L.C. MATHS

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Past Papers

## Probability

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Elementary Probability

Permutations

Combinations

Sample Spaces

Tree Diagrams

Conditional Probability

Expected Values

Bernoulli Trials

Independence of Two Events

## 2022 Paper 2 Question 1

(a) The table below gives some details on the number of different types of student in a
university. There are $$22{,}714$$ students in the university in total.

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

Total

$$8{,}576$$

$$22{,}714$$

(i) Fill in the three missing values to complete the table above.

(ii) One student is picked at random from the students in the university.
Let $$O$$ be the event that the student is $$24$$ years old, or older.
Let $$U$$ be the event that the student is an undergraduate.
Are the events $$O$$ and $$U$$ independent? Justify your answer.

(i)

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

$$5{,}654$$

$$7{,}007$$

Total

$$14{,}138$$

$$8{,}576$$

$$22{,}714$$

(ii) Since $$P(O)\times P(U)\neq P(O\cap U)$$, they are not independent.

Solution

(i)

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

$$5{,}654$$

$$7{,}007$$

Total

$$14{,}138$$

$$8{,}576$$

$$22{,}714$$

(ii)

\begin{align}P(O)=\frac{8{,}576}{22{,}714}\end{align}

and

\begin{align}P(U)=\frac{15{,}707}{22{,}714}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(O)\times P(U)&=\frac{8{,}576}{22{,}714}\times\frac{15{,}707}{22{,}714}\\&\approx0.261\end{align}

and

\begin{align}P(O\cap U)&=\frac{2{,}922}{22{,}714}\\&\approx0.129\end{align}

Since $$P(O)\times P(U)\neq P(O\cap U)$$, they are not independent.

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(b) Three people are picked at random from a class.
Find the probability that all three were born on the same day of the week.
Assume that the probability of being born on each day is the same.

$$\dfrac{1}{49}$$

Solution

\begin{align}7\times\left(\frac{1}{7}\times\frac{1}{7}\times\frac{1}{7}\right)=\frac{1}{49}\end{align}

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(c) There are $$b$$ boys and $$g$$ girls in a class, where $$b,g\in\mathbb{N}$$.
$$\dfrac{3}{5}$$ of the students in the class are girls.

$$4$$ boys and $$4$$ girls join the class.
One student is then picked at random from the whole class.
The probability that this student is a girl is now $$\dfrac{4}{7}$$.

Find the value of $$b$$ and the value of $$g$$.

$$b=8$$ and $$g=12$$

Solution

\begin{align}\frac{g}{b+g}=\frac{3}{5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3(b+g)=5g\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3b-2g=0\end{align}

and

\begin{align}\frac{g+4}{(b+4)+(g+4)}=\frac{4}{7}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4(b+g+8)=7(g+4)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4b-3g=-4\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}3b-2g=0\end{align}

\begin{align}4b-3g=-4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12b-8g=0\end{align}

\begin{align}12b-9g=-12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-g=-12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}g=12\end{align}

and

\begin{align}3b-2g=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b&=\frac{2g}{3}\\&=\frac{2(12)}{3}\\&=8\end{align}

Video Walkthrough

## 2022 Paper 2 Question 8(d)

(d) John bought a car a number of years ago. The table below gives an estimate of the probability that each of the following three events happens to John’s car in the next year.

Event Probability

$$0.095$$

Timing belt goes

$$0.041$$

Air filters break

$$0.073$$

(i) If the head gasket blows, John will have to replace his car, at an estimated cost of €$$20{,}000$$. If the head gasket is replaced now, it will cost €$$1{,}450$$, and the probability that it blows in the next year will be reduced to $$0.005$$.

Based on these figures, use expected values to work out if it is worth replacing the head gasket now, or not.

(ii) Work out the probability that at least one of the events in the table above happens to John’s car this year, taking these events to be independent. Give your answer correct to $$3$$ decimal places.

(i) It is worth replacing it now.

(ii) $$0.195$$

Solution

(i)

Doesn’t replace

\begin{align}E(X)&=(0.095)(20{,}000)\\&=1{,}900\mbox{ euro}\end{align}

$\,$

Replaces

\begin{align}E(X)&=(0.005)(20{,}000)+1{,}450\\&=1{,}550\mbox{ euro}\end{align}

It is therefore worth replacing it now.

(ii)

\begin{align}P(\mbox{at least one})&=1-P(\mbox{none}
)\\&=1-(1-0.095)\times(1-0.041)\times(1-0.073)\\&\approx0.195\end{align}

Video Walkthrough

## 2022 Paper 2 Question 10(b)-(d)

(b) Sally takes part in a number of different races in the competition.
The probability that she makes a false start in any given race is $$5\%$$.
Find the probability that she makes her first false start in her fourth race.
Give your answer correct to $$4$$ decimal places.

$$0.0429$$

Solution

\begin{align}P&=0.95\times0.95\times0.95\times(1-0.95)\\&\approx0.0429\end{align}

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(c) $$20$$ relay teams took part in the competition. For any particular team, the probability that they drop the baton at some point during the competition is $$0.1$$.

Find the probability that at most 𝟐 teams drop the baton during the competition.
Give your answer correct to $$4$$ decimal places.

$$0.6769$$

Solution

\begin{align}P(\mbox{at most 2})&=P(\mbox{0 or 1 or 2)}\\&=P(0)\mbox{ or }P(1)\mbox{ or }P(2)\\&={20\choose0}0.9^{20}+{20\choose1}0.1^10.9^{19}+{20\choose2}0.1^20.9^{18}\\&\approx0.6769\end{align}

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(d) $$300$$ runners take part in a road race.
Each runner has a number, from $$1$$ to $$300$$ inclusive. No two runners have the same number.

Two runners are picked at random from the runners in this race.
Work out the probability that the sum of their numbers is $$101$$.

$$\dfrac{1}{897}$$

Solution

\begin{align}P&=\frac{100}{300}\times\frac{1}{299}\\&=\frac{1}{897}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 1

In a particular population $$15\%$$ of the people are left footed.
A soccer team of $$11$$ players, including $$1$$ goalkeeper, is picked at random from the population.

(a) Find the probability that there is exactly one left footed player on the team.
Give you answer correct to three decimal places.

$$0.325$$

Solution

\begin{align}P(\mbox{one left-footed})&={11\choose1}\times0.15^1\times0.85^{10}\\&\approx0.325\end{align}

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(b) Find the probability that less than three players on the team are left footed.
Give you answer correct to two decimal places.

$$0.78$$

Solution

\begin{align}P(\mbox{less than three are left-footed})&=P(\mbox{none are left-footed}) \mbox{ or } P(\mbox{one is left-footed}) \mbox{ or } P(\mbox{two are left-footed})\\&={11\choose0}\times0.85^{11}+{11\choose1}\times0.15^1\times0.85^{10}+{11\choose2}\times0.15^2\times0.85^{9}\\&\approx0.78\end{align}

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(c) The goalkeeper is left footed.
Find the probability that at least eight of the remainder of the team are right footed.
Give you answer correct to two decimal places.

$$0.82$$

Solution

\begin{align}P(\mbox{at least eight are right-footed})&=P(\mbox{less than three are left-footed})\\&=P(\mbox{none are left-footed})\mbox{ or }P(\mbox{one is left-footed})\mbox{ or }P(\mbox{two are left-footed})\\&={10\choose0}\times0.85^{10}+{10\choose1}\times0.15^1\times0.85^{9}+{10\choose2}\times0.15^2\times0.85^{8}\\&\approx0.82\end{align}

Video Walkthrough

## 2021 Paper 2 Question 8(c)

(c) The school caretaker has a box with $$23$$ room keys in it.
$$12$$ of the keys are for general classrooms, $$6$$ for science labs and $$5$$ for offices.

(i) Four keys are drawn at random from the box.
What is the probability that the 4th key drawn is the first office key drawn?
Give your answer correct to $$4$$ decimal places.

(ii) All the keys are returned to the box. Then $$3$$ keys are drawn at random from the box one after the other, without replacement. What is the probability that one of them is for a general classroom, one is for a science lab and one is for an office?
Give your answer correct to $$4$$ decimal places.

(i) $$0.1152$$

(ii) $$0.2033$$

Solution

(i)

\begin{align}P&=\frac{18}{23}\times\frac{17}{22}\times\frac{16}{21}\times\frac{5}{20}\\&\approx0.1152\end{align}

(ii)

\begin{align}P&=3!\times\frac{12}{23}\times\frac{6}{22}\times\frac{5}{21}\\&\approx0.2033\end{align}

Video Walkthrough

## 2021 Paper 2 Question 10

People with O-negative blood type are called “universal donors” because their blood can be given to anyone else. In Ireland approximately $$8\%$$ of the population have O-negative blood type (source: Blood Transfusion Service).

(a)

(i) At a blood donation clinic, ten donors give blood, one after the other.
Find the probability that the tenth person is the third O-negative donor.

(ii) At a blood donation clinic, five donors give blood.
What is the probability that at least one of the five donates O-negative blood?

(iii) Find the minimum number of blood donors required, so that the probability that at least one of them is type O-negative is greater than $$0.97$$.

(i) $$0.0103$$

(ii) $$0.3409$$

(iii) $$43$$

Solution

(i)

\begin{align}P(\mbox{tenth is third O neg})&=P(\mbox{exactly two from first nine are O neg})\times P(\mbox{tenth is O neg})\\&=\left[{9\choose2}\left(\frac{8}{100}\right)^2\left(\frac{92}{100}\right)^7\right]\times\frac{8}{100}\\&\approx0.0103\end{align}

(ii)

\begin{align}P(\mbox{at least one is O neg})&=1-P(\mbox{none are O neg})\\&=1-\left(\frac{92}{100}\right)^5\\&\approx0.3409\end{align}

(iii)

\begin{align}1-\left(\frac{92}{100}\right)^n=0.97\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.92^n=0.03\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln(0.92^n)=\ln(0.03)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n\ln(0.92)=\ln(0.03)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n&=\frac{\ln(0.03)}{\ln(0.92)}\\&42.05…\end{align}

Therefore, the minimum number of blood donors is $$43$$.

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(b) A homeowner has a problem with the heating system in her house. A plumber has identified the problem as a faulty part. The homeowner knows that in $$80\%$$ of cases a repair of the part will fix the problem and this repair will cost €$$70$$. If the repair does not work then a new part will have to be bought costing €$$150$$ and there will be an additional labour cost of €$$80$$ to replace the old part with the new.
Find the expected value of the cost of fixing this faulty system.

$$116\mbox{ euro}$$

Solution

\begin{align}E&=(0.8)(70)+(0.2)(70+15+80)\\&=116\mbox{ euro}\end{align}

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(c) A life insurance policy pays out €$$120{,}000$$ if the policy holder dies and €$$40{,}000$$ if the policy holder becomes disabled. The insurance company has calculated that in general, in any given year, the probability of death is $$0.0001$$ and the probability of disability is $$0.002$$.
The company has $$18{,}000$$ policy holders on its books at present who are all charged the same premium. The company’s goal is to make €$$900{,}000$$ profit in a particular year.
Find the annual premium it should charge its customers which in an average year would generate this level of profit.

$$142\mbox{ euro}$$

Solution

The average amount that will have to pay to each customer is:

\begin{align}(120{,}000)(0.0001)+(40{,}000)(0.002)=92\mbox{ euro}\end{align}

The amount of profit that they wish to make from each customer is instead

\begin{align}\frac{900{,}000}{18{,}000}=50\mbox{ euro}\end{align}

To make this profit, they there need to charge each person a premium of $$92+50=142\mbox{ euro}$$.

Video Walkthrough

## 2020 Paper 2 Question 5

(a) Two events $$A$$ and $$B$$ are such that $$P(A)=\dfrac{3}{4}$$ and $$P(A\cap B)=\dfrac{1}{2}$$.

(i) Find $$P(B|A)$$. Give your answer as a fraction in its simplest form

(ii) $$P(A\cup B)=\dfrac{11}{12}$$. Investigate if the events $$A$$ and $$B$$ are independent.

(i) $$\dfrac{2}{3}$$

(ii) The events are independent.

Solution

(i)

\begin{align}P(B|A)&=\frac{P(A\cap B)}{P(A)}\\&=\frac{\frac{1}{2}}{\frac{3}{4}}\\&=\frac{2}{3}\end{align}

(ii)

\begin{align}P(A\cup B)=P(A)+P(B)-P(A\cap B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{11}{12}=\frac{3}{4}+P(B)-\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(B)&=\frac{11}{12}-\frac{3}{4}+\frac{1}{2}\\&=\frac{2}{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(A)\times P(B)&=\frac{3}{4}\times\frac{2}{3}\\&=\frac{1}{2}\\&=P(A\cup B)\end{align}

Therefore, the events are independent.

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(b) A spinner consists of $$4$$ segments, as shown.
Each segment is equally likely to be landed on.
Liam, Sorcha and Lee play a game in which the spinner is spun twice and the numbers landed on are added together.
The result is divided by $$3$$ and the remainder is recorded.

If the remainder is $$0$$ then Liam wins the game.
If the remainder is $$1$$ then Sorcha wins the game.
If the remainder is $$2$$ then Lee wins the game.

Is this a fair game? (i.e. Are all $$3$$ participants equally likely to win?)

It is an unfair game.

Solution

There are $$3$$ players and $$4\times4=16$$ outcomes.

Therefore, all three people cannot have the same number of outcomes and it is therefore an unfair game.

Video Walkthrough

## 2020 Paper 2 Question 6

(a) A class group carried out a study of the makes and fuel types of cars in a large carpark.
It found that $$30\%$$ of the cars ran on diesel and $$70\%$$ of these diesel cars were Volkswagen.
It found that $$60\%$$ of the cars ran on petrol and $$25\%$$ of these petrol cars were Volkswagen.
It found that $$10\%$$ of the cars were hybrid/electric and $$9\%$$ of these cars were Volkswagen.
One car is selected at random from the car park.
Find the probability that it is a Volkswagen car.

$$0.369$$

Solution

\begin{align}P&=0.3\times0.7+0.6\times0.25+0.1\times0.09\\&=0.369\end{align}

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(b) The Road Safety Authority has data on driving test pass rates at all its test centres.

(i) In a particular Driving Test Centre the probability that a person taking the test for the first time will pass is $$\dfrac{1}{4}$$. All of the test results are independent.
In this centre on a particular day Joe, along with $$5$$ others, takes the test.
All six are taking the test for the first time.
Find the probability that Joe passes the test along with exactly $$2$$ others.

(ii) The overall pass rate for all drivers at another centre is $$\dfrac{1}{2}$$
(Whether it is their first attempt
or a subsequent attempt).
On a particular day, $$n$$ people take the test in this centre.
The probability that two people or less than two people pass the test can be written in the form

\begin{align}\frac{an^2+bn+c}{2^{n+1}}\end{align}

where $$a,b,c\in\mathbb{N}$$.

Find the value of $$a$$, the value of $$b$$, and the value of $$c$$.

(i) $$\dfrac{135}{2048}$$

(ii) $$a=1$$, $$b=1$$ and $$c=2$$

Solution

(i)

\begin{align}P&={5\choose 2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)\\&=\frac{135}{2048}\end{align}

(ii)

\begin{align}P(\mbox{at most two pass})&=P(\mbox{none pass})+P(\mbox{one passes})+P(\mbox{two pass})\\&=\left(\frac{1}{2}\right)^n+{n\choose1}\left[\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{n-1}\right]+{n\choose2}\left[\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{n-2}\right]\\&=\frac{1}{2^n}+\frac{n}{2^n}+\frac{n(n-1)}{2^{n+1}}\\&=\frac{2+2n+n^2-n}{2^{n+1}}\\&=\frac{n^2+n+2}{2^{n+1}}\end{align}

\begin{align}\downarrow\end{align}

$$a=1$$, $$b=1$$ and $$c=2$$

Video Walkthrough

## 2019 Paper 2 Question 1

(a) A class consists of $$12$$ boys and $$8$$ girls.

(i) Two students are selected at random from the class. What is the probability that the two students selected will be a boy and a girl in any order?

(ii) Four students are selected, one at a time, at random from the class.
What is the probability that the first three students selected will be boys and the fourth will be a girl?

(i) $$\dfrac{48}{95}$$

(ii) $$\dfrac{88}{969}$$

Solution

(i)

\begin{align}P&=\frac{12}{20}\times\frac{8}{19}+\frac{8}{20}\times\frac{12}{19}\\&=\frac{48}{95}\end{align}

(ii)

\begin{align}P&=\frac{12}{20}\times\frac{11}{19}\times\frac{10}{18}\times\frac{8}{17}\\&=\frac{88}{969}\end{align}

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(b) An examination paper is made up of two sections, Section $$A$$ consisting of $$7$$ questions
and Section $$B$$ consisting of $$8$$ questions. The paper contains the following instruction:

“From section $$A$$ you must answer question $$1$$ and any three other questions.
From Section $$B$$ you must also answer any four questions.”

Find how many different combinations of questions may be answered if a candidate follows
this instruction.

$$1{,}400$$

Solution

\begin{align}N&={1\choose1}\times{6\choose3}\times{8\choose4}\\&=1{,}400\end{align}

Video Walkthrough

## 2019 Paper 2 Question 6

(a) Two independent events $$F$$ and $$S$$ are represented in the Venn diagram shown below.
$$P(F\backslash S)=\dfrac{1}{4}$$, $$P(F\cap S)=\dfrac{1}{5}$$, $$P(S\backslash F)=x$$, and $$P(F\cup S)’=y$$, where $$x,y\neq0$$.
Find the value of $$x$$ and the value of $$y$$.

$$x=\dfrac{11}{45}$$ and $$y=\dfrac{11}{36}$$

Solution

\begin{align}P(F\cap S)=P(F)\times P(S)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{5}=\left(\frac{1}{4}+\frac{1}{5}\right)\times P(S)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{5}=\frac{9}{20}\times P(S)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(S)&=\frac{20}{45}\\&=\frac{4}{9}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{5}+x=\frac{4}{9}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{4}{9}-\frac{1}{5}\\&=\frac{20-9}{45}\\&=\frac{11}{45}\end{align}

and

\begin{align}\frac{1}{4}+\frac{1}{5}+x+y=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{4}+\frac{1}{5}+\frac{11}{45}+y=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=1-\frac{1}{4}-\frac{1}{5}-\frac{11}{45}\\&=\frac{11}{36}\end{align}

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(b) In a club there are German, Irish and Spanish children only.
There are $$10$$ Spanish children.
There are twice as many Irish children as German children.

They are all in a group waiting to get on a swing.
One child will be selected at random to go first and will not re-join the group.
Then a second child will be selected at random to go next.

The probability that the first child selected will be German and that the second child selected will not be German is $$\dfrac{1}{6}$$.

Find how many children are in the club.

$$25$$

Solution

Let $$x$$ be the number of German children.

• There are $$10$$ Spanish children
• There are $$x$$ German children
• There are $$2x$$ Irish children

In total, there are therefore $$10+x+2x=10+3x$$ children.

\begin{align}\frac{x}{10+3x}\times\frac{2x+10}{9+3x}=\frac{1}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{2x^2+10x}{90+30x+27x+9x^2}=\frac{1}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x^2+60x=90+57x+9x^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+3x-90=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2+x-30=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+6)(x-5)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=5\end{align}

(as $$x>0$$). Therefore, the total number of children is

\begin{align}10+3x&=10+3(5)\\&=25\end{align}

Video Walkthrough

## 2018 Paper 2 Question 1

In a competition Mary has a probability of $$\dfrac{1}{20}$$ of winning, a probability of $$\dfrac{1}{10}$$ of finishing in second place, and a probability of $$\dfrac{1}{4}$$ of finishing in third place. If she wins the competition she gets €$$9000$$. If she comes second she gets €$$7000$$ and if she comes third she gets €$$3000$$. In all other cases she gets nothing. Each participant in the competition must pay €$$2000$$ to enter.

(a) Find the expected value of Mary’s loss if she enters the competition.

$$100\mbox{ euro}$$

Solution

\begin{align}E(X)&=2000-\left[\frac{1}{20}(9000)+\frac{1}{10}(7000)+\frac{1}{4}(3000)\right]\\&=100\mbox{ euro}\end{align}

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(b) Each of the 3 prizes in the competition above is increased by the same amount (€$$x$$) but the entry fee is unchanged.
For example, if Mary wins the competition now, she would get €($$9000+x$$).
Mary now expects to break even.
Find the value of $$x$$.

$$250\mbox{ euro}$$

Solution

\begin{align}\frac{1}{20}(9000+x)+\frac{1}{10}(7000+x)+\frac{1}{4}(3000+x)=2000\mbox{ euro}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1900+\frac{8x}{20}=2000\mbox{ euro}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{8x}{20}=100\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8x=2000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{2000}{8}\\&=250\mbox{ euro}\end{align}

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## 2018 Paper 2 Question 3(a)

(a) A security code consists of six digits chosen at random from the digits $$0$$ to $$9$$.
The code may begin with zero and digits may be repeated.
For example $$071737$$ is a valid code.

(i) Find how many of the possible codes will end with a zero.

(ii) Find how many of the possible codes will contain the digits $$2$$ $$0$$ $$1$$ $$8$$ together and in this order.

(i) $$100{,}000$$

(ii) $$300$$

Solution

(i)

\begin{align}N&=10\times10\times10\times10\times10\times1\\&=100{,}000\end{align}

(ii)

\begin{align}N&=3\times1\times10\times10\\&=300\end{align}

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## 2017 Paper 2 Question 1

When Conor rings Ciara’s house, the probability that Ciara answers the phone is $$\dfrac{1}{5}$$.

(a) Conor rings Ciara’s house once every day for $$7$$ consecutive days. Find the probability that she will answer the phone on the $$2$$nd, $$4$$th, and $$6$$th days but not on the other days.

$$\dfrac{256}{78{,}125}$$

Solution

\begin{align}P&=\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\\&=\frac{256}{78{,}125}\end{align}

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(b) Find the probability that she will answer the phone for the $$4$$th time on the $$7$$th day.

$$\dfrac{1280}{78{,}125}$$

Solution

\begin{align}P&={6\choose3}\times\left(\frac{1}{5}\right)^3\times\left(\frac{4}{5}\right)^3\times\frac{1}{5}\\&=\frac{1280}{78{,}125}\end{align}

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(c) Conor rings her house once every day for $$n$$ days. Write, in terms of $$n$$, the probability that Ciara will answer the phone at least once.

$$1-\left(\dfrac{4}{5}\right)^n$$

Solution

\begin{align}P(\mbox{at least once})&=1-P(\mbox{never})\\&=1-\left(\frac{4}{5}\right)^n\end{align}

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(d) Find the minimum value of $$n$$ for which the probability that Ciara will answer the phone at least once is greater than $$99\%$$.

$$n=21$$

Solution

\begin{align}1-\left(\frac{4}{5}\right)^n>0.99\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{4}{5}\right)^n<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n\ln\left(\frac{4}{5}\right)<\ln(0.01)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n<\frac{\ln(0.01)}{\ln\left(\frac{4}{5}\right)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n<20.637…\end{align}

Therefore, the minimum value of $$n$$ is $$21$$.

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## 2017 Paper 2 Question 8(b)

(b) In Galway, rain falls in the morning on $$\dfrac{1}{3}$$ of the school days in the year.

When it is raining the probability of heavy traffic is $$\dfrac{1}{2}$$.

When it is not raining the probability of heavy traffic is $$\dfrac{1}{4}$$.

When it is raining and there is heavy traffic, the probability of being late for school is $$\dfrac{1}{2}$$.

When it is not raining and there is no heavy traffic, the probability of being late for school is $$\dfrac{1}{8}$$.

In any other situation the probability of being late for school is $$\dfrac{1}{5}$$.

Some of this information is shown in the tree diagram below.

(i) Write the probability associated with each branch of the tree diagram and the probability of each outcome into the blank boxes provided.
Give each answer in the form $$\dfrac{a}{b}$$, where $$a,b\in\mathbb{N}$$.

(ii) On a random school day in Galway, find the probability of being late for school.

(iii) On a random school day in Galway, find the probability that it rained in the morning, given that you were late for school.

(i)

(ii) $$\dfrac{17}{80}$$

(iii) $$\frac{28}{51}$$

Solution

(i)

(ii)

\begin{align}P&=\frac{1}{12}+\frac{1}{30}+\frac{2}{60}+\frac{6}{96}\\&=\frac{17}{80}\end{align}

(iii)

\begin{align}P(R|L)&=\frac{P(R\cap L)}{P(L)}\\&=\frac{\frac{1}{12}+\frac{1}{30}}{\frac{17}{80}}\\&=\frac{28}{51}\end{align}

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## 2017 Paper 2 Question 9(f)

Conor’s property is bounded by the straight bank of a river, as shown in Figure 1 above.
$$T$$ is the base of a vertical tree that is growing near the opposite bank of the river.
$$|TE|$$ is the height of the tree, as shown in Figure 2 above.
From the point $$C$$, which is due west of the tree, the angle of elevation of $$E$$, the top of the tree, is $$60^{\circ}$$.
From the point $$D$$, which is $$15\mbox{ m}$$ due north of $$C$$, the angle of elevation of $$E$$ is $$30^{\circ}$$ (see Figure 2).
The land on both sides of the river is flat and at the same level.

The tree falls across the river and hits the bank at Conor’s side at the point $$F$$. The maximum size of the angle $$FTC$$ can be found to be $$54.7^{\circ}$$.

(f) If the tree was equally likely to fall in any direction, find the probability that it would hit the bank at Conor’s side, when it falls.
Give your answer as a percentage, correct to $$1$$ decimal place

$$30.4\%$$

Solution

\begin{align}P&=\frac{(54.7)(2)}{360}\\&=0.3038…\\&\approx30.4\%\end{align}

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## 2016 Paper 2 Question 5

(a)

(i) In an archery competition, the team consisting of John, David, and Mike will win $$1$$st prize if at least two of them hit the bullseye with their last arrows. From past experience, they know that the probability that John, David, and Mike will hit the
bullseye on their last arrow is $$\dfrac{1}{5}$$, $$\dfrac{1}{6}$$ and $$\dfrac{1}{4}$$ respectively.
Complete the table below to show all the ways in which they could win $$1$$st prize.

$$t$$ Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii) Hence or otherwise find the probability that they will win the competition.

(i)

$$t$$ Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii) $$\dfrac{13}{120}$$

Solution

(i)

$$t$$ Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii)

\begin{align}P&=\frac{1}{5}\times\frac{1}{6}\times\frac{3}{4}+\frac{1}{5}\times\frac{5}{6}\times\frac{1}{4}+\frac{4}{5}\times\frac{1}{6}\times\frac{1}{4}+\frac{1}{5}\times\frac{1}{6}\times\frac{1}{4}\\&=\frac{13}{120}\end{align}

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(b) Two events, A and B, are represented in the diagram.
$$P(A\cap B)=0.1$$, $$P(B\backslash A)=0.3$$ and $$P(A\backslash B)=x$$.
Write $$P(A)$$ in terms of $$x$$ and hence, or otherwise, find the value of $$x$$ for which the events $$A$$ and $$B$$ are independent.

$$P(A)=x+0.1$$ and $$x=0.15$$

Solution

$$P(A)=x+0.1$$.

\begin{align}P(A\cap B)=P(A)\times P(B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.1=(x+0.1)\times(0.3+0.1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{0.1}{0.3+0.1}-0.1\\&=0.15\end{align}

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## 2016 Paper 2 Question 6

A local sports club is planning to run a weekly lotto. To win the Jackpot of €$$1000$$, contestants must match one letter chosen from the $$26$$ letters in the alphabet and two numbers chosen, in the correct order, from the numbers $$0$$ to $$9$$. In this lotto, repetition of numbers is allowed (e.g. $$M$$, $$3$$, $$3$$ is an outcome).

(a) Calculate the probability that $$M,3,3$$ would be the winning outcome in a particular week.

$$\dfrac{1}{2600}$$

Solution

\begin{align}P&=\frac{1}{26}\times\frac{1}{10}\times\frac{1}{10}\\&=\frac{1}{2600}\end{align}

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(b) If a contestant matches the letter only, or the letter and one number (but not both numbers), they will win €$$50$$. Using the table below, or otherwise, find how much the club should expect to make or lose on each play, correct to the nearest cent, if they charge €$$2$$ per play.

Event Payout ($$x$$) Probability ($$P(x)$$) $$xP(x)$$

Win Jackpot

Match letter and first number only

Match letter and second number only

Match letter and neither number

Fail to win

Event Payout ($$x$$) Probability ($$P(x)$$) $$xP(x)$$

Win Jackpot

$$1{,}000$$

$$\dfrac{1}{2{,}600}$$

$$\dfrac{1{,}000}{2{,}600}$$

Match letter and first number only

$$50$$

$$\dfrac{9}{2{,}600}$$

$$\dfrac{450}{2{,}600}$$

Match letter and second number only

$$50$$

$$\dfrac{9}{2{,}600}$$

$$\dfrac{450}{2{,}600}$$

Match letter and neither number

$$50$$

$$\dfrac{81}{2{,}600}$$

$$\dfrac{4{,}050}{2{,}600}$$

Fail to win

$$0$$

$$0$$