Free worked solutions · 2026 Edition

Chapter 13: Probability worked solutions

Worked solutions to the practice questions in Chapter 13 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (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

[7 marks]

Events A and B are independent. P(B) = P(A) + 0.2 and P(A \cap B) = 0.15.

(a) Let P(A) = x. Find x.

[4]

(b) Find P(A \cup B).

[2]

(c) Find the probability that exactly one of A and B occurs.

[1]

Show solutionHide solution

(a) x(x + 0.2) = 0.15

x^2 + 0.2x - 0.15 = 0

(x + 0.5)(x - 0.3) = 0, \ x = 0.3 or x = -0.5

x = 0.3, because a probability cannot be negative

(b) P(B) = 0.5; \ P(A \cup B) = 0.3 + 0.5 - 0.15

= 0.65

(c) 0.65 - 0.15 = 0.5

Note: in (c) accept 0.3 \times 0.5 + 0.7 \times 0.5 = 0.5 from the tree. FT from (a).

Final answer

(a) x=0.3
(b) 0.65
(c) 0.5

Question 2

[8 marks]

At a language school, G is the event that a randomly chosen student learns German and K is the event that the student learns Korean. One third of the Korean learners also learn German, half of the German learners also learn Korean, and 80\% of the students learn at least one of these two languages. Let P(G \cap K) = k.

(a) Write P(G) and P(K) in terms of k.

[2]

(b) Find the value of k.

[2]

(c) Determine whether G and K are independent.

[2]

(d) Find the probability that a student learns German, given that the student does not learn Korean.

[2]

Show solutionHide solution

(a) P(G \mid K) = \dfrac{k}{P(K)} = \dfrac{1}{3} \Rightarrow P(K) = 3k

P(K \mid G) = \dfrac{k}{P(G)} = \dfrac{1}{2} \Rightarrow P(G) = 2k

(b) 2k + 3k - k = 0.8

k = 0.2

(c) P(G)P(K) = 0.4 \times 0.6 = 0.24

0.24 \ne 0.2 = P(G \cap K), so G and K are not independent

(d) P(G \mid K') = \dfrac{P(G) - k}{1 - P(K)} = \dfrac{0.4 - 0.2}{0.4}

= 0.5

Note: FT from (b) in (c) and (d).

Final answer

(a) P(G)=2k, P(K)=3k
(b) k=0.2
(c) not independent: P(G)P(K)=0.24\ne0.2
(d) 0.5

Question 3

[7 marks]

A bag contains n red counters and 5 blue counters. Two counters are taken at random without replacement. The probability that both are blue is \dfrac{5}{33}.

(a) Show that (n + 5)(n + 4) = 132.

[3]

(b) Find the value of n.

[2]

(c) Find the probability that both counters are red.

[2]

Show solutionHide solution

(a) P(\text{both blue}) = \dfrac{5}{n + 5} \times \dfrac{4}{n + 4}

\dfrac{20}{(n + 5)(n + 4)} = \dfrac{5}{33}

(n + 5)(n + 4) = \dfrac{20 \times 33}{5}

= 132

(b) n^2 + 9n - 112 = 0, \ (n + 16)(n - 7) = 0, so n = 7 or n = -16

n = 7, since a number of counters cannot be negative

(c) \dfrac{7}{12} \times \dfrac{6}{11}

= \dfrac{42}{132} = \dfrac{7}{22}

Final answer

(b) n=7
(c) \frac{7}{22}

Question 4

[4 marks]

X \sim \mathrm{B}(n,\ p) with \mathrm{E}(X) = 12 and \mathrm{Var}(X) = 7.2.

(a) Find p.

[3]

(b) Find n.

[1]

Show solutionHide solution

(a) np = 12 and np(1 - p) = 7.2

dividing: 1 - p = \dfrac{7.2}{12} = 0.6

p = 0.4

(b) n = \dfrac{12}{0.4} = 30

Final answer

(a) p=0.4
(b) n=30

Question 5

[10 marks]

A mountain cable car is closed on any day of thick fog. During the winter each day, independently, has thick fog with probability 0.23. Lena stays at the resort for 14 days. Let F be the number of days of her stay on which the cable car is closed.

(a) Find the expected number of days on which the cable car is closed.

[1]

(b) Find the probability that the cable car is closed on more than 4 days.

[2]

(c) Given that the cable car is closed on at least 2 days, find the probability that it is closed on at most 4 days.

[4]

(d) Given that the cable car is closed on the first day of her stay, find the probability that it is closed on exactly 3 days in total.

[3]

Show solutionHide solution

(a) 14 \times 0.23 = 3.22

(b) F \sim B(14,\ 0.23), \ P(F \ge 5) = 1 - P(F \le 4) (M1)

= 0.202\ (0.202336\ldots)

(c) P(F \le 4 \mid F \ge 2) = \dfrac{P(2 \le F \le 4)}{P(F \ge 2)}

P(2 \le F \le 4) = 0.664\ (0.664202\ldots), \ P(F \ge 2) = 0.867\ (0.866539\ldots) A1A1

= 0.767\ (0.766500\ldots)

(d) the other 13 days are independent of the first, so she needs exactly 2 closed days among them

Y \sim B(13,\ 0.23), \ P(Y = 2) (M1)

= 0.233\ (0.232781\ldots)

Note: in (c) the numerator is P(F \le 4 \cap F \ge 2) = P(2 \le F \le 4), not P(F \le 4). In (d) accept \dfrac{P(\text{first closed} \cap F = 3)}{0.23} = \dfrac{0.23 \times P(Y = 2)}{0.23}.

Final answer

(a) 3.22
(b) 0.202
(c) 0.767
(d) 0.233

Question 6

[8 marks]

The depths, D metres, of the feeding dives made by the penguins of one colony are normally distributed with mean 64 m and standard deviation 12 m.

(a) Find P(D > 80).

[2]

(b) A dive is deeper than 50 m. Find the probability that it is deeper than 80 m.

[3]

(c) Find P(D < 70 \mid D > 64).

[3]

Show solutionHide solution

(a) D \sim \text{N}(64,\ 12^2), P(D > 80) (M1)

= 0.0912\ (0.0912112\ldots)

(b) P(D > 80 \mid D > 50) = \dfrac{P(D > 80 \cap D > 50)}{P(D > 50)} = \dfrac{P(D > 80)}{P(D > 50)} (M1)

P(D > 50) = 0.878\ (0.878327\ldots) (A1)

= \dfrac{0.0912112\ldots}{0.878327\ldots} = 0.104\ (0.103846\ldots)

(c) P(D > 64) = 0.5 (the mean) and \{D < 70\} \cap \{D > 64\} = \{64 < D < 70\} (M1)

P(64 < D < 70) = 0.191\ (0.191462\ldots) (A1)

\dfrac{0.191462\ldots}{0.5} = 0.383\ (0.382924\ldots)

Note: the denominator is P(D > 64) = 0.5; dividing by 2 instead of by 0.5 earns A0.

Final answer

(a) 0.0912
(b) 0.104
(c) 0.383

Question 7

[8 marks]

The length, L cm, of the fronds of a seaweed growing on a rocky shore is normally distributed with mean \mu and standard deviation \sigma. A marine biologist finds that P(L > \mu + 5) = 0.27 and P(L < 30) = 0.07.

(a) Find the value of \sigma.

[3]

(b) Find the value of \mu.

[3]

(c) Find the probability that a frond is longer than 50 cm.

[2]

Show solutionHide solution

(a) the standardized value of \mu + 5 is \dfrac{5}{\sigma}, and P\left(Z > \dfrac{5}{\sigma}\right) = 0.27

\dfrac{5}{\sigma} = 0.612812\ldots

\sigma = 8.16\ (8.15909\ldots) cm

(b) P(L < 30) = 0.07, standardized value -1.47579\ldots (A1)

\dfrac{30 - \mu}{8.15909\ldots} = -1.47579\ldots

\mu = 42.0\ (42.0411\ldots) cm

(c) L \sim \text{N}(42.0411\ldots,\ 8.15909\ldots^2), P(L > 50) (M1)

= 0.165\ (0.164665\ldots)

Final answer

(a) \sigma=8.16
(b) \mu=42.0
(c) 0.165

Question 8

[7 marks]

A continuous random variable X has probability density function

f(x) = \begin{cases} k\left( 8 - x^3 \right), & 0 \le x \le 2 \\ 0, & \text{otherwise} \end{cases}

where k is a constant.

(a) Show that k = \dfrac{1}{12}.

[3]

(b) Find P(X < 1).

[2]

(c) Write down P(X \ge 1).

[1]

(d) Explain why P(X \le 1) = P(X < 1).

[1]

Show solutionHide solution

(a) \displaystyle\int_{0}^{2} k\left( 8 - x^3 \right) dx = 1

k\left[ 8x - \dfrac{x^4}{4} \right]_{0}^{2} = k(16 - 4) = 12k

12k = 1 \Rightarrow k = \dfrac{1}{12} A1AG

(b) \dfrac{1}{12}\displaystyle\int_{0}^{1} \left( 8 - x^3 \right) dx

= \dfrac{1}{12}\left( 8 - \dfrac{1}{4} \right) = \dfrac{31}{48}

(c) 1 - \dfrac{31}{48} = \dfrac{17}{48}

(d) for a continuous random variable P(X = 1) = 0, so including the end point adds nothing

Final answer

(b) \frac{31}{48}
(c) \frac{17}{48}
(d) P(X=1)=0 for a continuous variable

Question 9

[7 marks]

A continuous random variable X has probability density function

f(x) = \begin{cases} \dfrac{x + 1}{4}, & 0 \le x \le 2 \\ 0, & \text{otherwise.} \end{cases}

(a) Show that f is a valid probability density function.

[2]

(b) Write down the mode of X, giving a reason.

[1]

(c) Show that the median m satisfies m^2 + 2m - 4 = 0.

[2]

(d) Hence find the exact value of m.

[2]

Show solutionHide solution

(a) \displaystyle\int_{0}^{2} \dfrac{x + 1}{4}\,dx = \dfrac{1}{4}\left[ \dfrac{x^2}{2} + x \right]_{0}^{2} = \dfrac{1}{4}(2 + 2) = 1

f(x) \ge \dfrac{1}{4} > 0 on 0 \le x \le 2 and the area is 1, so f is valid

(b) f is increasing, so the mode is the right END POINT, x = 2

(c) \displaystyle\int_{0}^{m} \dfrac{x + 1}{4}\,dx = \dfrac{1}{2}

\dfrac{1}{4}\left( \dfrac{m^2}{2} + m \right) = \dfrac{1}{2} \Rightarrow m^2 + 2m - 4 = 0 A1AG

(d) m = \dfrac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5}

m = \sqrt{5} - 1; reject -1 - \sqrt{5} BECAUSE it is outside 0 \le x \le 2

Final answer

(b) mode 2 (the greatest value of f is at the end point)
(d) m=\sqrt5-1

Tutorial

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

Question 10

[7 marks]

The discrete random variable X has the probability distribution shown, where a and b are constants.

\begin{array}{c|cccc} x & \text{0} & \text{1} & \text{2} & \text{3} \\ \hline P(X=x) & a & b & 2a & \text{0.3} \end{array}

It is given that E(X)=1.7.

(a) Determine a and b.

[4]

(b) Find \mathrm{Var}(X).

[2]

(c) Find \mathrm{Var}(2X-1).

[1]

Show solutionHide solution

(a) Sum of probabilities: a+b+2a+0.3=1, so 3a+b=0.7

[1]

E(X)=0+b+2(2a)+3(0.3)=1.7, so 4a+b=0.8

[1]

Subtracting: a=0.1

[1]

b=0.7-0.3=0.4

[1]

(b) E(X^2)=1(0.4)+4(0.2)+9(0.3)=3.9

[1]

\mathrm{Var}(X)=3.9-1.7^2=3.9-2.89=1.01

[1]

(c) \mathrm{Var}(2X-1)=2^2\times1.01=4.04

[1]

Final answer

(a) a=0.1, b=0.4
(b) 1.01
(c) 4.04

Question 11

[6 marks]

An archer hits the target with each shot with probability 0.7, independently of other shots. The archer takes n shots.

Find the least value of n for which the probability of at least 5 hits is greater than 0.97.

Show solutionHide solution

Let X\sim\mathrm{B}(n,0.7). We need P(X\ge5)>0.97, that is 1-P(X\le4)>0.97

[1]

No simple algebraic route exists, so use a table of values on the GDC: y=1-\text{binomCdf}(x,0.7,0,4)

[1]

n=9: 0.9012\ldots

[1]

n=10: 0.9527\ldots

[1]

n=11: 0.9784\ldots

[1]

The values increase with n and first exceed 0.97 at n=11, so the least value is n=11.

[1]

Final answer

n=11

Question 12

[6 marks]

The random variable X\sim\mathrm{N}(\mu,\sigma^2) is such that P(X<28)=0.1 and P(X>46)=0.05.

(a) Determine \mu and \sigma.

[4]

(b) Hence find P(X<30).

[2]

Show solutionHide solution

(a) P(X<28)=0.1: \dfrac{28-\mu}{\sigma}=\text{invNorm}(0.1)=-1.2816\ldots

[1]

P(X<46)=0.95: \dfrac{46-\mu}{\sigma}=\text{invNorm}(0.95)=1.6449\ldots

[1]

Subtracting: \dfrac{18}{\sigma}=2.9265\ldots, so \sigma=6.1509\ldots=6.15 (3 s.f.)

[1]

\mu=28+1.2816\ldots\times6.1509\ldots=35.88\ldots=35.9 (3 s.f.)

[1]

(b) P(X<30)=\text{normCdf}(-10^{99},30,35.88\ldots,6.1509\ldots)

[1]

=0.16944\ldots=0.169

[1]

Final answer

(a) \mu=35.9, \sigma=6.15
(b) 0.169

Question 13

[6 marks]

The lifetime, L hours, of a type of resistor is normally distributed with mean 1200 and standard deviation 150. A resistor is said to fail early if its lifetime is less than 975 hours.

(a) Find the probability that a randomly chosen resistor fails early.

[2]

A batch contains n randomly chosen resistors.

(b) Find the least value of n for which the probability that at least one resistor in the batch fails early is greater than 0.99.

[4]

Show solutionHide solution

(a) P(L<975)=\text{normCdf}(-10^{99},975,1200,150)

[1]

=0.06680\ldots=0.0668

[1]

(b) Let p=0.06680\ldots. P(\text{at least one fails early})=1-(1-p)^n>0.99

[1]

(1-p)^n<0.01

[1]

n>\dfrac{\ln0.01}{\ln(1-p)}=66.60\ldots (the inequality reverses because \ln(1-p)<0)

[1]

The least value is n=67.

[1]

Final answer

(a) 0.0668
(b) n=67

Question 14

[7 marks]

Two coins are in a bag. One is fair. The other is biased so that it lands heads with probability 0.7. A coin is chosen at random from the bag and tossed 10 times. It lands heads exactly 8 times.

(a) Find the probability of obtaining exactly 8 heads in 10 tosses, given that the coin is (i) the fair coin; (ii) the biased coin.

[2]

(b) Hence find the probability that the coin chosen was the biased coin, given that it landed heads exactly 8 times.

[3]

(c) The same coin is tossed once more. Find the probability that it lands heads.

[2]

Show solutionHide solution

(a) (i) \text{binomPdf}(10,0.5,8)=\dfrac{45}{1024}=0.04394\ldots

[1]

(ii) \text{binomPdf}(10,0.7,8)=0.23347\ldots

[1]

(b) Let E be “8 heads”. The two coins are equally likely a priori:

P(\text{biased}\mid E)=\dfrac{0.5(0.23347\ldots)}{0.5(0.23347\ldots)+0.5(0.04394\ldots)} [1]

[1]

=0.84159\ldots=0.842

[1]

(c) Given E: P(\text{fair})=0.15840\ldots

[1]

P(\text{heads})=0.15840\ldots(0.5)+0.84159\ldots(0.7)=0.66831\ldots=0.668

[1]

Final answer

(a)(i) \frac{45}{1024}=0.0439
(ii) 0.233
(b) 0.842
(c) 0.668

Question 15

[7 marks]

A fisher catches two species of fish, A and B, in the proportions 55\% and 45\%. The mass of a species A fish is normally distributed with mean 500 g and standard deviation 40 g. The mass of a species B fish is normally distributed with mean 560 g and standard deviation 50 g.

One fish is picked from the catch.

(a) Find the probability that the fish has mass greater than 560 g.

[3]

(b) Given that the fish has mass greater than 560 g, find the probability that it is species A.

[3]

(c) Hence write down the probability that it is species B.

[1]

Show solutionHide solution

(a) P(M>560\mid A)=\text{normCdf}(560,10^{99},500,40)=0.06680\ldots

[1]

P(M>560\mid B)=0.5 (560 is the mean of B)

[1]

P(M>560)=0.55(0.06680\ldots)+0.45(0.5)=0.03674\ldots+0.225=0.26174\ldots=0.262

[1]

(b) P(A\mid M>560)=\dfrac{0.03674\ldots}{0.26174\ldots} [1]

[1]

=0.14038\ldots=0.140

[1]

(c) 1-0.14038\ldots=0.860

[1]

Final answer

(a) 0.262
(b) 0.140
(c) 0.860

Question 16

[8 marks]

The fraction X of a day's quota that a machine completes before its first stoppage is modelled by a variable whose density is

f(x)=\begin{cases}\dfrac{k}{1+x^2},&0\le x\le1\\0,&\text{otherwise.}\end{cases}

(a) Determine the exact value of k.

[2]

(b) Calculate the probability that more than half of the quota is completed.

[2]

(c) Determine \mathrm{E}(X) exactly.

[2]

(d) Determine the median of X, in exact form.

[2]

Show solutionHide solution

(a) \displaystyle\int_0^1\dfrac{k}{1+x^2}\,dx=k\left[\arctan x\right]_0^1

[1]

=\dfrac{k\pi}{4}=1, so k=\dfrac4\pi

[1]

(b) P\left(X>\tfrac12\right)=\dfrac4\pi\left[\arctan x\right]_{1/2}^{1}=\dfrac4\pi\left(\dfrac\pi4-\arctan\dfrac12\right)

[1]

=1-\dfrac4\pi\arctan\dfrac12=0.410 (3 s.f.)

[1]

(c) E(X)=\dfrac4\pi\displaystyle\int_0^1\dfrac{x}{1+x^2}\,dx=\dfrac4\pi\left[\tfrac12\ln(1+x^2)\right]_0^1

[1]

=\dfrac{2\ln2}{\pi} (=0.441 to 3 s.f.)

[1]

(d) The median m satisfies \dfrac4\pi\arctan m=\dfrac12, so \arctan m=\dfrac\pi8

[1]

m=\tan\dfrac\pi8=\sqrt2-1

[1]

Final answer

(a) k=\frac4\pi
(b) 0.410
(c) \frac{2\ln2}{\pi}
(d) \sqrt2-1

Question 17

[9 marks]

For a constant a>0, the continuous random variable X has probability density function

f(x)=\begin{cases}kx(a-x),&0\le x\le a\\0,&\text{otherwise.}\end{cases}

(a) Show that k=\dfrac6{a^3}.

[3]

(b) Write down \mathrm{E}(X) in terms of a, giving a reason.

[1]

(c) Show that \mathrm{Var}(X)=\dfrac{a^2}{20}.

[4]

(d) Given that \mathrm{Var}(X)=5, find the value of a.

[1]

Show solutionHide solution

(a) \displaystyle\int_0^ak(ax-x^2)\,dx=k\left[\dfrac{ax^2}2-\dfrac{x^3}3\right]_0^a

[1]

=k\left(\dfrac{a^3}2-\dfrac{a^3}3\right)

[1]

=\dfrac{ka^3}{6}

[1]

\dfrac{ka^3}6=1, so k=\dfrac6{a^3}

(b) f(x)=\dfrac{6x(a-x)}{a^3} is symmetric about x=\dfrac a2 (it is a downward parabola with roots 0 and a), so E(X)=\dfrac a2.

[1]

(c) E(X^2)=\displaystyle\int_0^a x^2\cdot\dfrac{6x(a-x)}{a^3}\,dx=\dfrac6{a^3}\left[\dfrac{ax^4}4-\dfrac{x^5}5\right]_0^a

[1]

=\dfrac6{a^3}\cdot a^5\left(\dfrac14-\dfrac15\right)=\dfrac{3a^2}{10}

[1]

\mathrm{Var}(X)=E(X^2)-[E(X)]^2=\dfrac{3a^2}{10}-\left(\dfrac a2\right)^2 [1]

[1]

=\dfrac{6a^2-5a^2}{20}=\dfrac{a^2}{20}

(d) \dfrac{a^2}{20}=5, so a^2=100 and a=10 (as a>0).

[1]

Final answer

(b) \frac a2, by symmetry about x=\frac a2
(d) a=10

Mastery

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

Question 18

[5 marks]

Events A and B have P(A) = 0.40, P(A \mid B) = 0.20 and P(A \cup B) = 0.52.

Calculate P(B \mid A').

[5]

Show solutionHide solution

The definition of conditional probability gives P(A \cap B) = 0.20\,P(B). Putting this into the addition rule:

0.52 = 0.40 + P(B) - 0.20\,P(B)

0.12 = 0.80\,P(B) \implies P(B) = 0.15

Then P(A \cap B) = 0.20\times 0.15 = 0.03, so

P(B \cap A') = P(B) - P(A \cap B) = 0.15 - 0.03 = 0.12

With P(A') = 0.60,

P(B \mid A') = \frac{0.12}{0.60} = \frac{1}{5}

P(B \mid A') = 0.2 \left(= \dfrac{1}{5}\right)

[5]

Final answer

0.2

Question 19

[6 marks]

For events A and B, P(A) = 0.4 and P(B) = 0.8.

(a) Determine P(A \cap B) on the assumption that A and B are independent.

[1]

(b) Determine the smallest value P(A \cap B) could take.

[3]

(c) Determine the largest value P(A \cap B) could take, justifying your answer.

[2]

Show solutionHide solution

(a) For independent events the probabilities multiply:

P(A \cap B) = 0.4 \times 0.8 = 0.32

[1]

(b) By the addition rule,

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1.2 - P(A \cap B)

and any probability satisfies P(A \cup B) \leq 1:

1.2 - P(A \cap B) \leq 1 \implies P(A \cap B) \geq 0.2

This value is attained when A \cup B is the whole sample space.

minimum= 0.2

[3]

(c) Since A \cap B \subseteq A, we have P(A \cap B) \leq P(A) = 0.4, and this is attained when A \subseteq B (possible here because P(A) < P(B)).

maximum= 0.4

[2]

Final answer

(a) 0.32
(b) 0.2
(c) 0.4

Question 20

[8 marks]

A discrete random variable X has this distribution.

\begin{array}{r|lll} x & 1 & 2 & 3 \\ \hline P(X = x) & 0.5 - 2a & 5a & 0.5 - 3a \\ \end{array}

(a) State the set of values that the constant a may take.

[2]

(b) When a = 0.1, calculate \mathrm{Var}(7 - 3X).

[6]

Show solutionHide solution

(a) The three entries already sum to 1 for every a, so the only requirement is that each be non-negative:

0.5 - 2a \geq 0 \implies a \leq 0.25

5a \geq 0 \implies a \geq 0

0.5 - 3a \geq 0 \implies a \leq \tfrac{1}{6}

The strictest pair of bounds gives

0 \leq a \leq \tfrac{1}{6}

[2]

(b) With a = 0.1 the probabilities are 0.3, 0.5 and 0.2:

E(X) = 1(0.3) + 2(0.5) + 3(0.2) = 0.3 + 1.0 + 0.6 = 1.9

E\!\left(X^{2}\right) = 1(0.3) + 4(0.5) + 9(0.2) = 0.3 + 2.0 + 1.8 = 4.1

\mathrm{Var}(X) = E\!\left(X^{2}\right) - \left[E(X)\right]^{2} = 4.1 - 3.61 = 0.49

Adding a constant leaves variance unchanged, and a multiplier is squared:

\mathrm{Var}(7 - 3X) = (-3)^{2}\,\mathrm{Var}(X) = 9 \times 0.49 = 4.41

\mathrm{Var}(7 - 3X) = 4.41

[6]

Final answer

(a) 0\le a\le\frac16
(b) 4.41

Question 21

[5 marks]

The random variable X has X \sim B(25,\, p) with \mathrm{Var}(X) = 4.

(a) Determine the possible values of p.

[3]

Let Y = 5 - 3X.

(b) Determine \mathrm{Var}(Y).

[2]

Show solutionHide solution

(a) For a binomial distribution \mathrm{Var}(X) = np(1 - p):

25p(1 - p) = 4 \implies p(1 - p) = 0.16

p^{2} - p + 0.16 = 0

Factorising (or using the quadratic formula):

(p - 0.2)(p - 0.8) = 0

p = 0.2 \quad \text{or} \quad p = 0.8

p = 0.2 or p = 0.8

[3]

(b) Adding a constant does not change variance, and a multiplier is squared:

\mathrm{Var}(Y) = \mathrm{Var}(5 - 3X) = (-3)^{2}\,\mathrm{Var}(X) = 9 \times 4 = 36

\mathrm{Var}(Y) = 36

[2]

Final answer

(a) p=0.2 or p=0.8
(b) 36

Question 22

[7 marks]

Two presses are used in a mint to produce commemorative coins. Press A produces faulty coins 3\% of the time, while press B produces faulty coins 7\% of the time.

One press is chosen at random, each equally likely, and produces fifteen coins.

(a) Calculate the probability that exactly one coin is faulty.

[4]

(b) Given that exactly one coin is faulty, calculate the probability that the coins were produced by press B.

[3]

Show solutionHide solution

(a)

Recognising the binomial distribution B(15, p) with \text{P}(X=1) = \binom{15}{1}p(1-p)^{14}.

Considering both presses using p = 0.03 and p = 0.07:

0.5 \times (0.293776\ldots) + 0.5 \times (0.380146\ldots) = 0.336961\ldots

0.337

[4]

(b)

Using conditional probability:

\text{P}(B \mid 1 \text{ faulty}) = \frac{\text{P}(B \cap 1 \text{ faulty})}{\text{P}(1 \text{ faulty})} = \frac{0.5\times 0.380146\ldots}{0.336961\ldots} = \frac{0.190073\ldots}{0.336961\ldots} = 0.564080\ldots

0.564

[3]

Final answer

(a) 0.337
(b) 0.564

Question 23

[8 marks]

A waiting time X, in minutes, is modelled by a continuous distribution with density

f(x) = \begin{cases} a\,x\,e^{-x/2}, & 0 \leq x \leq b \\ 0, & \text{otherwise,} \end{cases}

for positive constants a and b.

(a) Show that a = \dfrac{1}{4 - (2b+4)e^{-b/2}}.

[5]

(b) For the case b = 6, use your graphic display calculator to find the median of X.

[3]

Show solutionHide solution

(a)

Since f is a probability density function, \displaystyle\int_0^b a\,x\,e^{-x/2}\,dx = 1.

Integrating by parts with u = x and dv = e^{-x/2}\,dx, so v = -2e^{-x/2}:

\int_0^b x\,e^{-x/2}\,dx = \left[-2x\,e^{-x/2}\right]_0^b + 2\int_0^b e^{-x/2}\,dx \qquad \textbf{M1A1}

= -2b\,e^{-b/2} + \left[-4e^{-x/2}\right]_0^b = -2b\,e^{-b/2} - 4e^{-b/2} + 4 \qquad \textbf{A1}

Hence a\left(4 - (2b+4)e^{-b/2}\right) = 1, giving

a = \frac{1}{4 - (2b+4)e^{-b/2}} \qquad \textbf{AG}

[5]

(b)

When b = 6: a = \dfrac{1}{4 - 16e^{-3}} = 0.312167\ldots

The median m satisfies \displaystyle\int_0^m a\,x\,e^{-x/2}\,dx = \dfrac{1}{2}, that is

a\left(4 - (2m+4)e^{-m/2}\right) = \frac{1}{2} \qquad \textbf{(M1)(A1)}

Solving on the GDC: m = 2.7552\ldots

m = 2.76

A1[3]

Final answer

(b) 2.76

Question 24

[6 marks]

The density of a continuous random variable X is

f(x) = \begin{cases} ax^2, & 0 \le x \le 1 \\ a(2 - x), & 1 < x \le 2 \\ 0, & \text{elsewhere,} \end{cases}

where a is a positive constant.

(a) Show that a = \dfrac{6}{5}.

[2]

(b) Find the median of X.

[4]

Show solutionHide solution

(a)

\int_0^1 ax^2\,dx + \int_1^2 a(2 - x)\,dx = 1

\left[\frac{ax^3}{3}\right]_0^1 + \left[2ax - \frac{ax^2}{2}\right]_1^2 = 1

\frac{a}{3} + \left(4a - 2a\right) - \left(2a - \frac{a}{2}\right) = \frac{a}{3} + \frac{a}{2} = \frac{5a}{6} = 1

\implies a = \frac{6}{5}

[2]

(b) The median m satisfies \displaystyle\int_0^m f(x)\,dx = 0.5.

\int_0^1 \frac{6}{5}x^2\,dx = \frac{2}{5} = 0.4 < 0.5

so m > 1 and

\int_1^m \frac{6}{5}(2 - x)\,dx = 0.5 - 0.4 = 0.1

\frac{6}{5}\left[2x - \frac{x^2}{2}\right]_1^m = \frac{1}{10} \implies 2m - \frac{m^2}{2} - \frac{3}{2} = \frac{1}{12}

m^2 - 4m + \frac{19}{6} = 0 \implies m = 2 - \sqrt{\frac{5}{6}} = 1.0871\ldots

(the root 2 + \sqrt{5/6} exceeds 2 and is rejected).

m = 1.09

[4]

Final answer

(b) m=2-\sqrt{\frac56}\approx1.09

Question 25

[18 marks]

Marcus checks electronic circuits in a laboratory, and each check is independent of the others. Records show that he logs 6\% of circuits wrongly.

(a) One day Marcus checks 80 circuits.

(i) Calculate the probability that he logs at most three of them wrongly.

(ii) Given that he logs at most three wrongly, calculate the probability that he logs exactly two wrongly.

[5]

On another day he checks n circuits and the probability of at most three wrong logs is about 0.066.

(b) Calculate n.

[3]

Nia and Omar work in the same laboratory and log 4\% and 10\% of their circuits wrongly, respectively; here too each check is independent. Of all circuits handed out, Marcus checks 45\%, Nia 35\% and Omar 20\%.

(c) A circuit is picked at random. Find the probability that it was

(i) logged wrongly;

(ii) checked by Marcus, given that it was logged wrongly.

[6]

A year later Marcus and Nia are as accurate as before and the shares of circuits are unchanged, but Omar has improved: he now logs x\% of his circuits wrongly. The overall chance that a randomly chosen circuit is logged wrongly now equals the chance that Omar logs a circuit wrongly.

(d) Calculate x.

[4]

Show solutionHide solution

(a)(i)

E \sim B(80,\ 0.06), \text{ where } E \text{ is the number of incorrectly recorded circuits}

P(E \leq 3) = 0.285788\ldots = 0.286

[2]

(ii)

P(E = 2 \mid E \leq 3) = \frac{P(E = 2)}{P(E \leq 3)} = \frac{0.091193\ldots}{0.285788\ldots}

= 0.319094\ldots = 0.319

[3]

(b)

Using the cumulative binomial distribution to find n such that P(E \leq 3) \approx 0.066:

n = 119:\ 0.069, \qquad n = 120:\ P(E \leq 3) = 0.066042\ldots, \qquad n = 121:\ 0.063

n = 120

[3]

(c)(i)

Let I be the event “recorded incorrectly”.

P(I) = P(I \cap M) + P(I \cap N) + P(I \cap O)

= 0.45 \times 0.06 + 0.35 \times 0.04 + 0.20 \times 0.10 = 0.061

0.061 (6.1%)

(ii)

P(M \mid I) = \frac{P(M \cap I)}{P(I)} = \frac{0.45 \times 0.06}{0.061} = 0.442623\ldots

0.443 (44.3%)

[6]

(d)

Let P(I \mid O) = \dfrac{x}{100}.

P(I) = 0.45 \times 0.06 + 0.35 \times 0.04 + 0.20 \times \frac{x}{100}

Setting that overall probability equal to Omar's own error rate \dfrac{x}{100}:

\frac{x}{100} = 0.027 + 0.014 + 0.20 \times \frac{x}{100}

0.8\times\frac{x}{100} = 0.041 \implies x = 5.125

x = 5.125 (5.125%)

[4]

Final answer

(a)(i) 0.286
(ii) 0.319
(b) n=120
(c)(i) 0.061
(ii) 0.443
(d) x=5.125

Question 26

[17 marks]

At Greenway Orchard the masses of kiwi fruit, W, in grams, follow a normal distribution with a mean of 95 grams and a standard deviation of 7 grams.

(a) Calculate the probability that a randomly chosen kiwi fruit has a mass of less than 90 grams.

[2]

(b) It is found that 12% of the kiwi fruit have a mass of more than w grams. Calculate w, correct to four significant figures.

[2]

Every orchard classifies a kiwi fruit as premium when its mass is between 85 and 105 grams.

(c) Calculate the percentage of kiwi fruit that are classified as premium at Greenway Orchard.

[2]

After orders are completed, there are many kiwi fruit left over. Trays are filled with randomly chosen left-over kiwi fruit. Each tray contains 24 kiwi fruit.

(d) Calculate the probability that a randomly chosen tray contains at least 21 premium kiwi fruit.

[3]

(e) If 6 of these trays are randomly selected, calculate the probability that exactly 3 trays have at least 21 premium kiwi fruit.

[2]

At a neighbouring orchard the masses of kiwi fruit, M, in grams, follow a normal distribution with mean \mu and standard deviation \sigma. Further information about that orchard:

\bullet 88% of their kiwi fruit are classified as premium,

\bullet the percentage of kiwi fruit that weigh less than 85 grams is three times the percentage of kiwi fruit that weigh more than 105 grams.

(f) Calculate the value of \mu.

[6]

Show solutionHide solution

(a)

Finding P(W < 90):

P(W < 90) = 0.237525\ldots

0.238

[2]

(b)

Solving P(W > w) = 0.12:

w = 103.2249\ldots

w = 103.2 grams

[2]

(c)

Finding P(85 < W < 105):

P(85 < W < 105) = 0.846873\ldots

84.7%

[2]

(d)

Using a binomial probability with X \sim B(24,\ 0.846873\ldots):

P(X \geq 21) = 1 - P(X \leq 20) = 0.487590\ldots

0.488

[3]

(e)

With Y \sim B(6,\ 0.487590\ldots):

P(Y = 3) = 0.311923\ldots

0.312

[2]

(f)

Since 88\% are premium, P(M<85)+P(M>105)=0.12. With P(M<85)=3P(M>105):

P(M < 85) = 0.09, \qquad P(M > 105) = 0.03

The corresponding z-scores are

z = -1.340755\ldots, \qquad z = 1.880794\ldots

Solving the system:

-1.340755\ldots = \frac{85 - \mu}{\sigma}

1.880794\ldots = \frac{105 - \mu}{\sigma}

Subtracting gives \sigma = 6.20819\ldots, and so

\mu = 93.3237\ldots

\mu = 93.3 grams

[6]

Final answer

(a) 0.238
(b) 103.2
(c) 84.7\%
(d) 0.488
(e) 0.312
(f) \mu=93.3

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}

=\dfrac{364\times363\times362}{365^3}=0.984

(b) p_n=\dfrac{365\times364\times\cdots\times(366-n)}{365^n}=\dfrac{{}^{365}P_n}{365^n}

a shared birthday is more likely than not when p_n<0.5

p_{22}=0.524, p_{23}=0.493, so n=23

(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

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)

n=253

(d) in (b) any two of the n people can match, and 23 people form {}^{23}C_2=253 pairs; in (c) only the n pairs that include you count

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

n=13

(b) \ln p_n=\displaystyle\sum_{k=1}^{n-1}\ln\left(1-\frac kN\right)

<-\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)

\ln is increasing, so p_n<\mathrm e^{-n(n-1)/(2N)}

(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

N=365: 2N\ln2=505.997; 22\times21=462, 23\times22=506, so n=23, again exact (by a margin of 0.003)

(d) 2N\ln2=1\,386\,294.4; 1177\times1176=1\,384\,152, 1178\times1177=1\,386\,506, so 1178 accounts are enough

for large N, the least such n is large, and n(n-1)\approx n^2

so n^2\approx2N\ln2 and n\approx\sqrt{2N\ln2} (\approx1177.4 for N=10^6)

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

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}

(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})

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

d_0\left(1+\frac23+\frac49+\frac8{27}\right)=d_0\times\dfrac{65}{27}=1, so d_0=\dfrac{27}{65}=p_1

p_2=\dfrac{27+18}{65}=\dfrac{9}{13}

p_3=\dfrac{45+12}{65}=\dfrac{57}{65}

(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

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, 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

P=\dfrac14\left(1+\dfrac14+\dfrac1{16}+\cdots\right)=\dfrac{1/4}{1-1/4}=\dfrac13

(b) e_1=1+\frac12e_2 and e_2=1+\frac12e_1

e_1=e_2=2

(c) e_1=1+\frac12e_2, e_2=1+\frac12e_1+\frac12e_3, e_3=1+\frac12e_2

e_1=3, e_2=4, e_3=3

1\times2=2, 2\times1=2; 1\times3=3, 2\times2=4: conjecture e_k=k(N-k)

(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

e_k=\displaystyle\sum_{j=0}^{k-1}(N-1-2j)=k(N-1)-k(k-1)=k(N-k)

(e) N=100, k=10: e_{10}=10\times90=900 rounds

P(\text{reach }100)=\frac{10}{100}, so the probability of ending with none is 0.9

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. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Normal distribution	`2nd VARS (DISTR), 2:normalcdf(lower, upper, mean, sd); 3:invNorm(area, mean, sd)`
Binomial	`2nd VARS (DISTR), binomcdf(n, p, k) then 1 minus binomcdf(n, p, k-1) for P(Y≥k)`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 13 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