6\right) &=& 1-\text{P}\left(F\leq 6\right) \nonumber \\ &=& 0.169. \end{eqnarray}$$

Question

The lifts in the office buildings of a small city have occasional breakdowns. The breakdowns at any given time are independent of one another and can be modelled using a Poisson Distribution with mean 0.2 per day.

(a) Determine the probability that there will be exactly four breakdowns during the month of June (June has 30 days).

(b) Determine the probability that there are more than 3 breakdowns during the month of June.

(d) Find the probability that the first breakdown in June occurs on June ${3^{{\text{rd}}}}$.

(e) It costs 1850 Euros to service the lifts when they have breakdowns. Find the expected cost of servicing lifts for the month of June.

(f) Determine the probability that there will be no breakdowns in exactly 4 out of the first 5 days in June.

Answer/Explanation

Markscheme

(a) mean for 30 days: $30 \times 0.2 = 6$ . (A1)

${\text{P}}(X = 4) = \frac{{{6^4}}}{{4!}}{{\text{e}}^{ – 6}} = 0.134$ (M1)A1 N3

[3 marks]

(b) ${\text{P}}(X > 3) = 1 – {\text{P}}(X \leqslant 3) = 1 – {{\text{e}}^{ – 6}}(1 + 6 + 18 + 36) = 0.849$ (M1)A1 N2

[2 marks]

mean for five days: $5 \times 0.2 = 1$ (A1)

${\text{P}}(X = 0) = {{\text{e}}^{ – 1}}\,\,\,\,\,( = 0.368)$ A1 N2

mean for one day: 0.2 (A1)

${\text{P}}(X = 0) = {({{\text{e}}^{ – 0.2}})^5} = {{\text{e}}^{ – 1}}\,\,\,\,\,( = 0.368)$ A1 N2

[2 marks]

(d) Required probability $ = {{\text{e}}^{ – 0.2}} \times {{\text{e}}^{ – 0.2}} \times (1 – {{\text{e}}^{ – 0.2}})$ M1A1

= 0.122 A1 N3

[3 marks]

(e) Expected cost is $1850 \times 6 = {\text{11}}\,{\text{100 Euros}}$ A1

[1 mark]

(f) On any one day ${\text{P}}(X = 0) = {{\text{e}}^{ – 0.2}}$

Therefore, $\left( {\begin{array}{*{20}{c}}
5 \\
1
\end{array}} \right){({{\text{e}}^{ – 0.2}})^4}(1 – {{\text{e}}^{ – 0.2}}) = 0.407$ M1A1 N2

[2 marks]

Total [13 marks]

Question

Over a one month period, Ava and Sven play a total of n games of tennis.

The probability that Ava wins any game is 0.4. The result of each game played is independent of any other game played.

Let X denote the number of games won by Ava over a one month period.

(a) Find an expression for P(X = 2) in terms of n.

(b) If the probability that Ava wins two games is 0.121 correct to three decimal places, find the value of n.

Answer/Explanation

Markscheme

(a) $X \sim {\text{B}}(n,{\text{ }}0.4)$ (A1)

Using ${\text{P}}(X = x) = \left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right){(0.4)^x}{(0.6)^{n – x}}$ (M1)

${\text{P}}(X = 2) = \left( {\begin{array}{*{20}{c}}
n \\
2
\end{array}} \right){(0.4)^2}{(0.6)^{n – 2}}$ $\left( { = \frac{{n(n – 1)}}{2}{{(0.4)}^2}{{(0.6)}^{n – 2}}} \right)$ A1 N3

(b) P(X = 2) = 0.121 A1

Using an appropriate method (including trial and error) to solve their equation. (M1)

n = 10 A1 N2

Note: Do not award the last A1 if any other solution is given in their final answer.

[6 marks]

Examiners report

Part (a) was generally well done. The most common error was to omit the binomial coefficient i.e. not identifying that the situation is described by a binomial distribution.

Finding the correct value of n in part (b) proved to be more elusive. A significant proportion of candidates attempted algebraic approaches and seemingly did not realise that the equation could only be solved numerically. Candidates who obtained n = 10 often accomplished this by firstly attempting to solve the equation algebraically before ‘resorting’ to a GDC approach. Some candidates did not specify their final answer as an integer while others stated n = 1.76 as their final answer.

Question

(a) A box of biscuits is considered to be underweight if it weighs less than 228 grams. It is known that the weights of these boxes of biscuits are normally distributed with a mean of 231 grams and a standard deviation of 1.5 grams.

What is the probability that a box is underweight?

(b) The manufacturer decides that the probability of a box being underweight should be reduced to 0.002.

(i) Bill’s suggestion is to increase the mean and leave the standard deviation unchanged. Find the value of the new mean.

(ii) Sarah’s suggestion is to reduce the standard deviation and leave the mean unchanged. Find the value of the new standard deviation.

(c) After the probability of a box being underweight has been reduced to 0.002, a group of customers buys 100 boxes of biscuits. Find the probability that at least two of the boxes are underweight.

[11]

Part A.

There are six boys and five girls in a school tennis club. A team of two boys and two girls will be selected to represent the school in a tennis competition.

(a) In how many different ways can the team be selected?

(b) Tim is the youngest boy in the club and Anna is the youngest girl. In how many different ways can the team be selected if it must include both of them?

(d) Fred is the oldest boy in the club. Given that Fred is selected for the team, what is the probability that the team includes Tim or Anna, but not both?

[10]

Part B.

Answer/Explanation

Markscheme

(a) $X \sim {\text{N(231, 1.}}{{\text{5}}^2})$

${\text{P}}(X < 228) = 0.0228$ (M1)A1

Note: Accept 0.0227.

[2 marks]

(b) (i) $X \sim {\text{N(}}\mu {\text{, 1.}}{{\text{5}}^2})$

${\text{P}}(X < 228) = 0.002$

$\frac{{228 – \mu }}{{1.5}} = – 2.878…$ M1A1

$\mu = 232{\text{ grams}}$ A1 N3

(ii) $X \sim {\text{N(231, }}{\sigma ^2})$

$\frac{{228 – 231}}{\sigma } = – 2.878…$ M1A1

$\sigma = 1.04{\text{ grams}}$ A1 N3

[6 marks]

${\text{P}}(X \leqslant 1) = 0.982…$ (A1)

${\text{P}}(X \geqslant 2) = 1 – {\text{P}}(X \leqslant 1) = 0.0174$ A1

[3 marks]

Total [11 marks]

Part A.

(a) Boys can be chosen in $\frac{{6 \times 5}}{2} = 15$ ways (A1)

Girls can be chosen in $\frac{{5 \times 4}}{2} = 10$ ways (A1)

Total $ = 15 \times 10 = 150$ ways A1

[3 marks]

(b) Number of ways $ = 5 \times 4 = 20$ (M1)A1

[2 marks]

[1 mark]

(d) METHOD 1

${\text{P}}(T) = \frac{1}{5};{\text{ P}}(A) = \frac{2}{5}$ A1

P(T or A but not both) $ = {\text{P}}(T) \times {\text{P}}(A’) + {\text{P}}(T’) \times {\text{P}}(A)$ M1A1

$ = \frac{1}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{2}{5} = \frac{{11}}{{25}}$ A1

METHOD 2

Number of selections including Fred $ = 5 \times \left( {\begin{array}{*{20}{c}}
5 \\
2
\end{array}} \right) = 50$ A1

Number of selections including Tim but not Anna $ = \left( {\begin{array}{*{20}{c}}
4 \\
2
\end{array}} \right) = 6$ A1

Number of selections including Anna but not Tim $ = 4 \times 4 = 16$

Note: Both statements are needed to award A1.

P(T or A but not both) $ = \frac{{6 + 16}}{{50}} = \frac{{11}}{{25}}$ M1A1

[4 marks]

Total [10 marks]

Part B.

Examiners report

Part A was well done by many candidates although an arithmetic penalty was often awarded in (b)(i) for giving the new value of the mean to too many significant figures.

Part A.

Candidates are known, however, to be generally uncomfortable with combinatorial mathematics and Part B caused problems for many candidates. Even some of those candidates who solved (a) and (b) correctly were then unable to deduce the answer to (c), sometimes going off on some long-winded solution which invariably gave the wrong answer. Very few correct solutions were seen to (d).

Part B.

Question

Testing has shown that the volume of drink in a bottle of mineral water filled by Machine A at a bottling plant is normally distributed with a mean of $998$ ml and a standard deviation of $2.5$ ml.

(a) Show that the probability that a randomly selected bottle filled by Machine A contains more than $1000$ ml of mineral water is $0.212$.

(b) A random sample of $5$ bottles is taken from Machine A. Find the probability that exactly $3$ of them each contain more than $1000$ ml of mineral water.

(c) Find the minimum number of bottles that would need to be sampled to ensure that the probability of getting at least one bottle filled by Machine A containing more than $1000$ ml of mineral water, is greater than $0.99$.

(d) It has been found that for Machine B the probability of a bottle containing less than $996$ ml of mineral water is $0.1151$. The probability of a bottle containing more than $1000$ ml is $0.3446$. Find the mean and standard deviation for the volume of mineral water contained in bottles filled by Machine B.

(e) The company that makes the mineral water receives, on average, m phone calls every $10$ minutes. The number of phone calls, $X$ , follows a Poisson distribution such that ${\text{P}}(X = 2) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$ .

(i) Find the value of $m$ .

(ii) Find the probability that the company receives more than two telephone calls in a randomly selected $10$ minute period.

Answer/Explanation

Markscheme

(a) $X \sim {\text{N}}$($998$, ${2.5^2}$ ) M1

${\text{P}}(X > 1000) = 0.212$ AG

[1 mark]

(b) $X \sim {\text{B}}$($5$, $0.2119…$)

evidence of binomial (M1)

${\text{P}}(X = 3) = \left( {\begin{array}{*{20}{c}}
5 \\
3
\end{array}} \right){\left( {0.2119…} \right)^3}{\left( {0.7881…} \right)^2} = 0.0591$ (accept $0.0592$) (M1)A1

[3 marks]

$1 – {\left( {0.7881…} \right)^n} > 0.99$

${\left( {0.7881…} \right)^n} < 0.01$ A1

Note: Award A1 for line 2 or line 3 or equivalent.

$n > 19.3$ (A1)
minimum number of bottles required is $20$ A1N2

[4 marks]

(d) $\frac{{996 – \mu }}{\sigma } = – 1.1998$ (accept $1.2$) M1A1

$\frac{{1000 – \mu }}{\sigma } = 0.3999$ (accept $0.4$) M1A1

$\mu = 999$(ml), $\sigma = 2.50$(ml) A1A1

[6 marks]

(e) (i) $\frac{{{{\text{e}}^{ – m}}{m^2}}}{{2!}} = \frac{{{{\text{e}}^{ – m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}}$ M1A1

$\frac{{{m^2}}}{2} = \frac{{{m^3}}}{6} + \frac{{{m^4}}}{{24}}$

$12{m^2} – 4{m^3} – {m^4} = 0$ (A1)

$m = – 6$, $0$, $2$

$ \Rightarrow m = 2$ A1N2

(ii) ${\text{P}}(X > 2) = 1 – {\text{P}}(X \leqslant 2)$ (M1)

$ = 1 – {\text{P}}(X = 0) – {\text{P}}(X = 1) – {\text{P}}(X = 2)$

$ = 1 – {{\text{e}}^{ – 2}} – 2{{\text{e}}^{ – 2}} – \frac{{{2^2}{{\text{e}}^{ – 2}}}}{{2!}}$

$ = 0.323$ A1

[6 marks]

Total [20 marks]

Examiners report

This was the best done of the section B questions, with the majority of candidates making the correct choice of probability distribution for each part. The main sources of errors: (b) missing out the binomial coefficient in the calculation; (c) failure to rearrange ‘at least one bottle’ in terms of the probability of obtaining no bottles; (d) using $1.2$ rather than $ – 1.2$ in the inverse Normal or not performing an inverse Normal at all; (e)(ii) misinterpreting ‘more than two’.

Question

Casualties arrive at an accident unit with a mean rate of one every 10 minutes. Assume that the number of arrivals can be modelled by a Poisson distribution.

(a) Find the probability that there are no arrivals in a given half hour period.

(b) A nurse works for a two hour period. Find the probability that there are fewer than ten casualties during this period.

(c) Six nurses work consecutive two hour periods between 8am and 8pm. Find the probability that no more than three nurses have to attend to less than ten casualties during their working period.

(d) Calculate the time interval during which there is a 95 % chance of there being at least two casualties.

Answer/Explanation

Markscheme

Note: Accept exact answers in parts (a) to (c).

(a) number of patients in 30 minute period = X (A1)

$X \sim {\text{Po(3)}}$ (M1)A1

[3 marks]

(b) number of patients in working period = Y (A1)

$Y \sim {\text{Po(12)}}$ (M1)A1

[3 marks]

$W \sim {\text{B}}(6,{\text{ }}0.2424 \ldots )$ (M1)A1

[4 marks]

(d) number of patients in t minute interval = X

$X \sim {\text{Po}}(T)$

${\text{P}}(X \geqslant 2) = 0.95$

${\text{P}}(X = 0) + {\text{P}}(X = 1) = 0.05$ (M1)(A1)

${{\text{e}}^{ – T}}(1 + T) = 0.05$ (M1)

$T = 4.74$ (A1)

t = 47.4 minutes A1

[5 marks]

Total [15 marks]

Examiners report

Parts (a) and (b) were well answered, but many students were unable to recognise the Binomial distribution in part (c) and were unable to form the correct equation in part (d). There were many accuracy errors in this question.

Question

A student arrives at a school $X$ minutes after 08:00, where X may be assumed to be normally distributed. On a particular day it is observed that 40% of the students arrive before 08:30 and 90% arrive before 08:55.

Consider the function $f(x) = \frac{{\ln x}}{x}$ , $0 < x < {{\text{e}}^2}$ .

Find the mean and standard deviation of $X$.

[5]

The school has 1200 students and classes start at 09:00. Estimate the number of students who will be late on that day.

[3]

Maelis had not arrived by 08:30. Find the probability that she arrived late.

[2]

At 15:00 it is the end of the school day and it is assumed that the departure of the students from school can be modelled by a Poisson distribution. On average 24 students leave the school every minute.

Find the probability that at least 700 students leave school before 15:30.

[3]

There are 200 days in a school year. Given that $Y$ denotes the number of days in the year that at least 700 students leave before 15:30, find

(i) ${\text{E}}(Y)$ ;

(ii) $P(Y > 150)$ .

[4]

Answer/Explanation

Markscheme

${\text{P}}(X < 30) = 0.4$

${\text{P}}(X < 55) = 0.9$

or relevant sketch (M1)

given $Z = \frac{{X – \mu }}{\sigma }$

${\text{P}}(Z < z) = 0.4 \Rightarrow \frac{{30 – \mu }}{\sigma } = – 0.253…$ (A1)

${\text{P}}(Z < z) = 0.9 \Rightarrow \frac{{55 – \mu }}{\sigma } = 1.28…$ (A1)

$\mu = 30 + \left( {0.253…} \right) \times \sigma = 55 – \left( {1.28…} \right) \times \sigma $ M1

$\sigma = 16.3$ , $\mu = 34.1$ A1

Note: Accept 16 and 34.

Note: Working with 830 and 855 will only gain the two M marks.

[5 marks]

$X{\text{ ~ N}}$($34.12…$, $16.28…{^2}$ )

late to school $ \Rightarrow X > 60$

${\text{P}}(X > 60) = 0.056…$ (A1)

number of students late ${\text{ = 0}}{\text{.0560}}… \times {\text{1200}}$ (M1)

$= 67$ (to nearest integer) A1

Note: Accept $62$ for use of $34$ and $16$.

[3 marks]

${\text{P}}(X > 60|X > 30) = \frac{{{\text{P}}(X > 60)}}{{{\text{P}}(X > 30)}}$ M1

$ = 0.0935$ (accept anything between $0.093$ and $0.094$) A1

Note: If $34$ and $16$ are used $0.0870$ is obtained. This should be accepted.

[2 marks]

let $L$ be the random variable of the number of students who leave school in a 30 minute interval

since $24 \times 30 = 720$ A1

${\text{L ~ Po(720)}}$

${\text{P(}}L \geqslant 700) = 1 – {\text{P}}(L \leqslant 699)$ (M1)

$ = 0.777$ A1

Note: Award M1A0 for $P(L > 700) = 1 – P(L \leqslant 700)$ (this leads to $0.765$).

[3 marks]

(i) $Y{\text{ ~ B}}$($200$, $0.7767…$) (M1)

${\text{E}}(Y) = 200 \times 0.7767… = 155$ A1

Note: On ft, use of $0.765$ will lead to $153$.

(ii) ${\text{P}}(Y > 150) = 1 – {\text{P}}(Y \leqslant 150)$ (M1)

$ = 0.797$ A1

Note: Accept $0.799$ from using rounded answer.

Note: On ft, use of $0.765$ will lead to $0.666$.

[4 marks]

Examiners report

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Question

The fish in a lake have weights that are normally distributed with a mean of 1.3 kg and a standard deviation of 0.2 kg.

Determine the probability that a fish which is caught weighs less than 1.4 kg.

[1]

John catches 6 fish. Calculate the probability that at least 4 of the fish weigh more than 1.4 kg.

[3]

Determine the probability that a fish which is caught weighs less than 1 kg, given that it weighs less than 1.4 kg.

[2]

Answer/Explanation

Markscheme

${\text{P}}(x < 1.4) = 0.691\,\,\,\,\,$(accept 0.692) A1

[1 mark]

METHOD 1

$y \sim {\text{B(6, 0.3085…)}}$ (M1)

${\text{P}}(Y \geqslant 4) = 1 – {\text{P}}(Y \leqslant 3)$ (M1)

$ = 0.0775\,\,\,\,\,$(accept 0.0778 if 3sf approximation from (a) used) A1

METHOD 2

$X \sim {\text{B(6, 0.6914…)}}$ (M1)

${\text{P}}(X \leqslant 2)$ (M1)

$ = 0.0775\,\,\,\,\,$(accept 0.0778 if 3sf approximation from (a) used) A1

[3 marks]

${\text{P}}(x < 1|x < 1.4) = \frac{{{\text{P}}(x < 1)}}{{{\text{P}}(x < 1.4)}}$ M1

$ = \frac{{0.06680…}}{{0.6914…}}$

$ = 0.0966\,\,\,\,\,$(accept 0.0967) A1

[2 marks]

Examiners report

Part (a) was almost universally correctly answered, albeit with an accuracy penalty in some cases. In (b) it was generally recognised that the distribution was binomial, but with some wavering about the correct value of the parameter p. Part (c) was sometimes answered correctly, but not with much confidence.

Question

The probability that the 08:00 train will be delayed on a work day (Monday to Friday) is $\frac{1}{{10}}$. Assuming that delays occur independently,

find the probability that the 08:00 train is delayed exactly twice during any period of five work days;

[2]

find the minimum number of work days for which the probability of the 08:00 train being delayed at least once exceeds 90 %.

[3]

Answer/Explanation

Markscheme

$X \sim {\text{B(5, 0.1)}}$ (M1)

${\text{P}}(X = 2) = 0.0729$ A1

[2 marks]

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ (M1)

$0.9 < 1 – {\left( {\frac{9}{{10}}} \right)^n}$ (M1)

$n > \frac{{\ln 0.1}}{{\ln 0.9}}$

n = 22 days A1

[3 marks]

Examiners report

This question was generally answered successfully. Many candidates used the tabular feature of their GDC for (b) thereby avoiding potential errors in the algebraic manipulation of logs and inequalities.

Question

The probability that the 08:00 train will be delayed on a work day (Monday to Friday) is $\frac{1}{{10}}$. Assuming that delays occur independently,

find the probability that the 08:00 train is delayed exactly twice during any period of five work days;

[2]

find the minimum number of work days for which the probability of the 08:00 train being delayed at least once exceeds 90 %.

[3]

Answer/Explanation

Again this proved to be an accessible question for students with many students gaining full marks. Most candidates used the calculator to find the answers to parts (b) and (c) which is what was intended, but candidates should be aware that there are often marks for recognising what needs to be found, even if the candidate does not obtain the final correct answer. It is suggested that in this style of question, candidates should indicate what they are trying to find as well as giving the final answer.

Answer/Explanation

Markscheme

(i) ${\text{P}}(X > 225) = 0.158…$ (M1)(A1)

expected number $ = 450 \times 0.158… = 71.4$ A1

(ii) ${\text{P}}(X < m) = 0.7$ (M1)

$ \Rightarrow m = 213{\text{ (grams)}}$ A1

[5 marks]

$\frac{{270 – \mu }}{\sigma } = 1.40…$ (M1)A1

$\frac{{250 – \mu }}{\sigma } = – 1.03…$ A1

Note: These could be seen in graphical form.

solving simultaneously (M1)

$\mu = 258,{\text{ }}\sigma = 8.19$ A1A1

[6 marks]

$X \sim {\text{N}}({80,4^2})$

${\text{P}}(X > 82) = 0.3085…$ A1

recognition of the use of binomial distribution. (M1)

$X \sim {\text{B}}(5,\,0.3085…)$

${\text{P}}(X = 3) = 0.140$ A1

[3 marks]

Examiners report

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

Question

Emily walks to school every day. The length of time this takes can be modelled by a normal distribution with a mean of 11 minutes and a standard deviation of 3 minutes. She is late if her journey takes more than 15 minutes.

Find the probability she is late next Monday.

[2]

Find the probability she is late at least once during the next week (Monday to Friday).

[3]

Answer/Explanation

Markscheme

Let X represent the length of time a journey takes on a particular day.

$P(X > 15) = 0.0912112819 \ldots = 0.0912$ (M1)A1

Use of correct Binomial distribution (M1)

$N \sim B(5,0.091 \ldots )$

$1 – 0.0912112819 \ldots = 0.9087887181 \ldots $

$1 – {(0.9087887181 \ldots )^5} = 0.380109935 \ldots = 0.380$ (M1)A1

Note: Allow answers to be given as percentages.

[5 marks]

Examiners report

There were many good answers to this question. Some students lost accuracy marks by early rounding. Some students struggled with the Binomial distribution.

Question

(i) Express the sum of the first n positive odd integers using sigma notation.

(ii) Show that the sum stated above is ${n^2}$.

(iii) Deduce the value of the difference between the sum of the first 47 positive odd integers and the sum of the first 14 positive odd integers.

[4]

A number of distinct points are marked on the circumference of a circle, forming a polygon. Diagonals are drawn by joining all pairs of non-adjacent points.

(i) Show on a diagram all diagonals if there are 5 points.

(ii) Show that the number of diagonals is $\frac{{n(n – 3)}}{2}$ if there are n points, where $n > 2$.

(iii) Given that there are more than one million diagonals, determine the least number of points for which this is possible.

[7]

The random variable $X \sim B(n,{\text{ }}p)$ has mean 4 and variance 3.

(i) Determine n and p.

(ii) Find the probability that in a single experiment the outcome is 1 or 3.

[8]

Answer/Explanation

Markscheme

(i) $\sum\limits_{k = 1}^n {(2k – 1)} $ (or equivalent) A1

Note: Award A0 for $\sum\limits_{n = 1}^n {(2n – 1)} $ or equivalent.

(ii) EITHER

$2 \times \frac{{n(n + 1)}}{2} – n$ M1A1

$\frac{n}{2}\left( {2 + (n – 1)2} \right){\text{ (using }}{S_n} = \frac{n}{2}\left( {2{u_1} + (n – 1)d} \right))$ M1A1

$\frac{n}{2}(1 + 2n – 1){\text{ (using }}{S_n} = \frac{n}{2}({u_1} + {u_n}))$ M1A1

THEN

$ = {n^2}$ AG

(iii) ${47^2} – {14^2} = 2013$ A1

[4 marks]

(i) EITHER

a pentagon and five diagonals A1

five diagonals (circle optional) A1

(ii) Each point joins to n – 3 other points. A1

a correct argument for ${n(n – 3)}$ R1

a correct argument for $\frac{{n(n – 3)}}{2}$ R1

(iii) attempting to solve $\frac{1}{2}n(n – 3) > 1\,000\,000$ for n. (M1)

$n > 1415.7$ (A1)

$n = 1416$ A1

[7 marks]

(i) np = 4 and npq = 3 (A1)

attempting to solve for n and p (M1)

$n = 16$ and $p = \frac{1}{4}$ A1

(ii) $X \sim B(16,0.25)$ (A1)

$P(X = 1) = 0.0534538…( = \left( {\begin{array}{*{20}{c}}
{16} \\
1
\end{array}} \right)(0.25){(0.75)^{15}})$ (A1)

$P(X = 3) = 0.207876…( = \left( {\begin{array}{*{20}{c}}
{16} \\
3
\end{array}} \right){(0.25)^3}{(0.75)^{13}})$ (A1)

${\text{P}}(X = 1) + {\text{P}}(X = 3)$ (M1)

= 0.261 A1

[8 marks]

Examiners report

In part (a) (i), a large number of candidates were unable to correctly use sigma notation to express the sum of the first n positive odd integers. Common errors included summing $2n – 1$ from 1 to n and specifying sums with incorrect limits. Parts (a) (ii) and (iii) were generally well done.

Parts (b) (i) and (iii) were generally well done. In part (b) (iii), many candidates unnecessarily simplified their quadratic when direct GDC use could have been employed. A few candidates gave $n > 1416$ as their final answer. While some candidates displayed sound reasoning in part (b) (ii), many candidates unfortunately adopted a ‘proof by example’ approach.

Part (c) was generally well done. In part (c) (ii), some candidates multiplied the two probabilities rather than adding the two probabilities.

Question

Six customers wait in a queue in a supermarket. A customer can choose to pay with cash or a credit card. Assume that whether or not a customer pays with a credit card is independent of any other customers’ methods of payment.

It is known that 60% of customers choose to pay with a credit card.

(a) Find the probability that:

(i) the first three customers pay with a credit card and the next three pay with cash;

(ii) exactly three of the six customers pay with a credit card.

There are n customers waiting in another queue in the same supermarket. The probability that at least one customer pays with cash is greater than 0.995.

(b) Find the minimum value of n.

Answer/Explanation

Markscheme

(a) (i) ${0.6^3} \times {0.4^3}$ (M1)

Note: Award (M1) for use of the product of probabilities.

$ = 0.0138$ A1

(ii) binomial distribution $X:{\text{B(6, 0.6)}}$ (M1)

Note: Award (M1) for recognizing the binomial distribution.

${\text{P}}(X = 3) = $ $^6{C_3}{(0.6)^3}{(0.4)^3}$

$ = 0.276$ A1

Note: Award (M1)A1 for $^6{C_3} \times 0.0138 = 0.276$.

[4 marks]

(b) $Y:{\text{B(}}n,{\text{ 0.4)}}$

${\text{P}}(Y \geqslant 1) > 0.995$

$1 – {\text{P}}(Y = 0) > 0.995$

${\text{P}}(Y = 0) < 0.005$ (M1)

Note: Award (M1) for any of the last three lines. Accept equalities.

${0.6^n} < 0.005$ (M1)

Note: Award (M1) for attempting to solve ${0.6^n} < 0.005$ using any method, eg, logs, graphically, use of solver. Accept an equality.

$n > 10.4$

$\therefore n = 11$ A1

[3 marks]

Total [7 marks]

Examiners report

[N/A]

Question

A mosaic is going to be created by randomly selecting 1000 small tiles, each of which is either black or white. The probability that a tile is white is 0.1. Let the random variable $W$ be the number of white tiles.

State the distribution of $W$, including the values of any parameters.

[2]

Write down the mean of $W$.

[1]

Find ${\text{P}}(W > 89)$.

[2]

Answer/Explanation

Markscheme

$W \sim B(1000,{\text{ }}0.1)\;\;\;\left( {{\text{accept }}C_k^{1000}{{(0.1)}^k}{{(0.9)}^{1000 – k}}} \right)$ A1A1

Note: First A1 is for recognizing the binomial, second A1 for both parameters if stated explicitly in this part of the question.

[2 marks]

$\mu ( = 1000 \times 0.1) = 100$ A1

[1 mark]

${\text{P}}(W > 89) = {\text{P}}(W \ge 90)\;\;\;\left( { = 1 – {\text{P}}(W \le 89)} \right)$ (M1)

$ = 0.867$ A1

Notes: Award M0A0 for $0.889$

[2 marks]

Total [5 marks]

Examiners report

Overall this question was well answered. In part (a) a number of candidates did not mention the binomial distribution or failed to state its parameters although they could go on and do the next parts.

In part (b) most candidates could state the expected value.

In part (c) many candidates had problems with inequalities due to the discrete nature of the variable. Most candidates that could deal with the inequality were able to use the GDC to obtain the answer.

Question

State the distribution of $W$, including the values of any parameters.

[2]

Write down the mean of $W$.

[1]

Find ${\text{P}}(W > 89)$.

[2]

Answer/Explanation

Markscheme

$W \sim B(1000,{\text{ }}0.1)\;\;\;\left( {{\text{accept }}C_k^{1000}{{(0.1)}^k}{{(0.9)}^{1000 – k}}} \right)$ A1A1

Note: First A1 is for recognizing the binomial, second A1 for both parameters if stated explicitly in this part of the question.

[2 marks]

$\mu ( = 1000 \times 0.1) = 100$ A1

[1 mark]

${\text{P}}(W > 89) = {\text{P}}(W \ge 90)\;\;\;\left( { = 1 – {\text{P}}(W \le 89)} \right)$ (M1)

$ = 0.867$ A1

Notes: Award M0A0 for $0.889$

[2 marks]

Total [5 marks]

Examiners report

In part (b) most candidates could state the expected value.

Question

State the distribution of $W$, including the values of any parameters.

[2]

Write down the mean of $W$.

[1]

Find ${\text{P}}(W > 89)$.

[2]

Answer/Explanation

Markscheme

$W \sim B(1000,{\text{ }}0.1)\;\;\;\left( {{\text{accept }}C_k^{1000}{{(0.1)}^k}{{(0.9)}^{1000 – k}}} \right)$ A1A1

Note: First A1 is for recognizing the binomial, second A1 for both parameters if stated explicitly in this part of the question.

[2 marks]

$\mu ( = 1000 \times 0.1) = 100$ A1

[1 mark]

${\text{P}}(W > 89) = {\text{P}}(W \ge 90)\;\;\;\left( { = 1 – {\text{P}}(W \le 89)} \right)$ (M1)

$ = 0.867$ A1

Notes: Award M0A0 for $0.889$

[2 marks]

Total [5 marks]

Examiners report

In part (b) most candidates could state the expected value.

Question

Farmer Suzie grows turnips and the weights of her turnips are normally distributed with a mean of $122g$ and standard deviation of $14.7g$.

(i) Calculate the percentage of Suzie’s turnips that weigh between $110g$ and $130g$.

(ii) Suzie has $100$ turnips to take to market. Find the expected number weighing more than $130g$.

(iii) Find the probability that at least $30$ of the $100g$ turnips weigh more than $130g$.

[6]

Farmer Ray also grows turnips and the weights of his turnips are normally distributed with a mean of $144g$. Ray only takes to market turnips that weigh more than $130g$. Over a period of time, Ray finds he has to reject $1$ in $15$ turnips due to their being underweight.

(i) Find the standard deviation of the weights of Ray’s turnips.

(ii) Ray has $200$ turnips to take to market. Find the expected number weighing more than $150g$.

[6]

Answer/Explanation

Markscheme

(i) $P(110 < X < 130) = 0.49969 \ldots = 0.500 = 50.0\% $ (M1)A1

Note: Accept $50$

Note: Award M1A0 for $0.50$ ($0.500$)

(ii) $P(X > 130) = (1 – 0.707 \ldots ) = 0.293 \ldots $ M1

expected number of turnips $ = 29.3$ A1

Note: Accept $29$.

(iii) no of turnips weighing more than $130$ is $Y \sim B(100,{\text{ }}0.293)$ M1

$P(Y \ge 30) = 0.478$ A1

[6 marks]

(i) $X \sim N(144,{\text{ }}{\sigma ^2})$

$P(X \le 130) = \frac{1}{{15}} = 0.0667$ (M1)

$P\left( {Z \le \frac{{130 – 144}}{\sigma }} \right) = 0.0667$

$\frac{{14}}{\sigma } = 1.501$ (A1)

$\sigma = 9.33{\text{ g}}$ A1

(ii) $P(X > 150|X > 130) = \frac{{P(X > 150)}}{{P(X > 130)}}$ M1

$ = \frac{{0.26008 \ldots }}{{1 – 0.06667}} = 0.279$ A1

expected number of turnips $ = 55.7$ A1

[6 marks]

Total [12 marks]

Examiners report

[N/A]

Question

Farmer Suzie grows turnips and the weights of her turnips are normally distributed with a mean of $122g$ and standard deviation of $14.7g$.

(i) Calculate the percentage of Suzie’s turnips that weigh between $110g$ and $130g$.

(ii) Suzie has $100$ turnips to take to market. Find the expected number weighing more than $130g$.

(iii) Find the probability that at least $30$ of the $100g$ turnips weigh more than $130g$.

[6]

(i) Find the standard deviation of the weights of Ray’s turnips.

(ii) Ray has $200$ turnips to take to market. Find the expected number weighing more than $150g$.

[6]

Answer/Explanation

Markscheme

(i) $P(110 < X < 130) = 0.49969 \ldots = 0.500 = 50.0\% $ (M1)A1

Note: Accept $50$

Note: Award M1A0 for $0.50$ ($0.500$)

(ii) $P(X > 130) = (1 – 0.707 \ldots ) = 0.293 \ldots $ M1

expected number of turnips $ = 29.3$ A1

Note: Accept $29$.

(iii) no of turnips weighing more than $130$ is $Y \sim B(100,{\text{ }}0.293)$ M1

$P(Y \ge 30) = 0.478$ A1

[6 marks]

(i) $X \sim N(144,{\text{ }}{\sigma ^2})$

$P(X \le 130) = \frac{1}{{15}} = 0.0667$ (M1)

$P\left( {Z \le \frac{{130 – 144}}{\sigma }} \right) = 0.0667$

$\frac{{14}}{\sigma } = 1.501$ (A1)

$\sigma = 9.33{\text{ g}}$ A1

(ii) $P(X > 150|X > 130) = \frac{{P(X > 150)}}{{P(X > 130)}}$ M1

$ = \frac{{0.26008 \ldots }}{{1 – 0.06667}} = 0.279$ A1

expected number of turnips $ = 55.7$ A1

[6 marks]

Total [12 marks]

Examiners report

[N/A]

Question

A survey is conducted in a large office building. It is found that $30\% $ of the office workers weigh less than $62$ kg and that $25\% $ of the office workers weigh more than $98$ kg.

The weights of the office workers may be modelled by a normal distribution with mean $\mu $ and standard deviation $\sigma $.

(i) Determine two simultaneous linear equations satisfied by $\mu $ and $\sigma $.

(ii) Find the values of $\mu $ and $\sigma $.

[6]

Find the probability that an office worker weighs more than $100$ kg.

[1]

There are elevators in the office building that take the office workers to their offices.

Given that there are $10$ workers in a particular elevator,

find the probability that at least four of the workers weigh more than $100$ kg.

[2]

Given that there are $10$ workers in an elevator and at least one weighs more than $100$ kg,

find the probability that there are fewer than four workers exceeding $100$ kg.

[3]

The arrival of the elevators at the ground floor between $08:00$ and $09:00$ can be modelled by a Poisson distribution. Elevators arrive on average every $36$ seconds.

Find the probability that in any half hour period between $08:00$ and $09:00$ more than $60$ elevators arrive at the ground floor.

[3]

An elevator can take a maximum of $10$ workers. Given that $400$ workers arrive in a half hour period independently of each other,

find the probability that there are sufficient elevators to take them to their offices.

[3]

Answer/Explanation

Markscheme

Note: In Section B, accept answers that correctly round to 2 sf.

(i) let $W$ be the weight of a worker and $W \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}\left( {Z < \frac{{62 – \mu }}{\alpha }} \right) = 0.3$ and ${\text{P}}\left( {Z < \frac{{98 – \mu }}{\sigma }} \right) = 0.75$ (M1)

Note: Award M1 for a correctly shaded and labelled diagram.

$\frac{{62 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.3)\;\;\;( = – 0.524 \ldots )\;\;\;$and

$\frac{{98 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.75)\;\;\;( = 0.674 \ldots )$

or linear equivalents A1A1

Note: Condone equations containing the GDC inverse normal command.

(ii) attempting to solve simultaneously (M1)

$\mu = 77.7,{\text{ }}\sigma = 30.0$ A1A1

[6 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(W > 100) = 0.229$ A1

[1 mark]

Note: In Section B, accept answers that correctly round to 2 sf.

let $X$ represent the number of workers over $100$ kg in a lift of ten passengers

$X \sim {\text{B}}(10,{\text{ }}0.229 \ldots )$ (M1)

${\text{P}}(X \ge 4) = 0.178$ A1

[2 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(X < 4|X \ge 1) = \frac{{{\text{P}}(1 \le X \le 3)}}{{{\text{P}}(X \ge 1)}}$ M1(A1)

Note: Award the M1 for a clear indication of a conditional probability.

$ = 0.808$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$L \sim {\text{Po}}(50)$ (M1)

${\text{P}}(L > 60) = 1 – {\text{P}}(L \le 60)$ (M1)

$ = 0.0722$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$400$ workers require at least $40$ elevators (A1)

${\text{P}}(L \ge 40) = 1 – {\text{P}}(L \le 39)$ (M1)

$ = 0.935$ A1

[3 marks]

Total [18 marks]

Examiners report

[N/A]

Question

Students sign up at a desk for an activity during the course of an afternoon. The arrival of each student is independent of the arrival of any other student and the number of students arriving per hour can be modelled as a Poisson distribution with a mean of $\lambda $.

The desk is open for 4 hours. If exactly 5 people arrive to sign up for the activity during that time find the probability that exactly 3 of them arrived during the first hour.

Answer/Explanation

Markscheme

P(3 in the first hour) $ = \frac{{{\lambda ^3}{e^{ – \lambda }}}}{{3!}}$ A1

number to arrive in the four hours follows $Po(4\lambda )$ M1

P(5 arrive in total) $ = \frac{{{{(4\lambda )}^5}{e^{ – 4\lambda }}}}{{5!}}$ A1

attempt to find P(2 arrive in the next three hours) M1

$ = \frac{{{{(3\lambda )}^2}{e^{ – 3\lambda }}}}{{2!}}$ A1

use of conditional probability formula M1

P(3 in the first hour given 5 in total) $ = \frac{{\frac{{{\lambda ^3}{e^{ – \lambda }}}}{{3!}} \times \frac{{{{(3\lambda )}^2}{e^{ – 3\lambda }}}}{{2!}}}}{{\frac{{{{(4\lambda )}^5}{e^{ – 4\lambda }}}}{{5!}}}}$ A1

$\frac{{\left( {\frac{9}{{2!3!}}} \right)}}{{\left( {\frac{{{4^5}}}{{5!}}} \right)}} = \frac{{45}}{{512}} = 0.0879$ A1

[8 marks]

Examiners report

A more difficult question, but it was still surprising how many candidates were unable to make a good start with it. Many were using $\lambda = \frac{5}{4}$ and consequently unable to progress very far. Many students failed to recognise that a conditional probability should be used.

Question

Six balls numbered 1, 2, 2, 3, 3, 3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

A company produces rectangular sheets of glass of area 5 square metres. During manufacturing these glass sheets flaws occur at the rate of 0.5 per 5 square metres. It is assumed that the number of flaws per glass sheet follows a Poisson distribution.

Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.

This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.

A larger glass sheet is chosen at random.

Find the probability that a randomly chosen glass sheet contains at least one flaw.

[3]

Find the expected profit, $P$ dollars, per glass sheet.

[3]

Find the probability that it contains no flaws.

[2]

Answer/Explanation

Markscheme

${\text{X}} \sim {\text{Po}}(0.5)$ (A1)

${\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 – {{\text{e}}^{ – 0.5}})$ (M1)A1

[3 marks]

${\text{P}}(X = 0) = 0.607 \ldots $ (A1)

Part (a) was reasonably well done. Some candidates calculated ${\text{P}}(X = 1)$.

Part (c) was very well done.

Question

Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.

This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.

A larger glass sheet is chosen at random.

Find the probability that a randomly chosen glass sheet contains at least one flaw.

[3]

Find the expected profit, $P$ dollars, per glass sheet.

[3]

Find the probability that it contains no flaws.

[2]

Answer/Explanation

Markscheme

${\text{X}} \sim {\text{Po}}(0.5)$ (A1)

${\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 – {{\text{e}}^{ – 0.5}})$ (M1)A1

[3 marks]

${\text{P}}(X = 0) = 0.607 \ldots $ (A1)

${\text{E}}(P) = (0.607 \ldots \times 5) – (0.393 \ldots \times 3)$ (M1)

the expected profit is $1.85 per glass sheet A1

[3 marks]

$Y \sim {\text{Po}}(2)$ (M1)

${\text{P}}(Y = 0) = 0.135{\text{ }}( = {{\text{e}}^{ – 2}})$ A1

[2 marks]

Examiners report

Part (a) was reasonably well done. Some candidates calculated ${\text{P}}(X = 1)$.

Part (c) was very well done.

Question

A discrete random variable $X$ follows a Poisson distribution ${\text{Po}}(\mu )$.

Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$.

[3]

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$, use part (a) to find the value of $\mu $.

[3]

Answer/Explanation

Markscheme

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = {\text{P}}(X = x + 1)$ AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}}}$ (M1)

$ = \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$ A1

$ = \frac{\mu }{{x + 1}}$ A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

[3 marks]

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$ A1

attempting to solve for $\mu $ (M1)

$\mu = 1.40$ A1

[3 marks]

Examiners report

[N/A]

Question

A discrete random variable $X$ follows a Poisson distribution ${\text{Po}}(\mu )$.

Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$.

[3]

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$, use part (a) to find the value of $\mu $.

[3]

Answer/Explanation

Markscheme

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = {\text{P}}(X = x + 1)$ AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}}}$ (M1)

$ = \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$ A1

$ = \frac{\mu }{{x + 1}}$ A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

[3 marks]

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$ A1

attempting to solve for $\mu $ (M1)

$\mu = 1.40$ A1

[3 marks]

Examiners report

[N/A]

Question

The number of bananas that Lucca eats during any particular day follows a Poisson distribution with mean 0.2.

Find the probability that Lucca eats at least one banana in a particular day.

[2]

Find the expected number of weeks in the year in which Lucca eats no bananas.

[4]

Answer/Explanation

Markscheme

let $X$ be the number of bananas eaten in one day

$X \sim {\text{Po}}(0.2)$

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ (M1)

$ = 0.181{\text{ }}( = 1 – {{\text{e}}^{ – 0.2}})$ A1

[2 marks]

EITHER

let $Y$ be the number of bananas eaten in one week

${\text{Y}} \sim {\text{Po}}(1.4)$ (A1)

${\text{P}}(Y = 0) = 0.246596 \ldots {\text{ }}( = {{\text{e}}^{ – 1.4}})$ (A1)

let $Z$ be the number of days in one week at least one banana is eaten

$Z \sim {\text{B}}(7,{\text{ }}0.181 \ldots )$ (A1)

${\text{P}}(Z = 0) = 0.246596 \ldots $ (A1)

THEN

$52 \times 0.246596 \ldots $ (M1)

$ = 12.8{\text{ }}( = 52{{\text{e}}^{ – 1.4}})$ A1

[4 marks]

Examiners report

[N/A]

Question

The number of bananas that Lucca eats during any particular day follows a Poisson distribution with mean 0.2.

Find the probability that Lucca eats at least one banana in a particular day.

[2]

Find the expected number of weeks in the year in which Lucca eats no bananas.

[4]

Answer/Explanation

Markscheme

let $X$ be the number of bananas eaten in one day

$X \sim {\text{Po}}(0.2)$

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ (M1)

$ = 0.181{\text{ }}( = 1 – {{\text{e}}^{ – 0.2}})$ A1

[2 marks]

EITHER

let $Y$ be the number of bananas eaten in one week

${\text{Y}} \sim {\text{Po}}(1.4)$ (A1)

${\text{P}}(Y = 0) = 0.246596 \ldots {\text{ }}( = {{\text{e}}^{ – 1.4}})$ (A1)

let $Z$ be the number of days in one week at least one banana is eaten

$Z \sim {\text{B}}(7,{\text{ }}0.181 \ldots )$ (A1)

${\text{P}}(Z = 0) = 0.246596 \ldots $ (A1)

THEN

$52 \times 0.246596 \ldots $ (M1)

$ = 12.8{\text{ }}( = 52{{\text{e}}^{ – 1.4}})$ A1

[4 marks]

Examiners report

[N/A]

Question

The random variable X has a binomial distribution with parameters n and p.
It is given that E(X) = 3.5.

Find the least possible value of n.

[2]

It is further given that P(X ≤ 1) = 0.09478 correct to 4 significant figures.

Determine the value of n and the value of p.

[5]

Answer/Explanation

Markscheme

np = 3.5 (A1)

p ≤ 1 ⇒ least n = 4 A1

[2 marks]

(1 − p)ⁿ + np(1 − p)ⁿ⁻¹ = 0.09478 M1A1

attempt to solve above equation with np = 3.5 (M1)

n = 12, p = $\frac{7}{{24}}$ (=0.292) A1A1

Note: Do not accept n as a decimal.

[5 marks]

Examiners report

[N/A]

Question

The random variable X has a binomial distribution with parameters n and p.
It is given that E(X) = 3.5.

Find the least possible value of n.

[2]

It is further given that P(X ≤ 1) = 0.09478 correct to 4 significant figures.

Determine the value of n and the value of p.

[5]

Answer/Explanation

Markscheme

np = 3.5 (A1)

p ≤ 1 ⇒ least n = 4 A1

[2 marks]

(1 − p)ⁿ + np(1 − p)ⁿ⁻¹ = 0.09478 M1A1

attempt to solve above equation with np = 3.5 (M1)

n = 12, p = $\frac{7}{{24}}$ (=0.292) A1A1

Note: Do not accept n as a decimal.

[5 marks]

Examiners report

[N/A]

Question

The weights, in kg, of male birds of a certain species are modelled by a normal distribution with mean $\mu $ and standard deviation $\sigma $ .

Given that 70 % of the birds weigh more than 2.1 kg and 25 % of the birds weigh more than 2.5 kg, calculate the value of $\mu $ and the value of $\sigma $ .

[4]

A random sample of ten of these birds is obtained. Let X denote the number of birds in the sample weighing more than 2.5 kg.

(i) Calculate ${\text{E}}(X)$ .

(ii) Calculate the probability that exactly five of these birds weigh more than 2.5 kg.

(iii) Determine the most likely value of X .

[5]

The number of eggs, Y , laid by female birds of this species during the nesting season is modelled by a Poisson distribution with mean $\lambda $ . You are given that ${\text{P}}(Y \geqslant 2) = 0.80085$ , correct to 5 decimal places.

(i) Determine the value of $\lambda $ .

(ii) Calculate the probability that two randomly chosen birds lay a total of

two eggs between them.

(iii) Given that the two birds lay a total of two eggs between them, calculate the probability that they each lay one egg.

[8]

Answer/Explanation

Markscheme

we are given that

$2.1 = \mu – 0.5244\sigma $

$2.5 = \mu + 0.6745\sigma $ M1A1

$\mu = 2.27{\text{ , }}\sigma = 0.334$ A1A1

[4 marks]

(i) let X denote the number of birds weighing more than 2.5 kg

then X is B(10, 0.25) A1

${\text{E}}(X) = 2.5$ A1

(ii) 0.0584 A1

(iii) to find the most likely value of X , consider

${p_0} = 0.0563 \ldots ,{\text{ }}{p_1} = 0.1877 \ldots ,{\text{ }}{p_2} = 0.2815 \ldots ,{\text{ }}{p_3} = 0.2502 \ldots $ M1

therefore, most likely value = 2 A1

[5 marks]

(i) we solve $1 – {\text{P}}(Y \leqslant 1) = 0.80085$ using the GDC M1

$\lambda = 3.00$ A1

(ii) let ${X_1}{\text{, }}{X_2}$ denote the number of eggs laid by each bird

${\text{P}}({X_1} + {X_2} = 2) = {\text{P}}({X_1} = 0){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 1){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 2){\text{P}}({X_2} = 0)$ M1A1

$ = {{\text{e}}^{ – 3}} \times {{\text{e}}^{ – 3}} \times \frac{9}{2} + {({{\text{e}}^{ – 3}} \times 3)^2} + {{\text{e}}^{ – 3}} \times \frac{9}{2} \times {{\text{e}}^{ – 3}} = 0.0446$ A1

(iii) ${\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1|{X_1} + {X_2} = 2) = \frac{{{\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1)}}{{{\text{P}}({X_1} + {X_2} = 2)}}$ M1A1

$ = 0.5$ A1

[8 marks]

Examiners report

[N/A]