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Probability

Practice exam-style IB Math AI questions for Probability, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A city council surveys 600 households. Of these, 147 households expect to install solar panels during the next five years.

A

the relative frequency of households that expect to install solar panels.

[1]
B

The city contains 2200 households. Using the survey, the expected number of households that will install solar panels.

[2]
C

An earlier model predicted that the probability a household would install solar panels was 0.250.25. whether the survey result supports this model. Give a reason for your answer.

[2]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A computer simulation models a game in which the theoretical probability of winning is 0.250.25. In 8000 simulated games, there are 1964 wins.

A

the experimental probability of winning.

[1]
B

Using the experimental probability, the expected number of wins in 300 further games.

[2]
C

why increasing the number of simulated games may improve the , but may not correct an unsuitable simulation model.

[2]
Question 3
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

Spinner AA has four equal sectors numbered 11 to 44. Spinner BB has six equal sectors numbered 11 to 66. Each spinner is spun once. The spins are independent.

A

the number of outcomes in the sample space.

[1]
B

the probability that the sum of the two numbers is 77.

[2]
C

Given that the sum is 77, the probability that the number on spinner AA is even.

[2]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

At a school, 58%58\% of students study a language, 47%47\% study a creative arts subject and 26%26\% study both. Let LL be the event that a randomly selected student studies a language and CC the event that the student studies a creative arts subject.

A

P(L∪C)P(L\cup C).

[2]
B

the probability that the student studies neither subject.

[1]
C

Given that the student studies a creative arts subject, the probability that the student also studies a language.

[2]

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Question 5
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A shop receives 65%65\% of its headphones from supplier AA and the remainder from supplier BB. The probability that a headphone from supplier AA is defective is 0.030.03. The probability that a headphone from supplier BB is defective is 0.080.08.

A

the probability that a randomly selected headphone is defective.

[3]
B

Given that a selected headphone is defective, the probability that it came from supplier BB.

[2]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

For two events AA and BB, P(A)=0.40P(A)=0.40, P(B)=0.55P(B)=0.55 and P(A∪B)=0.73P(A\cup B)=0.73.

A

P(A∩B)P(A\cap B).

[2]
B

whether AA and BB are independent. your answer.

[2]
C

why AA and BB are not mutually exclusive.

[1]
Question 7
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A device contains two independent components, AA and BB. The probability that component AA fails during its first year is 0.080.08, and the probability that component BB fails is 0.050.05.

A

the probability that neither component fails during the first year.

[2]
B

the probability that at least one component fails.

[1]
C

A company operates 250 such devices. the expected number of devices for which at least one component fails during the first year.

[2]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Customers of a streaming service are classified as having either a basic subscription, BB, or a premium subscription, PP. Each month, 18%18\% of basic customers change to premium and 8%8\% of premium customers change to basic. matrices are column matrices in the order B,PB,P.

A

the transition matrix TT.

[2]
B

Initially, 70%70\% of customers have basic subscriptions. the probability matrix after one month.

[2]
C

There are 2000 customers. the expected number with premium subscriptions after one month.

[1]

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Question 9
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A bag contains five red counters, four blue counters and three green counters. Two counters are selected at random without replacement.

A

the probability that both counters are the same colour.

[3]
B

the probability that at least one counter is green.

[2]
C

Given that both counters are the same colour, the probability that both are red.

[1]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A Markov chain has states AA, BB and CC, with transition matrix

T=(0.60.20.10.30.50.20.10.30.7)T=\begin{pmatrix}0.6&0.2&0.1\\0.3&0.5&0.2\\0.1&0.3&0.7\end{pmatrix}

The system starts in AA.

A

the initial matrix s0s_0.

[1]
B

the probability matrix after two transitions.

[2]
C

the entry in row 22, column 33 of T4T^4 and it in context.

[2]
Question 11
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Each day, a commuter travels by car, bus or bicycle. If the commuter travels by car, the probabilities of using car, bus and bicycle the next day are 0.700.70, 0.200.20 and 0.100.10, respectively. From bus, the corresponding probabilities are 0.250.25, 0.600.60 and 0.150.15. From bicycle, they are 0.400.40, 0.100.10 and 0.500.50.

A

the transition matrix TT, using the order car, bus, bicycle.

[3]
B

Given that the commuter travels by bus today, the probability that the commuter travels by car two days from now.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A wildlife reserve classifies an animal as being either inside a protected zone, II, or outside it, OO. The monthly transition matrix is

T=(0.750.300.250.70)T=\begin{pmatrix}0.75&0.30\\0.25&0.70\end{pmatrix}

where the order is I,OI,O.

A

the exact steady- probability matrix by solving a system of equations.

[3]
B

In the long term, 1200 animals are monitored. the expected number outside the protected zone.

[2]

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Question 13
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The weekly movement of customers between three supermarkets, AA, BB and CC, is modelled by

T=(0.700.150.200.200.700.300.100.150.50)T=\begin{pmatrix}0.70&0.15&0.20\\0.20&0.70&0.30\\0.10&0.15&0.50\end{pmatrix}

The chain is regular.

A

Using repeated multiplication, the steady- probability matrix.

[3]
B

the second entry of the steady- matrix.

[1]
C

why the long-term probabilities do not depend on the initial matrix.

[1]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A Markov chain has transition matrix

T=(0.500.20.50.60.300.40.5)T=\begin{pmatrix}0.5&0&0.2\\0.5&0.6&0.3\\0&0.4&0.5\end{pmatrix}
A

T2T^2.

[3]
B

that the Markov chain is regular.

[1]
C

one consequence of regularity for the long-term probabilities.

[1]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A two- Markov chain has transition matrix

T=(0.8p0.21−p)T=\begin{pmatrix}0.8&p\\0.2&1-p\end{pmatrix}

Its steady- probability matrix is

s=(3525)s=\begin{pmatrix}\frac35\\\frac25\end{pmatrix}
A

the value of pp.

[3]
B

The system starts in 22. the probability matrix after three transitions.

[3]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A research task can be active, completed or abandoned. At the end of each week, an active task remains active with probability 0.600.60, is completed with probability 0.300.30, and is abandoned with probability 0.100.10. Once completed or abandoned, its does not change.

A

the transition matrix TT using the order active, completed, abandoned.

[2]
B

A task is initially active. the probability that it is still active after five weeks.

[1]
C

the probability that the task is completed within five weeks.

[2]
D

the probability that the task is eventually completed.

[1]

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Question 17
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

At an outdoor market, the probability that it rains is 0.300.30. If it rains, the probability that a stallholder arrives late is 0.550.55. If it does not rain, the probability that a stallholder arrives late is 0.120.12.

Let RR denote the event that it rains and LL the event that a stallholder arrives late.

A
I.

(a)(i) P(L∣R′)P(L\mid R').

[1]
II.

(a)(ii) the probability that it rains and the stallholder arrives late.

[2]
B
I.

(b)(i) the probability that a stallholder arrives late.

[2]
II.

(b)(ii) Given that a stallholder arrives late, the probability that it rained.

[2]
C

On one market day, 400400 stallholders are expected. the expected number who arrive late and your answer.

[2]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A survey of 240240 residents records whether they use a community swimming pool, SS, and whether they use a community gym, GG. There are 138138 residents who use the pool, 112112 who use the gym and 6464 who use both.

A
I.

the number of residents who use the pool only and the number who use the gym only.

[2]
II.

the number of residents who use neither facility.

[1]
B
I.

A resident is selected at random. P(G∣S)P(G\mid S).

[2]
II.

whether SS and GG are independent. your answer.

[2]
C

Two residents are selected at random without replacement. the probability that neither resident uses either facility.

[2]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A game has a theoretical probability of success of 0.250.25. Three computer simulations of the game produce the results shown.

Number of simulated games

Number of successes

Relative frequency of success

200

57

0.285

1000

263

0.263

5000

1255

A
I.

the relative frequency of success in the simulation of 50005000 games.

[1]
II.

Using this relative frequency, the expected number of successes in a further 12001200 games.

[2]
B
I.

the theoretical probability of at least one success in six independent games.

[2]
II.

why the relative frequencies do not exactly equal the theoretical probability.

[1]
C

Evaluate the claim that increasing the number of simulated games will always produce an accurate of the true probability.

[2]
Question 20
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A plant produces seeds that are classified as round, RR, or non-round, R′R'. A seed may also germinate, GG. For a randomly selected seed, P(R)=0.55P(R)=0.55, P(G∣R)=0.72P(G\mid R)=0.72 and P(G∣R′)=0.38P(G\mid R')=0.38.

A
I.

P(G)P(G).

[2]
II.

Given that a seed germinates, the probability that it is round.

[2]
B

A tray contains 350350 seeds selected from this population. the expected number that germinate.

[2]
C

The value of P(G∣R′)P(G\mid R') is changed to qq, while the other probabilities originally given in the stem remain unchanged. For this part, P(G′∣R′)=1−qP(G'\mid R')=1-q. qq so that RR and GG are independent.

[2]

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Question 21
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

At an international conference, 62%62\% of delegates attend an artificial intelligence session, AA, and 48%48\% attend a sustainability session, BB. A total of 31%31\% attend both sessions.

A
I.

P(A∪B)P(A\cup B).

[1]
II.

the probability that a delegate attends neither session.

[1]
B

Given that a delegate attends the sustainability session, the probability that they also attend the artificial intelligence session.

[2]
C

the probability that a delegate attends exactly one of the two sessions.

[1]
D

whether attending the two sessions is independent. your answer.

[2]
E

There are 12501250 delegates. the expected number who attend neither session.

[1]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A container holds 66 gold tokens, 55 silver tokens and 44 black tokens. Three tokens are selected at random without replacement.

A
I.

the probability that no black token is selected.

[2]
II.

the probability that at least one black token is selected.

[2]
B
I.

the probability that exactly one black token is selected.

[2]
II.

Given that at least one black token is selected, the probability that exactly one black token is selected. If you did not obtain an answer to part (a)(ii), use 0.6370.637.

[1]
C

The selection process is repeated 9090 times, with all tokens returned after each selection of three. the expected number of selections containing at least one black token.

[2]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A fair four-sided die is numbered 11 to 44 and a fair eight-sided die is numbered 11 to 88. Both dice are rolled once. Let AA be the event that the sum is greater than 88, and let BB be the event that the product is a multiple of 66.

D4\D8

1

2

3

4

5

6

7

8

1

(1,1)

(1,2)

(1,3)

(1,4)

(1,5)

(1,6)

(1,7)

(1,8)

2

(2,1)

(2,2)

(2,3)

(2,4)

(2,5)

(2,6)

(2,7)

(2,8)

3

(3,1)

(3,2)

(3,3)

(3,4)

(3,5)

(3,6)

(3,7)

(3,8)

4

(4,1)

(4,2)

(4,3)

(4,4)

(4,5)

(4,6)

(4,7)

(4,8)

A
I.

the number of equally likely outcomes in the sample space.

[1]
II.

P(A)P(A).

[2]
B
I.

P(B)P(B).

[2]
II.

P(A∩B)P(A\cap B).

[1]
C

whether AA and BB are independent. your answer.

[2]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A screening test is used for a condition affecting 4%4\% of a population. For a person with the condition, the probability of a positive result is 0.920.92. For a person without the condition, the probability of a positive result is 0.070.07.

Let CC denote having the condition and TT denote receiving a positive test result.

A
I.

P(C∩T)P(C\cap T).

[2]
II.

P(T)P(T).

[2]
B
I.

Given that a person receives a positive result, the probability that the person has the condition.

[2]
II.

why a positive result does not mean that the person certainly has the condition.

[1]
C

The test is administered to 500500 people from this population. the expected number of people who do not have the condition but receive a positive result.

[2]

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Question 25
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

For a flight selected at random, let DD be the event that the flight is delayed and BB the event that its passengers experience a baggage-handling problem. It is known that

P(D)=0.28,P(B)=0.18,P(D∪B)=0.402P(D)=0.28,\qquad P(B)=0.18,\qquad P(D\cup B)=0.402
A
I.

P(D∩B)P(D\cap B).

[2]
II.

P(D∣B)P(D\mid B).

[1]
B
I.

whether DD and BB are independent. your answer.

[2]
II.

the expected number of flights, from 750750 flights, that are delayed and have a baggage-handling problem.

[1]
C

After a new baggage system is introduced, the probabilities P(D)=0.28P(D)=0.28 and P(B)=0.18P(B)=0.18 remain unchanged, but the events become independent. the new value of P(D∪B)P(D\cup B).

[2]
Question 26
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

For a particular plant species, the probability that a seed produces a red-flowered plant is 0.350.35. The probability that a plant is tall is 0.720.72 if it has red flowers and 0.480.48 if it does not have red flowers.

Let RR denote red flowers and TT denote a tall plant.

A
I.

P(R∩T)P(R\cap T).

[2]
II.

P(T)P(T).

[2]
B
I.

Given that a plant is tall, the probability that it has red flowers.

[2]
II.

whether RR and TT are independent. your answer.

[1]
C

Two seeds are selected independently. the probability that at least one produces a tall plant.

[2]
Question 27
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A library classifies each book as on a shelf, SS, on loan, LL, or being reshelved, RR. Daily movement is modelled by the transition matrix

T=(0.650.100.550.300.750.200.050.150.25)T=\begin{pmatrix} 0.65&0.10&0.55\\ 0.30&0.75&0.20\\ 0.05&0.15&0.25 \end{pmatrix}

where matrices are column matrices in the order S,L,RS,L,R.

A
I.

A book is on loan initially. its initial matrix.

[1]
II.

the probability matrix after two days.

[3]
B
I.

the probability that the book is on a shelf after four days.

[2]
II.

the entry in row 11, column 33 of T4T^4.

[1]
C

the exact steady- probability matrix and the expected number of books on loan in the long term if the library has 18001800 books.

[3]
Question 28
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A wetland is classified each year as healthy, HH, or degraded, DD. Its condition is modelled by

T=(0.820.350.180.65)T=\begin{pmatrix}0.82&0.35\\0.18&0.65\end{pmatrix}

where the order is H,DH,D. Initially, the probabilities that the wetland is healthy and degraded are 0.900.90 and 0.100.10, respectively.

A
I.

the probability matrix after one year.

[2]
II.

the probability that the wetland is healthy after six years.

[2]
B
I.

the exact steady- probability matrix by solving simultaneous equations.

[2]
II.

why the long-term probabilities do not depend on the initial matrix.

[1]
C

A conservation agency monitors 24002400 similar wetlands. the expected number that are degraded in the long term.

[2]

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Question 29
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A parcel is classified as in transit, TT, delivered, DD, or lost, LL. Each day, a parcel in transit remains in transit with probability 0.580.58, is delivered with probability 0.370.37, and is lost with probability 0.050.05. Delivered and lost are absorbing states. The order is T,D,LT,D,L.

A
I.

the transition matrix.

[2]
II.

why this Markov chain is not regular.

[1]
B
I.

A parcel is initially in transit. the probability that it is still in transit after four days.

[2]
II.

the probability that the parcel has been delivered within four days.

[2]
C

the probability that the parcel is eventually delivered.

[2]
Question 30
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Tourists in a country are classified as staying in a coastal region, CC, a mountain region, MM, or a lake region, LL. Their weekly movement is modelled by

T=(0.600.250.200.300.500.250.100.250.55)T=\begin{pmatrix} 0.60&0.25&0.20\\ 0.30&0.50&0.25\\ 0.10&0.25&0.55 \end{pmatrix}

Initially, the probability matrix is

s0=(0.500.300.20)s_0=\begin{pmatrix}0.50\\0.30\\0.20\end{pmatrix}
A
I.

the probability matrix after one week.

[2]
II.

the probability matrix after three weeks.

[2]
B
I.

the steady- probability matrix.

[2]
II.

why the chain is regular.

[1]
C

In the long term, there are 50005000 tourists. the expected number staying in the mountain region.

[2]
Question 31
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A coastal warning system issues an alert after severe rainfall with probability 0.910.91. The probability of severe rainfall on a particular day is 0.120.12. On a day without severe rainfall, the system issues a false alert with probability 0.030.03. A simulation of 10 00010\,000 days produced 14001400 alerts.

A
I.

the experimental probability of an alert.

[1]
II.

the theoretical probability of an alert.

[2]
B

the expected number of alerts during 50005000 future days under the theoretical model.

[1]
C

Suppose the false-alert probability is changed to xx. xx if the overall probability of an alert is to be 0.150.15.

[2]
D

Evaluate the claim that increasing the simulation to one million days guarantees an accurate of the real probability of an alert.

[2]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A container holds 66 amber beads, 55 blue beads and 44 clear beads. Three beads are selected at random.

A

The beads are selected without replacement.

I.

the probability that one bead of each colour is selected.

[2]
II.

the probability that at least one amber bead is selected.

[2]
B

Given that at least one amber bead is selected, the probability that one bead of each colour is selected.

[2]
C

The three beads are instead selected with replacement. the probability that one bead of each colour is selected, and it with the answer to part (a)(i).

[2]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

In a game, a fair eight-sector spinner numbered 11 to 88 is spun and a coin is tossed. The coin toss and spinner result are independent, both for the fair coin and for the biased coin in part b. A player wins if the spinner number is divisible by 33, or if the coin shows heads and the spinner number is odd. The word “or” is inclusive.

Coin / Spinner

1

2

3

4

5

6

7

8

H

H&O

—

Both

—

H&O

D

H&O

—

T

—

—

D

—

—

D

—

—

A
I.

the probability that the player wins when the coin is fair.

[3]
II.

Given that the player wins, the probability that the coin showed heads.

[2]
B

The coin is replaced by a biased coin for which P(H)=pP(H)=p. pp if the probability of winning is to be 0.550.55.

[2]
C

Using the biased coin, the expected number of wins in 240240 games.

[1]
Question 34
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A curator has 1212 visually identical keys, of which 44 open a secure cabinet. Keys are selected at random.

A

Three keys are selected without replacement.

I.

the probability that at least one selected key opens the cabinet.

[2]
II.

the probability that exactly one selected key opens the cabinet.

[2]
B

Given that at least one selected key opens the cabinet, the probability that exactly one does so.

[2]
C

Keys continue to be selected without replacement.

I.

the probability of opening the cabinet within three attempts.

[1]
II.

the minimum number of attempts required for the probability of opening the cabinet to be at least 0.950.95.

[2]
Question 35
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

An animal is classified each month as occupying forest, FF, edge habitat, EE, or open land, OO. From FF, the next-month probabilities for F,E,OF,E,O are 0.70,0.20,0.100.70,0.20,0.10. From EE, they are 0.25,0.60,0.150.25,0.60,0.15. From OO, they are 0.15,0.35,0.500.15,0.35,0.50. matrices use the order F,E,OF,E,O.

A
I.

the transition matrix TT.

[2]
II.

An animal is initially in forest. its probability matrix after two months.

[2]
B

the exact steady- probability matrix by solving a system of equations.

[3]
C

In the long term, 423423 animals are monitored. and the expected number occupying open land.

[2]
Question 36
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Daily weather at an observatory is modelled as sunny, SS, or cloudy, CC. After a sunny day, the next day is sunny with probability 0.750.75. After a cloudy day, the next day is sunny with probability 0.400.40. Let xnx_n be the probability of sunny weather after nn transitions, with x0=1x_0=1.

A
I.

the transition matrix TT using the order S,CS,C.

[1]
II.

the exact steady- probability matrix.

[3]
B

xn=813+513(0.35)nx_n=\frac8{13}+\frac5{13}(0.35)^n
[2]
C

the minimum value of nn for which xnx_n is within 0.0010.001 of its steady- value. If you did not obtain the result in part (b), use the expression given there.

[2]

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Question 37
HL • Paper 2
Hard
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HL • Paper 2
Hard
Calculator Permitted

In a game, a player's score is one of 0,1,2,3,40,1,2,3,4. From scores 11, 22 or 33, the score increases by one with probability 0.550.55 and decreases by one with probability 0.450.45. Scores 00 and 44 are absorbing states. matrices use the order 0,1,2,3,40,1,2,3,4.

A
I.

the transition matrix TT.

[3]
II.

why this chain is not regular.

[1]
B
I.

The player starts with score 22. the probability matrix after two moves.

[2]
II.

the probability that the game has ended after two moves.

[1]
C

Let qiq_i be the probability that the player reaches score 44 before score 00, given that the current score is ii. Using q0=0q_0=0 and q4=1q_4=1, q2q_2.

[3]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Each day, a patient is classified as stable, SS, in intensive care, II, or discharged, DD. A stable patient remains stable with probability 0.700.70, moves to intensive care with probability 0.200.20, and is discharged with probability 0.100.10. A patient in intensive care moves to stable with probability 0.250.25, remains in intensive care with probability 0.600.60, and is discharged with probability 0.150.15. Discharged is an absorbing .

A
I.

the transition matrix using the order S,I,DS,I,D.

[2]
II.

the initial matrix for a patient who is stable initially.

[1]
B
I.

the probability that the patient has been discharged after two days.

[2]
II.

the probability that the patient is in intensive care after three days.

[2]
C

the probability that the patient has been discharged after ten days. on the long-term probability of discharge.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Customers of an energy company have either a fixed-price contract, FF, or a variable-price contract, VV. Each month, a proportion pp of fixed-price customers changes to a variable-price contract, while 12%12\% of variable-price customers changes to a fixed-price contract. The steady- probability matrix is

s=(0.400.60)s=\begin{pmatrix}0.40\\0.60\end{pmatrix}

where the order is F,VF,V.

A
I.

the transition matrix in terms of pp.

[1]
II.

Using Ts=sTs=s, pp.

[3]
B
I.

Initially, 75%75\% of customers have fixed-price contracts. the probability matrix after three months.

[2]
II.

the probability that a customer has a variable-price contract after twelve months.

[1]
C

The company has 32003200 customers. the long-term expected number with fixed-price contracts and why this value does not depend on the initial contract distribution.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A screening test is used for a condition affecting 3.5%3.5\% of a population. For a person with the condition, the probability of a positive result is 0.940.94. For a person without the condition, the probability of a negative result is 0.890.89. Test results are assumed independent conditional on whether the person has the condition.

A
I.

the probability that a randomly selected person receives a positive result.

[2]
II.

Given a positive result, the probability that the person has the condition.

[2]
B

A person takes the test twice and receives two positive results. the probability that the person has the condition.

[3]
C

The sensitivity remains 0.940.94, but the specificity is changed to ss. the minimum value of ss for which a person receiving one positive result is at least as likely as not to have the condition.

[2]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A communication station contains two components, AA and BB. On a storm day, which occurs with probability 0.080.08, the failure probabilities are 0.300.30 and 0.200.20, respectively. On a calm day, the corresponding probabilities are 0.040.04 and 0.030.03. Component failures are independent conditional on the type of day.

A
I.

the probability that at least one component fails on a storm day.

[2]
II.

the overall probability that at least one component fails.

[2]
B

Given that at least one component fails, the probability that the day was stormy.

[2]
C

A technician claims that the component failures are independent without conditioning on the weather. why this claim is not justified.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A three- manufacturing process has transition matrix

T=(0.6p0.20.30.50.30.10.5−p0.5)T=\begin{pmatrix}0.6&p&0.2\\0.3&0.5&0.3\\0.1&0.5-p&0.5\end{pmatrix}

where 0≤p≤0.50\leq p\leq0.5. Its steady- probability matrix is (2372381136)T\begin{pmatrix}\dfrac{23}{72}&\dfrac{3}{8}&\dfrac{11}{36}\end{pmatrix}^{\mathsf T}.

A
I.

pp.

[2]
II.

The process begins in 33. the probability matrix after two transitions.

[2]
B

that the Markov chain is regular.

[2]
C

the long-term consequence of regularity for this process.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A player has between 00 and 33 tokens. On each round, the player gains one token with probability 0.550.55 and loses one token with probability 0.450.45. States 00 and 33 are absorbing. The order is 0,1,2,30,1,2,3. Use the column-vector convention, with Tij=P(next state i∣current state j)T_{ij}=P(\text{next state }i\mid\text{current state }j).

A
AI.

the transition matrix TT.

[2]
AII.

why this Markov chain is not regular.

[1]
B

Let hih_i be the probability of reaching 33 before 00, starting from ii. h1h_1.

[3]
C
CI.

the probability of reaching 00 before 33, starting from 11.

[1]
CII.

For 15001500 independent players starting in 11, the expected number who reach 33 first.

[1]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Members of a library are classified as active, AA, or inactive, II. Each month, 12%12\% of active members become inactive. An unknown proportion bb of inactive members become active. Initially, 80%80\% of members are active. After one month, 74%74\% are active. matrices use the order A,IA,I.

A
I.

the transition matrix in terms of bb.

[2]
II.

bb.

[2]
B

the exact steady- probability matrix.

[2]
C

Let xnx_n be the proportion of active members after nn months.

xn=0.6+0.2(0.7)nx_n=0.6+0.2(0.7)^n
[2]
D

the first month for which the proportion of active members is within 0.010.01 of its steady- value. If you did not obtain the result in part (c), use xn=0.6+0.2(0.7)nx_n=0.6+0.2(0.7)^n.

[2]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

A bee colony is classified each week as healthy, HH, under treatment, TT, recovered, RR, or lost, LL. From HH, the probabilities of moving to H,T,R,LH,T,R,L are 0.50,0.30,0.15,0.050.50,0.30,0.15,0.05. From TT, they are 0.10,0.40,0.40,0.100.10,0.40,0.40,0.10. States RR and LL are absorbing. The order is H,T,R,LH,T,R,L.

A
I.

the transition matrix TT.

[2]
II.

why there is no unique steady- probability matrix.

[1]
B

Let rHr_H and rTr_T be the probabilities of eventual recovery starting from HH and TT, respectively. rHr_H.

[3]
C

A conservation programme begins with 900 colonies, all initially in HH.

I.

the expected number that eventually recover.

[1]
II.

the probability that a healthy colony is eventually lost, and why the two eventual outcomes have probabilities summing to 11.

[2]
Question 46
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

An insect is observed once each hour and classified as being on a leaf, LL, on a stem, SS, or in the soil, GG. Its movement is modelled by

T=(0.500.200.100.400.500.300.100.300.60)T=\begin{pmatrix} 0.50&0.20&0.10\\ 0.40&0.50&0.30\\ 0.10&0.30&0.60 \end{pmatrix}

with order L,S,GL,S,G.

A
I.

An insect is initially in the soil. its initial matrix.

[1]
II.

the probability matrix after three hours.

[3]
B
I.

the exact steady- probability matrix.

[3]
II.

the long-term expected number of insects on stems in a population of 10001000 insects.

[1]
C

Researchers that, among insects currently on a stem, the proportion moving to a leaf is 0.180.18 when the insect was previously on a leaf, but 0.310.31 when it was previously in the soil. Evaluate whether the Markov model is appropriate.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Employees alternate between working on-site, OO, and remotely, RR. In each week, an on-site employee changes to remote work with probability aa, while a remote employee changes to on-site work with probability bb, where 0<a<10<a<1 and 0<b<10<b<1. Let xnx_n be the probability that an employee works on-site after nn weeks. Use the column-vector convention and the order (O,R)(O,R).

A
I.

the transition matrix TT.

[1]
II.

the exact steady- probability matrix.

[2]
B

Derive the recurrence relation

xn+1=b+(1−a−b)xnx_{n+1}=b+(1-a-b)x_n
[2]
C

xn=ba+b+(x0−ba+b)(1−a−b)nx_n=\frac{b}{a+b}+\left(x_0-\frac{b}{a+b}\right)(1-a-b)^n
[2]
D

the long-term behaviour when a+b>1a+b>1, and that the probabilities still converge.

[2]
E

why the Markov chain is regular for the stated range of aa and bb.

[1]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monitoring device is either calibrated, CC, or drifting, DD. Its daily transition matrix is

T=(0.900.300.100.70)T=\begin{pmatrix}0.90&0.30\\0.10&0.70\end{pmatrix}

using the order C,DC,D. When calibrated, it produces an alert with probability 0.040.04. When drifting, it produces an alert with probability 0.650.65. Conditional on the device , alert outcomes are independent.

A
I.

the exact steady- probability matrix of the device .

[2]
II.

Assuming the device is initially in steady , the probability that it produces an alert.

[2]
B

Given that the device produces an alert, the probability that it is drifting on that day.

[2]
C

No maintenance is performed after the alert. the probability that the device produces an alert on the following day.

[2]
D

For a separate diagnostic procedure, assume the device remains fixed while repeated independent readings are taken. Starting from the steady- probabilities, the minimum number of consecutive alerts required for the conditional probability that the device is drifting to exceed 0.990.99.

[2]

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Hypothesis Testing