Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 8: Probability worked solutions

Worked solutions to the practice questions in Chapter 8 of Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[5 marks]
A fair coin is tossed and a fair spinner with five equal sectors numbered 1 to 5 is spun. One outcome is H3 (head and 3).
(a) List the sample space and write down n(U).
[2]
(b) Find the probability that the coin shows a head and the spinner shows an odd number.
[2]
(c) Find the probability that the spinner does not show 5.
[1]
Show solutionHide solution
(a) U = \{\text{H}1, \text{H}2, \text{H}3, \text{H}4, \text{H}5, \text{T}1, \text{T}2, \text{T}3, \text{T}4, \text{T}5\}
A1
n(U) = 10
A1
(b) H1, H3, H5: \ n = 3 (M1)
\dfrac{3}{10}
A1
(c) 1 - \dfrac{2}{10} = \dfrac{8}{10} = \dfrac{4}{5}
A1

Final answer

(a) H1–H5, T1–T5; 10 (b) 3/10 (c) 4/5

Question 2

[4 marks]
A plastic bottle cap is flipped 400 times. It lands top up 132 times, top down 236 times and on its side 32 times.
(a) Estimate the probability that the cap lands top down.
[1]
(b) Estimate the probability that the cap does not land on its side.
[1]
(c) Explain why the probability of each way of landing is not \dfrac{1}{3}.
[1]
(d) Another student flips a similar cap 40 times. State, with a reason, whose estimate in (a) is more reliable.
[1]
Show solutionHide solution
(a) relative frequency \dfrac{236}{400} = 0.59
A1
(b) \dfrac{400 - 32}{400} = \dfrac{368}{400} = 0.92
A1
(c) the three ways of landing are not equally likely (the cap is not symmetrical), so \dfrac{n(A)}{n(U)} cannot be used; the data show very different frequencies
R1
(d) the estimate from 400 flips, because more trials give a more reliable relative frequency
R1

Final answer

(a) 0.59 (b) 0.92 (c) outcomes are not equally likely (d) 400 flips: more trials

Question 3

[4 marks]
(a) The probability that a particular bus is late on a school day is 0.15. Find the expected number of late days in a school year of 260 days.
[1]
(b) The probability that a new phone charger is faulty is 0.004. A shop receives 7500 chargers. Find the expected number of faulty chargers.
[1]
(c) Each ticket in a charity lucky dip wins a small prize with probability 0.3. Lin buys 45 tickets.
(i) Find the expected number of prizes Lin wins.
[1]
(ii) Explain why an expected number does not have to be a whole number.
[1]
Show solutionHide solution
(a) 260 \times 0.15 = 39
A1
(b) 7500 \times 0.004 = 30
A1
(c)(i) 45 \times 0.3 = 13.5
A1
(ii) it is a long-run average (the mean number over many repeats of buying 45 tickets), not a number that must happen on one occasion
R1
Note: do not accept “round it to 14”.

Final answer

(a) 39 (b) 30 (c) 13.5; a long-run average

Question 4

[5 marks]
For a visitor to a library chosen at random, N is the event that the visitor borrows a novel and D is the event that the visitor borrows a DVD. It is known that
P(N) = 0.45, \quad P(D) = 0.3, \quad P(N \cap D) = 0.12.
(a) Find P(N \cup D).
[2]
(b) Find the probability that the visitor borrows neither a novel nor a DVD.
[1]
(c) Find the probability that the visitor borrows a novel but not a DVD.
[2]
Show solutionHide solution
(a) P(N \cup D) = 0.45 + 0.3 - 0.12
M1
= 0.63
A1
(b) P(N' \cap D') = 1 - 0.63 = 0.37
A1
(c) P(N \cap D') = P(N) - P(N \cap D) = 0.45 - 0.12
M1
= 0.33
A1
Note: FT from (a) in (b). A correct Venn diagram (regions 0.33, 0.12, 0.18, 0.37) earns full marks.

Final answer

(a) 0.63 (b) 0.37 (c) 0.33

Question 5

[5 marks]
A drone has a front camera and a rear camera, which fail independently of each other. On a flight, the front camera works with probability 0.9 and the rear camera works with probability p. The probability that exactly one camera works on a flight is 0.26.
(a) Find p.
[3]
(b) Find the probability that at least one camera works.
[2]
Show solutionHide solution
(a) 0.9(1 - p) + 0.1p = 0.26
M1
0.9 - 0.8p = 0.26
A1
p = 0.8
A1
(b) 1 - P(\text{neither works}) = 1 - 0.1 \times 0.2
M1
= 0.98
A1
Note: 0.9 + 0.8 - 0.9 \times 0.8 = 0.98 is equally good. FT from (a).

Final answer

(a) p=0.8 (b) 0.98

Question 6

[6 marks]
In a board game a counter moves along a track. On each turn the number of squares it moves, X, has the following distribution, where a negative value means a move backwards.
x-2-1013
P(X = x)\frac{1}{12}\frac{1}{6}\frac{1}{4}\frac{1}{3}\frac{1}{6}
(a) Find the exact value of E(X).
[2]
(b) Find the probability that the counter moves forwards on a turn.
[1]
(c) X can never equal E(X). Explain what the value of E(X) tells you about the game.
[2]
(d) Write down the expected position of the counter after 40 turns, relative to its starting square.
[1]
Show solutionHide solution
(a) E(X) = -2\left(\dfrac{1}{12}\right) - 1\left(\dfrac{1}{6}\right) + 0 + 1\left(\dfrac{1}{3}\right) + 3\left(\dfrac{1}{6}\right) = \dfrac{-2 - 2 + 0 + 4 + 6}{12} (M1)
= \dfrac{1}{2}
A1
(b) P(X > 0) = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}
A1
(c) E(X) is the long-run mean of X: over many turns the counter moves forwards on average half a square per turn
R1
it is an average, so it does not have to be one of the values X can take (the moves are whole numbers of squares)
R1
(d) 40 \times \dfrac{1}{2} = 20 squares forwards of the start
A1

Final answer

(a) 1/2 (b) 1/2 (c) average forward move of half a square (d) 20 squares forwards

Question 7

[5 marks]
For a pair of events, the information is P(A) = 0.64, P(B) = 0.4 and P(A \cap B) = 0.16.
(a) Write down, in words, what P(A \mid B) means.
[1]
(b) Find P(A \mid B).
[2]
(c) Find P(B \mid A).
[2]
Show solutionHide solution
(a) the probability that A happens, given that B has happened
A1
(b) P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.16}{0.4} (M1)
= 0.4
A1
(c) P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{0.16}{0.64} (M1)
= 0.25
A1
Note: the event after the bar is the condition, so it gives the denominator; P(A \mid B) \ne P(B \mid A) in general.

Final answer

(a) the probability that A occurs, given that B has occurred (b) 0.4 (c) 0.25

Question 8

[5 marks]
Of the riders who use a city's ride-hailing app, 40\% are students (S). The probability that a rider gives the driver a tip (R) is 0.3 for students and is also 0.3 for riders who are not students. A rider is chosen at random.
(a) Find P(R).
[2]
(b) Hence show that R and S are independent.
[1]
(c) Find P(R \cup S).
[2]
Show solutionHide solution
(a) P(R) = P(S)P(R \mid S) + P(S')P(R \mid S') = 0.4 \times 0.3 + 0.6 \times 0.3 (M1)
= 0.3
A1
(b) P(R \mid S) = P(R) = P(R \mid S') = 0.3, so R and S are independent R1AG
(c) P(R \cap S) = 0.4 \times 0.3 = 0.12; \ P(R \cup S) = 0.3 + 0.4 - 0.12 (M1)
= 0.58
A1

Final answer

(a) 0.3 (b) P(R\mid S)=P(R)=0.3 (c) 0.58

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[5 marks]
A fair four-sided die (faces 1 to 4) and a fair six-sided die (faces 1 to 6) are rolled together. Let D be the positive difference between the two scores (D=0 if the scores are equal).
(a) Complete the sample space diagram showing the value of D for every outcome.
[2]
\begin{array}{c|cccccc} & \text{die 2} & & & & & \\ \text{die 1} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} \\ \hline \text{1} & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} \\ \text{2} & & & & & & \\ \text{3} & & & & & & \\ \text{4} & & & & & & \\ \end{array}
(b) Find P(D=2).
[1]
(c) Find P(D\geq 3).
[2]
Show solutionHide solution
(a) Each entry is the positive difference of the row and column scores.
[1]
\begin{array}{c|cccccc} & \text{six-sided die} & & & & & \\ \text{four-sided die} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} \\ \hline \text{1} & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} \\ \text{2} & \text{1} & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} \\ \text{3} & \text{2} & \text{1} & \text{0} & \text{1} & \text{2} & \text{3} \\ \text{4} & \text{3} & \text{2} & \text{1} & \text{0} & \text{1} & \text{2} \\ \end{array}
All 24 entries correct.
[1]
(b) D=2 occurs 6 times out of 24, so P(D=2)=\dfrac{6}{24}=\dfrac14
[1]
(c) D=3 occurs 4 times, D=4 twice, D=5 once: 7 outcomes
[1]
P(D\geq3)=\dfrac{7}{24}
[1]

Final answer

(a) Rows: 1,0,1,2,3,4; 2,1,0,1,2,3; 3,2,1,0,1,2 (b) 1/4 (c) 7/24

Question 10

[4 marks]
In a survey of 150 cyclists, 82 own a bell, 71 own a pump and 14 own neither.
(a) Use the addition rule (in terms of numbers of cyclists) to find how many own both a bell and a pump.
[3]
(b) A cyclist is chosen at random from the survey. Find the probability that the cyclist owns both items.
[1]
Show solutionHide solution
(a) Cyclists owning a bell or a pump: 150-14=136
[1]
n(B\cup P)=n(B)+n(P)-n(B\cap P), so 136=82+71-n(B\cap P)
[1]
n(B\cap P)=153-136=17
[1]
(b) \dfrac{17}{150}
[1]

Final answer

(a) 17 (b) 17/150

Question 11

[5 marks]
Events A and B are independent, with P(B)=0.2 and P(A\cup B)=0.68.
(a) Find P(A).
[3]
(b) Find P(A\cap B).
[1]
(c) Find the probability that neither A nor B occurs.
[1]
Show solutionHide solution
(a) Independence: P(A\cap B)=0.2\,P(A).
[1]
0.68=P(A)+0.2-0.2\,P(A)
[1]
0.8\,P(A)=0.48\ \Rightarrow\ P(A)=0.6
[1]
(b) P(A\cap B)=0.6\times0.2=0.12
[1]
(c) 1-P(A\cup B)=1-0.68=0.32
[1]

Final answer

(a) 0.6 (b) 0.12 (c) 0.32

Question 12

[5 marks]
A fruit bowl contains 6 apples and 4 pears. Three pieces of fruit are taken from the bowl at random, without replacement.
(a) Find the probability that all three pieces are apples.
[2]
(b) Find the probability that at least one piece is a pear.
[1]
(c) Find the probability that exactly two of the pieces are apples.
[2]
Show solutionHide solution
(a) P(AAA)=\dfrac6{10}\times\dfrac59\times\dfrac48
[1]
=\dfrac{120}{720}=\dfrac16
[1]
(b) 1-\dfrac16=\dfrac56
[1]
(c) Each of the orders AAP, APA, PAA has probability \dfrac{6\times5\times4}{10\times9\times8}=\dfrac16 (the numerators are the same product).
[1]
P(\text{exactly two apples})=3\times\dfrac16=\dfrac12
[1]

Final answer

(a) 1/6 (b) 5/6 (c) 1/2

Question 13

[4 marks]
The discrete random variable X has probability distribution P(X=x)=kx^2 for x=1,2,3, where k is a constant.
(a) Find the value of k.
[2]
(b) Find P(X<3).
[2]
Show solutionHide solution
(a) k(1^2+2^2+3^2)=1
[1]
14k=1\ \Rightarrow\ k=\dfrac1{14}
[1]
(b) P(X<3)=P(X=1)+P(X=2)=\dfrac1{14}+\dfrac4{14}
[1]
=\dfrac5{14}
[1]

Final answer

(a) 1/14 (b) 5/14

Question 14

[8 marks]
The four possible scores of X and their probabilities are listed below, where k is a constant.
\begin{array}{|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & 0.2 & k^2 & k & 0.24 \\ \hline \end{array}
(a) Find the value of k, giving a reason for rejecting any other solution.
[5]
(b) Find P(X<3\mid X>1).
[3]
Show solutionHide solution
(a) 0.2+k^2+k+0.24=1
[1]
k^2+k-0.56=0
[1]
(k-0.4)(k+1.4)=0
[1]
k=0.4 or k=-1.4
[1]
k=-1.4 would make P(X=3)=k negative, which is impossible, so k=0.4.
[1]
(b) The distribution is 0.2,\ 0.16,\ 0.4,\ 0.24. “X<3 and X>1” means X=2, so the numerator is 0.16.
[1]
P(X>1)=1-0.2=0.8
[1]
P(X<3\mid X>1)=\dfrac{0.16}{0.8}=0.2
[1]

Final answer

(a) k=0.4; reject -1.4 (b) 0.2

Question 15

[5 marks]
The table lists the probabilities of all four values of X. Two entries, p and q, are unknown.
\begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline P(X=x) & p & q & 0.4 & 0.05 \\ \hline \end{array}
It is given that \mathrm{E}(X)=1.25. Determine the two missing entries p and q.
[5]
Show solutionHide solution
Probabilities sum to 1: p+q+0.4+0.05=1, so p+q=0.55
[2]
\mathrm{E}(X)=0p+1q+2(0.4)+3(0.05)=q+0.95
[1]
q+0.95=1.25\ \Rightarrow\ q=0.3
[1]
p=0.55-0.3=0.25
[1]

Final answer

p=0.25, q=0.3

Question 16

[5 marks]
In a game, a fair six-sided die is rolled. If a 6 is rolled the player wins a dollars. If a 4 or a 5 is rolled the player wins 2 dollars. If a 1, 2 or 3 is rolled the player loses 3 dollars.
(a) Find an expression, in terms of a, for the expected gain of the player.
[3]
(b) The game is fair. Find the value of a.
[2]
Show solutionHide solution
(a) P(6)=\dfrac16, P(4\text{ or }5)=\dfrac26, P(1,2\text{ or }3)=\dfrac36.
[1]
\mathrm{E}(\text{gain})=a\left(\dfrac16\right)+2\left(\dfrac26\right)+(-3)\left(\dfrac36\right)
[1]
=\dfrac a6+\dfrac46-\dfrac96=\dfrac{a-5}6
[1]
(b) Fair: \dfrac{a-5}{6}=0
[1]
a=5
[1]

Final answer

(a) (a-5)/6 dollars (b) a=5

Question 17

[9 marks]
A bag contains 3 red counters and 2 green counters. In a game, two counters are drawn at random without replacement. The player wins 4 dollars if both counters are red, wins 3 dollars if both are green, and loses a dollars if the counters are different colours.
(a) Find the probability that the two counters are (i) both red, (ii) both green, (iii) different colours.
[3]
(b) Show that the expected gain for the player is \left(1.5-0.6a\right) dollars.
[2]
(c) Find the value of a for which the game is fair.
[2]
The game is changed so that a=3.
(d) Find the expected total gain of a player who plays the game 50 times.
[2]
Show solutionHide solution
(a) (i) P(RR)=\dfrac35\times\dfrac24=\dfrac3{10}
[1]
(ii) P(GG)=\dfrac25\times\dfrac14=\dfrac1{10}
[1]
(iii) 1-\dfrac3{10}-\dfrac1{10}=\dfrac6{10}=\dfrac35
[1]
(b) \mathrm{E}(\text{gain})=4\left(\dfrac3{10}\right)+3\left(\dfrac1{10}\right)-a\left(\dfrac6{10}\right)
[1]
=1.2+0.3-0.6a=1.5-0.6a
AG[1]
(c) Fair: 1.5-0.6a=0
[1]
a=2.5, i.e. \$2.50
[1]
(d) With a=3, the expected gain per game is 1.5-1.8=-\$0.30
[1]
Over 50 games: 50\times(-0.30)=-\$15 (an expected loss of \$15)
[1]

Final answer

(a) 3/10, 1/10, 3/5 (b) 1.5-0.6a dollars (c) a=2.5 (d) -15 dollars

Question 18

[6 marks]
The following conditions hold for A and B: P(A\mid B)=\dfrac13, P(B\mid A)=\dfrac12 and P(A\cup B)=\dfrac23.
(a) Find P(A\cap B).
[4]
(b) Determine whether A and B are independent.
[2]
Show solutionHide solution
(a) Let t=P(A\cap B). From P(A\cap B)=P(B)P(A\mid B): t=\dfrac13P(B), so P(B)=3t.
[1]
From t=P(A)P(B\mid A): t=\dfrac12P(A), so P(A)=2t.
[1]
Addition rule: \dfrac23=2t+3t-t=4t
[1]
t=\dfrac16
[1]
(b) P(A)=\dfrac13 and P(B)=\dfrac12, so P(A)P(B)=\dfrac16
[1]
This equals P(A\cap B), so A and B are independent.
[1]

Final answer

(a) 1/6 (b) independent: P(A)P(B)=1/6

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[10 marks]
A repair bench has four sensors: one works and three are faulty. Nia tests them in a random order without replacement and stops at the working sensor. Let T be the number of tests.
(a) Construct the probability distribution of T.
[4]
(b) Find \operatorname{E}(T).
[2]
(c) Given that more than one test is needed, find P(T=4\mid T>1).
[2]
(d) Each test costs 6 credits. Find the expected total testing cost.
[2]
Show solutionHide solution
(a) P(T=1)=1/4; P(T=2)=(3/4)(1/3)=1/4; P(T=3)=(3/4)(2/3)(1/2)=1/4; P(T=4)=(3/4)(2/3)(1/2)(1)=1/4. Each correct entry A1.
(b) E(T)=(1+2+3+4)/4=5/2 (M1 A1).
(c) (1/4)/(3/4)=1/3 (M1 A1).
(d) Cost values are 6,12,18,24 with probability 1/4 each; expected cost =(6+12+18+24)/4=15 credits (M1 A1).

Final answer

(a) T=1,2,3,4: each 1/4 (b) 5/2 (c) 1/3 (d) 15 credits

Question 20

[6 marks]
The discrete random variable X has probability distribution P(X=x)=\dfrac kx for x\in\{1,2,3,6\}.
(a) Find the value of k.
[2]
(b) Find \mathrm{E}(X).
[3]
A player pays c dollars to play a game in which they receive X dollars.
(c) Find the value of c for which the game is fair.
[1]
Show solutionHide solution
(a) k\left(1+\dfrac12+\dfrac13+\dfrac16\right)=1
[1]
2k=1\ \Rightarrow\ k=\dfrac12
[1]
(b) Probabilities: \dfrac12,\dfrac14,\dfrac16,\dfrac1{12}.
[1]
\mathrm{E}(X)=1\left(\dfrac12\right)+2\left(\dfrac14\right)+3\left(\dfrac16\right)+6\left(\dfrac1{12}\right)
[1]
=\dfrac12+\dfrac12+\dfrac12+\dfrac12=2
[1]
(c) Fair when c=\mathrm{E}(X)=2, i.e. \$2.
[1]

Final answer

(a) 1/2 (b) 2 (c) 2 dollars

Question 21

[6 marks]
A fair three-sided spinner has sectors numbered 1, 2 and 3, and a fair five-sided spinner has sectors numbered 1 to 5. Both are spun and the two scores are multiplied. Let E be the event “the product is even” and let G be the event “the product is greater than 6”.
(a) Find P(E).
[2]
(b) Find P(G).
[1]
(c) Find P(G\mid E).
[2]
(d) State, with a reason, whether E and G are independent.
[1]
Show solutionHide solution
There are 3\times5=15 equally likely outcomes.
(a) The product is odd only when both scores are odd: 2\times3=6 outcomes.
[1]
P(E)=1-\dfrac6{15}=\dfrac9{15}=\dfrac35
[1]
(b) Products greater than 6: (2,4)=8, (2,5)=10, (3,3)=9, (3,4)=12, (3,5)=15, five outcomes, so P(G)=\dfrac5{15}=\dfrac13
[1]
(c) Even products greater than 6: 8, 10 and 12, so 3 outcomes, out of the 9 even outcomes.
[1]
P(G\mid E)=\dfrac39=\dfrac13
[1]
(d) P(G\mid E)=P(G)=\dfrac13, so E and G are independent.
[1]

Final answer

(a) 3/5 (b) 1/3 (c) 1/3 (d) independent

Question 22

[7 marks]
A bag contains 2 green balls and 4 red balls. Two balls are selected at random, one after the other, and neither is returned. Let X be the number of green balls drawn.
(a) Find the probability distribution of X, presenting it in a table.
[3]
(b) Find \mathrm{E}(X).
[2]
A player pays c dollars to play, and receives 3 dollars for each green ball drawn.
(c) Find the value of c for which the game is fair.
[2]
Show solutionHide solution
(a) P(X=0)=\dfrac46\times\dfrac35=\dfrac25
[1]
P(X=2)=\dfrac26\times\dfrac15=\dfrac1{15}
[1]
P(X=1)=1-\dfrac25-\dfrac1{15}=\dfrac8{15} (or 2\times\dfrac46\times\dfrac25)
[1]
\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline P(X=x) & \tfrac25 & \tfrac8{15} & \tfrac1{15} \\ \hline \end{array}
(b) \mathrm{E}(X)=0\left(\dfrac25\right)+1\left(\dfrac8{15}\right)+2\left(\dfrac1{15}\right)
[1]
=\dfrac{10}{15}=\dfrac23
[1]
(c) Expected receipt =3\,\mathrm{E}(X)=3\times\dfrac23=2
[1]
Fair when c=\$2.
[1]

Final answer

(a) x=0,1,2: 2/5,8/15,1/15 (b) 2/3 (c) 2 dollars

Question 23

[10 marks]
(a) Events A and B satisfy 0<P(B)<1 and P(A\mid B)=P(A\mid B'). Prove that A and B are independent.
[4]
(b) Events C and D are independent, with P(C\cap D')=\tfrac14 and P(C'\cap D)=\tfrac16. Find the two possible values of P(C\cup D).
[6]
Show solutionHide solution
(a) Let b=P(B) and t=P(A\cap B). Then P(A\cap B')=P(A)-t.
[1]
\dfrac{t}{b}=\dfrac{P(A)-t}{1-b}
[1]
t(1-b)=b\bigl(P(A)-t\bigr), so t-tb=bP(A)-bt
[1]
t=b\,P(A), that is P(A\cap B)=P(A)P(B), so A and B are independent.
[1]
(b) Let c=P(C) and d=P(D). Then P(C\cap D')=c-cd=c(1-d) and P(C'\cap D)=d-cd=(1-c)d (using P(C\cap D)=cd).
[1]
So c(1-d)=\dfrac14 and (1-c)d=\dfrac16. Subtracting: c-d=\dfrac14-\dfrac16=\dfrac1{12}, so c=d+\dfrac1{12}
[1]
\left(d+\dfrac1{12}\right)(1-d)=\dfrac14\ \Rightarrow\ d+\dfrac1{12}-d^2-\dfrac d{12}=\dfrac14, which gives 12d^2-11d+2=0
[2]
(3d-2)(4d-1)=0, so d=\dfrac23 (then c=\dfrac34) or d=\dfrac14 (then c=\dfrac13). Both pairs lie in [0,1].
[1]
P(C\cup D)=c+d-cd: for \left(\dfrac34,\dfrac23\right) it is \dfrac34+\dfrac23-\dfrac12=\dfrac{11}{12}; for \left(\dfrac13,\dfrac14\right) it is \dfrac13+\dfrac14-\dfrac1{12}=\dfrac12.
[1]

Final answer

(a) P(A\cap B)=P(A)P(B) (b) 1/2 or 11/12

Question 24

[8 marks]
A bag contains 3 red counters and n blue counters, where n\geq1. Two counters are drawn at random without replacement. Let q be the probability that the two counters are different colours.
(a) Show that q=\dfrac{6n}{(n+2)(n+3)}.
[3]
(b) Use your GDC to find the two values of n for which q is greatest, and write down this greatest value of q.
[3]
(c) Describe, with a reason in terms of the counters, what happens to q as n becomes very large.
[2]
Show solutionHide solution
(a) There are n+3 counters. P(RB)=\dfrac3{n+3}\times\dfrac n{n+2} and P(BR)=\dfrac n{n+3}\times\dfrac3{n+2}
[2]
q=2\times\dfrac{3n}{(n+3)(n+2)}=\dfrac{6n}{(n+2)(n+3)}
AG[1]
(b) Tabulate (or graph) y=\dfrac{6x}{(x+2)(x+3)} on the GDC for whole x:
[1]
\begin{array}{c|ccccccc} n & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7} \\ \hline q & 0.500 & 0.600 & 0.600 & 0.571 & 0.536 & 0.500 & 0.467 \\ \end{array}
The greatest values occur at n=2 (\frac{12}{20}) and n=3 (\frac{18}{30}).
[1]
Greatest value of q=0.6.
[1]
(c) q=\dfrac{6n}{n^2+5n+6}=\dfrac{6}{n+5+6/n}, which tends to 0 as n\to\infty.
[1]
With very many blue counters, the two counters drawn are almost certainly both blue, so a mixed pair becomes very unlikely.
[1]

Final answer

(a) 6n/((n+2)(n+3)) (b) n=2 or 3, q=0.600 (c) tends to 0: almost all counters are blue

Question 25

[13 marks]
Of the 80 members of a climbing club, 40 do lead climbing (L) and 30 do bouldering (B). The events L and B are independent.
(a) Show that 15 members do both.
[3]
(b) Draw a Venn diagram showing the number of members in each region.
[2]
A member is chosen at random.
(c) Find P(B\mid L').
[2]
Two members are chosen at random, without replacement.
(d) Find the probability that neither of them does lead climbing or bouldering.
[3]
(e) Find the probability that at least one of them does lead climbing.
[3]
Show solutionHide solution
(a) P(L)=\dfrac{40}{80}=\dfrac12 and P(B)=\dfrac{30}{80}=\dfrac38
[1]
Independence: P(L\cap B)=\dfrac12\times\dfrac38=\dfrac3{16}
[1]
n(L\cap B)=80\times\dfrac3{16}=15
AG[1]
(b) Only L: 40-15=25; only B: 30-15=15; neither: 80-(25+15+15)=25.
[1]
The Venn regions are L only 25, overlap 15, B only 15, outside 25.
Diagram with all four regions correct.
[1]
(c) P(B\mid L')=\dfrac{n(L'\cap B)}{n(L')}=\dfrac{15}{40}
[1]
=\dfrac38 (equal to P(B), as expected from independence)
[1]
(d) Neither: 25 members, so P(\text{both in neither})=\dfrac{25}{80}\times\dfrac{24}{79}
[2]
=\dfrac{600}{6320}=\dfrac{15}{158}
[1]
(e) 1-P(\text{neither does lead climbing}); 40 of the 80 do not lead climb
[1]
P(\text{neither leads})=\dfrac{40}{80}\times\dfrac{39}{79}=\dfrac{39}{158}
[1]
1-\dfrac{39}{158}=\dfrac{119}{158}
[1]

Final answer

(a) 15 (b) L only 25, both 15, B only 15, neither 25 (c) 3/8 (d) 15/158 (e) 119/158

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 The birthday problem

G1 Worked example

[9 marks]
Assume that a person's birthday is equally likely to be any of 365 days, independently of everyone else (ignore 29 February).
(a) Show that the probability that four people have four different birthdays is \dfrac{364\times363\times362}{365^3}, and give its value to three significant figures.
[2]
(b) Write down an expression for p_n, the probability that n people all have different birthdays. Use your GDC to find the least n for which a shared birthday is more likely than not.
[3]
(c) Find the least n for which it is more likely than not that at least one of n other people has the same birthday as you.
[3]
(d) Explain briefly why the answers to (b) and (c) are so different.
[1]
Show solutionHide solution
(a) the first person's birthday can be any day; the second must avoid 1 day, the third 2, the fourth 3: 1\times\dfrac{364}{365}\times\dfrac{363}{365}\times\dfrac{362}{365}
M1
=\dfrac{364\times363\times362}{365^3}=0.984
A1
(b) p_n=\dfrac{365\times364\times\cdots\times(366-n)}{365^n}
A1
a shared birthday is more likely than not when p_n<0.5
M1
p_{22}=0.524, p_{23}=0.493, so n=23
A1
(c) P(\text{no one shares your birthday})=\left(\dfrac{364}{365}\right)^n, and we need 1-\left(\dfrac{364}{365}\right)^n>0.5
M1
n\ln\dfrac{364}{365}<\ln0.5, so n>\dfrac{\ln0.5}{\ln(364/365)}=252.65\ldots (the inequality turns because \ln\frac{364}{365}<0)
A1
n=253
A1
(d) in (b) any two of the n people can match, and 23 people form 1+2+\cdots+22=253 pairs; in (c) only the n pairs that include you count
R1

Final answer

  • (a) 0.984
  • (b) n=23
  • (c) n=253

G1 Investigation

[12 marks]
In a year of N equally likely days, let p_n=\left(1-\dfrac1N\right)\left(1-\dfrac2N\right)\cdots\left(1-\dfrac{n-1}{N}\right) be the probability that n people all have different birthdays. You may assume that \ln(1-x)<-x for 0<x<1.
(a) For N=100, use your GDC to find the least n with p_n<\frac12.
[2]
(b) Show that p_n<\mathrm e^{-n(n-1)/(2N)} for 2\le n\le N.
[3]
(c) Hence show that p_n<\frac12 whenever n(n-1)\ge2N\ln2. Find the least n that this condition gives when N=100 and when N=365, and compare each with the exact answer.
[4]
(d) A website gives each new account a code chosen at random from 10^6 possible codes. Use (c) to find a number of accounts for which a repeated code is more likely than not. Show that, for large N, the condition in (c) gives n\approx\sqrt{2N\ln2}.
[3]
Show solutionHide solution
(a) p_{12}=0.503, p_{13}=0.443
M1
n=13
A1
(b) \ln p_n=\displaystyle\sum_{k=1}^{n-1}\ln\left(1-\frac kN\right)
M1
<-\displaystyle\sum_{k=1}^{n-1}\frac kN=-\frac{n(n-1)}{2N}, using the given inequality with x=\frac kN and the arithmetic series 1+2+\cdots+(n-1)
A1
\ln is increasing, so p_n<\mathrm e^{-n(n-1)/(2N)}
R1
(c) if n(n-1)\ge2N\ln2 then \dfrac{n(n-1)}{2N}\ge\ln2, so p_n<\mathrm e^{-\ln2}=\frac12
M1A1
N=100: 2N\ln2=138.6; 12\times11=132, 13\times12=156, so n=13, the exact answer
A1
N=365: 2N\ln2=505.997; 22\times21=462, 23\times22=506, so n=23, again exact (by a margin of 0.003)
A1
(d) 2N\ln2=1\,386\,294.4; 1177\times1176=1\,384\,152, 1178\times1177=1\,386\,506, so 1178 accounts are enough
A1
for large N, the least such n is large, and n(n-1)\approx n^2
M1
so n^2\approx2N\ln2 and n\approx\sqrt{2N\ln2} (\approx1177.4 for N=10^6)
A1

Final answer

  • (a) 13
  • (c) 13 and 23, both exact
  • (d) 1178

G2 Fair games and ruin

G2 Worked example

[9 marks]
A player starts with k counters, where 0<k<4. In each round she wins one counter with probability 0.6 or loses one with probability 0.4, independently of earlier rounds. Play stops when she has 0 counters or 4 counters. Let p_k be the probability that she reaches 4 counters, so p_0=0 and p_4=1.
(a) Explain why p_k=0.6\,p_{k+1}+0.4\,p_{k-1} for k=1,2,3.
[2]
(b) Let d_k=p_{k+1}-p_k. Show that d_k=\frac23d_{k-1}.
[2]
(c) Hence find p_1, p_2 and p_3.
[4]
(d) Compare p_2 with its value when each round is won with probability 0.5.
[1]
Show solutionHide solution
(a) condition on the first round: with probability 0.6 she then has k+1 counters, with probability 0.4 she has k-1
M1
from there the game is the same game started again, so by total probability p_k=0.6\,p_{k+1}+0.4\,p_{k-1}
R1
(b) write p_k=0.6\,p_k+0.4\,p_k: then 0.6(p_{k+1}-p_k)=0.4(p_k-p_{k-1})
M1
d_k=\dfrac{0.4}{0.6}d_{k-1}=\dfrac23d_{k-1}
A1AG
(c) d_0+d_1+d_2+d_3=p_4-p_0=1, and the d_k form a geometric sequence with ratio \frac23
M1
d_0\left(1+\frac23+\frac49+\frac8{27}\right)=d_0\times\dfrac{65}{27}=1, so d_0=\dfrac{27}{65}=p_1
A1
p_2=\dfrac{27+18}{65}=\dfrac{9}{13}
A1
p_3=\dfrac{45+12}{65}=\dfrac{57}{65}
A1
(d) with probability 0.5 the differences are all equal, so p_2=\frac12; the edge of 0.6 per round lifts this to \frac{9}{13}\approx0.692
A1

Final answer

  • (c) p_1=\frac{27}{65}, p_2=\frac{9}{13}, p_3=\frac{57}{65}
  • (d) \frac12

G2 Investigation

[13 marks]
In this investigation each round is won or lost with probability \frac12. A player starts with k counters and plays until she has 0 or N counters.
(a) Let N=3 and k=1. Draw a tree for the first two rounds, and show that the probability of reaching 3 counters is \frac14+\frac14\times\frac14+\left(\frac14\right)^2\times\frac14+\cdots. Find this probability.
[2]
(b) Let e_k be the expected number of rounds the game lasts when the player starts with k counters, so e_0=e_N=0. You may assume that e_k=1+\frac12e_{k-1}+\frac12e_{k+1} for 0<k<N. For N=3, find e_1 and e_2.
[2]
(c) For N=4, use the symmetry e_1=e_3 to find e_1, e_2 and e_3. Conjecture a formula for e_k in terms of k and N.
[3]
(d) Let g_k=e_{k+1}-e_k. Show that g_k=g_{k-1}-2, and hence prove your conjecture.
[4]
(e) A player with 10 counters plays against an opponent with 90 until one of them has none. Find the expected number of rounds, and the probability that the player who starts with 10 is the one who ends with none.
[2]
Show solutionHide solution
(a) from 1: lose (end at 0), or win to 2; from 2: win (end at 3) or lose back to 1. So each visit to 1 leads to 3 within two rounds with probability \frac14 and back to 1 with probability \frac14
M1
P=\dfrac14\left(1+\dfrac14+\dfrac1{16}+\cdots\right)=\dfrac{1/4}{1-1/4}=\dfrac13
A1
(b) e_1=1+\frac12e_2 and e_2=1+\frac12e_1
M1
e_1=e_2=2
A1
(c) Symmetry gives e_1=e_3. Then e_1=1+\frac12e_2 and e_2=1+e_1
M1
e_1=3, e_2=4, e_3=3
A1
1\times2=2, 2\times1=2; 1\times3=3, 2\times2=4: conjecture e_k=k(N-k)
A1
(d) multiply the given equation by 2: 2e_k=2+e_{k-1}+e_{k+1}, so e_{k+1}-e_k=(e_k-e_{k-1})-2, that is g_k=g_{k-1}-2
M1A1
so g_0,g_1,\dots is arithmetic with difference -2: g_j=e_1-2j. Adding, e_N-e_0=\displaystyle\sum_{j=0}^{N-1}(e_1-2j)=Ne_1-N(N-1)=0, so e_1=N-1
M1
e_k=\displaystyle\sum_{j=0}^{k-1}(N-1-2j)=k(N-1)-k(k-1)=k(N-k)
A1
(e) N=100, k=10: e_{10}=10\times90=900 rounds
A1
P(\text{reach }100)=\frac{10}{100}, so the probability of ending with none is 0.9
A1

Final answer

  • (a) \frac13
  • (b) 2, 2
  • (c) 3,4,3; e_k=k(N-k)
  • (e) 900, 0.9

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. The TI-84 Plus CE keystrokes for the common calculator jobs are listed in Chapter 12.

From the book

Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma

Chapter 8 teaches probability from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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