IB Maths AI HL 4.6 Combined, conditional and independent events Question Bank

Practise IB Mathematics SL/HL 4.6 by applying combined, conditional and independent events methods to exam-style questions.

[Maximum number: 5]

This question compares possible designs for a new computer network between multiple school buildings, and whether they meet specific requirements.
A school's administration team decides to install new fibre-optic internet cables underground. The school has eight buildings that need to be connected by these cables. A map of the school is shown below, with the internet access point of each building labelled A-H.

Figure for Question SL 4.6—Combined, conditional and independent events question 1 — IB Maths AI HL

Jonas is planning where to install the underground cables. He begins by determining the distances, in metres, between the underground access points in each of the buildings.

He finds $\mathrm{AD}=89.2 \mathrm{~m}, \mathrm{DF}=104.9 \mathrm{~m}$ and $\mathrm{A} \hat{\mathrm{DF}}=83^{\circ}$ .

After more research, Jonas decides to install the cables as shown in the diagram below.

Each individual cable is installed such that each end of the cable is connected to a building's access point. The connection between each end of a cable and an access point has a 1.4 % probability of failing after a power surge.

For the network to be successful, each building in the network must be able to communicate with every other building in the network. In other words, there must be a path that connects any two buildings in the network. Jonas would like the network to have less than a 2 % probability of failing to operate after a power surge.

Show that Jonas's network satisfies the requirement of there being less than a 2 % probability of the network failing after a power surge.

METHOD 1 recognition of a binomial distribution
$X \sim \mathrm{~B}(2,0.014)$
finding the probability that a cable fails (at least one of its connections fails)
P(X>0)=0.027804 OR 1-P(X=0)=0.027804
recognition that two cables must fail for the network to go offline
recognition of binomial distribution for network, $Y \sim \mathrm{~B}(8,0.027804)$ $\mathrm{P}(Y \geq 2)=0.0194(0.0193602 \ldots) \quad$ OR $1-\mathrm{P}(Y<2)=0.0194(0.0193602 \ldots)$
therefore, the diagram satisfies the requirement since 1.94 %<2 %

Note: Evidence of binomial distribution may be seen as combinations.

METHOD 2 recognition of a binomial distribution
$X \sim \mathrm{~B}(16,0.014)$
finding the probability that at least two connections fail $\mathrm{P}(X \geq 2)=0.0206473 \ldots$ OR $1-\mathrm{P}(X<2)=0.0206473 \ldots$
recognition that the previous answer is an overestimate
finding probability of two ends of the same cable failing, $F \sim \mathrm{~B}(2,0.014)$ , and the ends of the other 14 cables not failing, $S \sim \mathrm{~B}(14,0.014) \mathrm{P}(F=2) \times \mathrm{P}(S=0)=0.0000160891 \ldots$ $0.0000160891 \ldots \times 8=0.00128713 \ldots$

therefore, the diagram satisfies the requirement since 1.94 %<2 %

METHOD 3 recognition of a binomial distribution
$X \sim \mathrm{~B}(16,0.014)$
finding the probability that the network remains secure if 0 or 1 connections fail or if 2 connections fail provided that the second failed connection occurs at the other end of the cable with the first failure
P( remains secure $)=\mathrm{P}(X \leq 1)+\frac{1}{15} \times \mathrm{P}(X=2)$
=0.9806397625
P( network fails $)=1-0.9806397625=0.0194(0.0193602 \ldots)$
therefore, the diagram satisfies the requirement since 1.94 %<2 %

METHOD 4

P(network failing)
=1-P(0 connections failing )-P(1 connection failing )
- P (2 connections on the same cable failing) M1
$=1-0.986{ }^{16}-{ }^{16} C_{1} \times 0.014 \times 0.986{ }^{15}-{ }^{8} C_{1} \times 0.014^{2} \times 0.986{ }^{14}$
A1A1A1

Note: Award A1 for each of $2^{\text {nd }}, 3^{\text {rd }}$ and last terms.
$=0.0194(0.0193602 \ldots)$
A1
therefore, the diagram satisfies the requirement since 1.94 %<2 %
AG

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SL 4.6—Combined, conditional and independent events question 1