Question
Consider the independent events A and B . Given that \({\rm{P}}(B) = 2{\rm{P}}(A)\) , and \({\rm{P}}(A \cup B) = 0.52\) , find \({\rm{P}}(B)\) .
Answer/Explanation
Markscheme
METHOD 1
for independence \({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)\) (R1)
expression for \({\rm{P}}(A \cap B)\) , indicating \({\rm{P}}(B) = 2{\rm{P}}(A)\) (A1)
e.g. \({\rm{P}}(A) \times 2{\rm{P}}(A)\) , \(x \times 2x\)
substituting into \({\rm{P}}(A \cup B) = {\rm{P}}(A) + {\rm{P}}(B) – {\rm{P}}(A \cap B)\) (M1)
correct substitution A1
e.g. \(0.52 = x + 2x – 2{x^2}\) , \(0.52 = {\rm{P}}(A) + 2{\rm{P}}(A) – 2{\rm{P}}(A){\rm{P}}(A)\)
correct solutions to the equation (A2)
e.g. \(0.2\), \(1.3\) (accept the single answer \(0.2\))
\({\rm{P}}(B) = 0.4\) A1 N6
[7 marks]
METHOD 2
for independence \({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)\) (R1)
expression for \({\rm{P}}(A \cap B)\) , indicating \({\rm{P}}(A) = \frac{1}{2}{\rm{P}}(B)\) (A1)
e.g. \({\rm{P}}(B) \times \frac{1}{2}{\rm{P}}(B)\) , \(x \times \frac{1}{2}x\)
substituting into \({\rm{P}}(A \cup B) = {\rm{P}}(A) + {\rm{P}}(B) – {\rm{P}}(A \cap B)\) (M1)
correct substitution A1
e.g. \(0.52 = 0.5x + x – 0.5{x^2}\) , \(0.52 = 0.5{\rm{P}}(B) + {\rm{P}}(B) – 0.5{\rm{P}}(B){\rm{P}}(B)\)
correct solutions to the equation (A2)
e.g. 0.4, 2.6 (accept the single answer 0.4)
\({\rm{P}}(B) = 0.4\) (accept \(x = 0.4\) if x set up as \({\rm{P}}(B)\) ) A1 N6
[7 marks]
Question
A company produces a large number of water containers. Each container has two parts, a bottle and a cap. The bottles and caps are tested to check that they are not defective.
A cap has a probability of 0.012 of being defective. A random sample of 10 caps is selected for inspection.
Find the probability that exactly one cap in the sample will be defective.
The sample of caps passes inspection if at most one cap is defective. Find the probability that the sample passes inspection.
The heights of the bottles are normally distributed with a mean of \(22{\text{ cm}}\) and a standard deviation of \(0.3{\text{ cm}}\).
(i) Copy and complete the following diagram, shading the region representing where the heights are less than \(22.63{\text{ cm}}\).
(ii) Find the probability that the height of a bottle is less than \(22.63{\text{ cm}}\).
(i) A bottle is accepted if its height lies between \(21.37{\text{ cm}}\) and \(22.63{\text{ cm}}\). Find the probability that a bottle selected at random is accepted.
(ii) A sample of 10 bottles passes inspection if all of the bottles in the sample are accepted. Find the probability that the sample passes inspection.
The bottles and caps are manufactured separately. A sample of 10 bottles and a sample of 10 caps are randomly selected for testing. Find the probability that both samples pass inspection.
Answer/Explanation
Markscheme
Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.
evidence of recognizing binomial (seen anywhere in the question) (M1)
e.g. \(_n{C_r}{p^r}{q^{n – r}}\) , \({\text{B}}(n{\text{, }}p)\) , \(^{10}{C_1}{(0.012)^1}{(0.988)^9}\)
\(p = 0.108\) A1 N2
[2 marks]
valid approach (M1)
e.g. \({\rm{P}}(X \le 1)\) , \(0.88627 \ldots + 0.10764 \ldots \)
\(p = 0.994\) A1 N2
[2 marks]
(i)
A1A1 N2
Note: Award A1 for vertical line to right of mean, A1 for shading to left of their vertical line.
(ii) valid approach (M1)
e.g. \({\rm{P}}(X < 22.63)\)
working to find standardized value (A1)
e.g. \(\frac{{22.63 – 22}}{{0.3}}\) , 2.1
\(p = 0.982\) A1 N3
[5 marks]
valid approach (M1)
e.g. \({\rm{P}}(21.37 < X < 22.63)\) , \({\rm{P}}( – 2.1 < z < 2.1)\)
correct working (A1)
e.g. \(0.982 – (1 – 0.982)\)
\(p = 0.964\) A1 N3
(ii) correct working (A1)
e.g. \(X \sim {\rm{B}}(10,0.964)\) , \({(0.964)^{10}}\)
\(p = 0.695\) (accept 0.694 from tables) A1 N2
[5 marks]
valid approach (M1)
e.g. \({\rm{P}}(A \cap B) = {\rm{P}}(A){\rm{P}}(B)\) , \((0.994) \times {(0.964)^{10}}\)
\(p = 0.691\) (accept \(0.690\) from tables) A1 N2
[2 marks]
Question
At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.
Find the percentage of students who learn both Spanish and French.
At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.
Find the percentage of students who learn Spanish, but not French.
At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.
At this school, \(52\% \) of the students are girls, and \(85\% \) of the girls learn Spanish.
A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish.
(i) Find \({\rm{P}}(G \cap S)\) .
(ii) Show that G and S are not independent.
At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.
At this school, \(52\% \) of the students are girls, and \(85\% \) of the girls learn Spanish.
A boy is chosen at random. Find the probability that he learns Spanish.
Answer/Explanation
Markscheme
valid approach (M1)
e.g. Venn diagram with intersection, union formula,
\({\rm{P}}(S \cap F) = 0.75 + 0.40 – 1\)
\(15\) (accept \(15\% \)) A1 N2
[2 marks]
valid approach involving subtraction (M1)
e.g. Venn diagram, \(75 – 15\)
60 (accept \(60\% \)) A1 N2
[2 marks]
(i) valid approach (M1)
e.g. tree diagram, multiplying probabilities, \({\rm{P}}(S|G) \times {\rm{P(}}G{\rm{)}}\)
correct calculation (A1)
e.g. \(0.52 \times 0.85\)
\({\rm{P}}(G \cap S) = 0.442\) (exact) A1 N3
(ii) valid reasoning, with words, symbols or numbers (seen anywhere) R1
e.g. \({\rm{P(}}G{\rm{)}} \times {\rm{P}}(S) \ne {\rm{P}}(G \cap S)\) , \({\rm{P}}(S|G) \ne {\rm{P(}}S{\rm{)}}\) , not equal,
one correct value A1
e.g. \({\rm{P}}(G) \times {\rm{P}}(S) = 0.39\) , \({\rm{P}}(S|G) = 0.85\) , \(0.39 \ne 0.442\)
G and S are not independent AG N0
[5 marks]
METHOD 1
\(48\% \) are boys (seen anywhere) A1
e.g. \({\rm{P}}(B) = 0.48\)
appropriate approach (M1)
e.g. \({\text{P(girl and Spanish)}} + {\text{P(boy and Spanish)}} = {\text{P(Spanish)}}\)
correct approach to find P(boy and Spanish) (A1)
e.g. \({\rm{P(}}B \cap S{\rm{) = P(}}S{\rm{)}} – {\rm{P}}(G \cap S)\) , \({\rm{P(}}B \cap S{\rm{) = P(}}S|B) \times {\rm{P}}(B)\) , 0.308
correct substitution (A1)
e.g. \(0.442 + 0.48x = 0.75\) , \(0.48x = 0.308\)
correct manipulation (A1)
e.g. \({\rm{P}}(S|B) = \frac{{0.308}}{{0.48}}\)
\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.641666 \ldots \) , \(0.641\bar 6\)
\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.642\) \([0.641{\text{, }}0.642]\) A1 N3
[6 marks]
METHOD 2
\(48\% \) are boys (seen anywhere) A1
e.g. 0.48 used in tree diagram
appropriate approach (M1)
e.g. tree diagram
correctly labelled branches on tree diagram (A1)
e.g. first branches are boy/girl, second branches are Spanish/not Spanish
correct substitution (A1)
e.g. \(0.442 + 0.48x = 0.75\)
correct manipulation (A1)
e.g. \(0.48x = 0.308\) , \({\rm{P}}(S|B) = \frac{{0.308}}{{0.48}}\)
\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.641666 \ldots \) , \(0.641\bar 6\)
\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.642\) \([0.641{\text{, }}0.642]\)
[6 marks]
Question
Two events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.2\) and \({\text{P}}(A \cup B) = 0.5\).
Given that \(A\) and \(B\) are mutually exclusive, find \({\text{P}}(B)\).
Given that \(A\) and \(B\) are independent, find \({\text{P}}(B)\).
Answer/Explanation
Markscheme
correct approach (A1)
eg \(0.5 = 0.2 + {\text{P}}(B),{\text{ P}}(A \cap B) = 0\)
\({\text{P}}(B) = 0.3\) A1 N2
[2 marks]
Correct expression for \({\text{P}}(A \cap B)\) (seen anywhere) A1
eg \({\text{P}}(A \cap B) = 0.2{\text{P}}(B),{\text{ }}0.2x\)
attempt to substitute into correct formula for \({\text{P}}(A \cup B)\) (M1)
eg \({\text{P}}(A \cup B) = 0.2 + {\text{P}}(B) – {\text{P}}(A \cap B),{\text{ P}}(A \cup B) = 0.2 + x – 0.2x\)
correct working (A1)
eg \(0.5 = 0.2 + {\text{P}}(B) – 0.2{\text{P}}(B),{\text{ }}0.8x = 0.3\)
\({\text{P}}(B) = \frac{3}{8}{\text{ }}( = 0.375,{\text{ exact}})\) A1 N3
[4 marks]
Question
Let \(A\) and \(B\) be independent events, where \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.6\).
Find \({\text{P}}(A \cap B)\).
Find \({\text{P}}(A \cup B)\).
On the following Venn diagram, shade the region that represents \(A \cap B’\).
Find \({\text{P}}(A \cap B’)\).
Answer/Explanation
Markscheme
correct substitution (A1)
eg \(0.3 \times 0.6\)
\({\text{P}}(A \cap B) = 0.18\) A1 N2
[2 marks]
correct substitution (A1)
eg \({\text{P}}(A \cup B) = 0.3 + 0.6 – 0.18\)
\({\text{P}}(A \cup B) = 0.72\) A1 N2
[2 marks]
A1 N1
appropriate approach (M1)
eg \(0.3 – 0.18,{\text{ P}}(A) \times {\text{P}}(B’)\)
\({\text{P}}(A \cap B’) = 0.12\) (may be seen in Venn diagram) A1 N2
[2 marks]
Question
A forest has a large number of tall trees. The heights of the trees are normally distributed with a mean of \(53\) metres and a standard deviation of \(8\) metres. Trees are classified as giant trees if they are more than \(60\) metres tall.
A tree is selected at random from the forest.
Find the probability that this tree is a giant.
A tree is selected at random from the forest.
Given that this tree is a giant, find the probability that it is taller than \(70\) metres.
Two trees are selected at random. Find the probability that they are both giants.
\(100\) trees are selected at random.
Find the expected number of these trees that are giants.
\(100\) trees are selected at random.
Find the probability that at least \(25\) of these trees are giants.
Answer/Explanation
Markscheme
valid approach (M1)
eg \({\text{P}}(G) = {\text{P}}(H > 60,{\text{ }}z = 0.875,{\text{ P}}(H > 60) = 1 – 0.809,{\text{ N}}\left( {53, {8^2}} \right)\)
\(0.190786\)
\({\text{P}}(G) = 0.191\) A1 N2
[3 marks]
finding \({\text{P}}(H > 70) = 0.01679\) (seen anywhere) (A1)
recognizing conditional probability (R1)
eg \({\text{P}}(A\left| {B),{\text{ P}}(H > 70\left| {H > 60)} \right.} \right.\)
correct working (A1)
eg \(\frac{{0.01679}}{{0.191}}\)
\(0.0880209\)
\({\text{P}}(X > 70\left| {G) = 0.0880} \right.\) A1 N3
[6 marks]
attempt to square their \({\text{P}}(G)\) (M1)
eg \({0.191^2}\)
\(0.0363996\)
\({\text{P}}({\text{both }}G) = 0.0364\) A1 N2
[2 marks]
correct substitution into formula for \({\text{E}}(X)\) (A1)
eg \(100(0.191)\)
\({\text{E}}(G) = 19.1{\text{ }}[19.0,{\text{ }}19.1]\) A1 N2
[3 marks]
recognizing binomial probability (may be seen in part (c)(i)) (R1)
eg \(X \sim {\text{B}}(n,{\text{ }}p)\)
valid approach (seen anywhere) (M1)
eg \({\text{P}}(X \geqslant 25) = 1 – {\text{P}}(X \leqslant 24),{\text{ }}1 – {\text{P}}(X < a)\)
correct working (A1)
eg \({\text{P}}(X \leqslant 24) = 0.913 \ldots ,{\text{ }}1 – 0.913 \ldots \)
\(0.0869002\)
\({\text{P}}(X \geqslant 25) = 0.0869\) A1 N2
[3 marks]
Question
Let \(C\) and \(D\) be independent events, with \({\text{P}}(C) = 2k\) and \({\text{P}}(D) = 3{k^2}\), where \(0 < k < 0.5\).
Write down an expression for \({\text{P}}(C \cap D)\) in terms of \(k\).
Given that \({\text{P}}(C \cap D) = 0.162\) find \(k\).
Find \({\text{P}}(C’|D)\).
Answer/Explanation
Markscheme
\({\text{P}}(C \cap D) = 2k \times 3{k^2}\) (A1)
\({\text{P}}(C \cap D) = 6{k^3}\) A1 N2
[2 marks]
their correct equation (A1)
eg\(\;\;\;2k \times 3{k^2} = 0.162,{\text{ }}6{k^3} = 0.162\)
\(k = 0.3\) A1 N2
METHOD 1
finding their \({\text{P}}(C’ \cap D)\) (seen anywhere) (A1)
eg \(0.4 \times 0.27,0.27 – 0.162,0.108\)
correct substitution into conditional probability formula (A1)
eg\(\;\;\;{\text{P}}(C’|D) = \frac{{{\text{P}}(C’ \cap D)}}{{0.27}},{\text{ }}\frac{{(1 – 2k)(3{k^2})}}{{3{k^2}}}\)
\({\text{P}}(C’|D) = 0.4\) A1 N2
METHOD 2
recognizing \({\text{P}}(C’|D) = {\text{P}}(C’)\) A1
finding their \({\text{P}}(C’) = 1 – {\text{P}}(C)\) (only if first line seen) (A1)
eg\(\;\;\;1 – 2k,{\text{ }}1 – 0.6\)
\({\text{P}}(C’|D) = 0.4\) A1 N2
[3 marks]
Total [7 marks]
Question
A factory has two machines, A and B. The number of breakdowns of each machine is independent from day to day.
Let \(A\) be the number of breakdowns of Machine A on any given day. The probability distribution for \(A\) can be modelled by the following table.
Let \(B\) be the number of breakdowns of Machine B on any given day. The probability distribution for \(B\) can be modelled by the following table.
On Tuesday, the factory uses both Machine A and Machine B. The variables \(A\) and \(B\) are independent.
Find \(k\).
(i) A day is chosen at random. Write down the probability that Machine A has no breakdowns.
(ii) Five days are chosen at random. Find the probability that Machine A has no breakdowns on exactly four of these days.
Find \({\text{E}}(B)\).
(i) Find the probability that there are exactly two breakdowns on Tuesday.
(ii) Given that there are exactly two breakdowns on Tuesday, find the probability that both breakdowns are of Machine A.
Answer/Explanation
Answer/Explanation
Markscheme
evidence of summing to 1 (M1)
eg\(\,\,\,\,\,\)\(0.55 + 0.3 + 0.1 + k = 1\)
\(k = 0.05{\text{ (exact)}}\) A1 N2
[2 marks]
(i) 0.55 A1 N1
(ii) recognizing binomial probability (M1)
eg\(\,\,\,\,\,\)\(X:{\text{ }}B(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right),{\text{ }}{(0.55)^4}(1 – 0.55),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n – r}}\)
\(P(X = 4) = 0.205889\)
\(P(X = 4) = 0.206\) A1 N2
[3 marks]
correct substitution into formula for \({\text{E}}(X)\) (A1)
eg\(\,\,\,\,\,\)\(0.2 + (2 \times 0.08) + (3 \times 0.02)\)
\({\text{E}}(B) = 0.42{\text{ (exact)}}\) A1 N2
[2 marks]
(i) valid attempt to find one possible way of having 2 breakdowns (M1)
eg\(\,\,\,\,\,\)\(2A,{\text{ }}2B,{\text{ }}1A\) and \(1B\), tree diagram
one correct calculation for 1 way (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\(0.1 \times 0.7,{\text{ }}0.55 \times 0.08,{\text{ }}0.3 \times 0.2\)
recognizing there are 3 ways of having 2 breakdowns (M1)
eg\(\,\,\,\,\,\)A twice or B twice or one breakdown each
correct working (A1)
eg\(\,\,\,\,\,\)\((0.1 \times 0.7) + (0.55 \times 0.08) + (0.3 \times 0.2)\)
\({\text{P(2 breakdowns)}} = 0.174{\text{ (exact)}}\) A1 N3
(ii) recognizing conditional probability (M1)
eg\(\,\,\,\,\,\)\({\text{P}}(A|B),{\text{ P}}(2A|{\text{2breakdowns}})\)
correct working (A1)
eg\(\,\,\,\,\,\)\(\frac{{0.1 \times 0.7}}{{0.174}}\)
\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402298\)
\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402\) A1 N2
[8 marks]
Question
A competition consists of two independent events, shooting at 100 targets and running for one hour.
The number of targets a contestant hits is the \(S\) score. The \(S\) scores are normally distributed with mean 65 and standard deviation 10.
The distance in km that a contestant runs in one hour is the \(R\) score. The \(R\) scores are normally distributed with mean 12 and standard deviation 2.5. The \(R\) score is independent of the \(S\) score.
Contestants are disqualified if their \(S\) score is less than 50 and their \(R\) score is less than \(x\) km.
A contestant is chosen at random. Find the probability that their \(S\) score is less than 50.
Given that 1% of the contestants are disqualified, find the value of \(x\).
Answer/Explanation
Markscheme
0.0668072
\({\text{P}}(S < 50) = 0.0668{\text{ }}({\text{accept P}}(S \leqslant 49) = 0.0548)\) A2 N2
[2 marks]
valid approach (M1)
Eg\(\,\,\,\,\,\)\({\text{P}}(S < 50) \times {\text{P}}(R < x)\)
correct equation (accept any variable) A1
eg\(\,\,\,\,\,\)\({\text{P}}(S < 50) \times {\text{P}}(R < x) = 1\% ,{\text{ }}0.0668072 \times p = 0.01,{\text{ P}}(R < x) = \frac{{0.01}}{{0.0668}}\)
finding the value of \({\text{P}}(R < x)\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{0.01}}{{0.0668}},{\text{ }}0.149684\)
9.40553
\(x = 9.41{\text{ }}({\text{accept }}x = 9.74{\text{ from }}0.0548)\) A1 N3
[4 marks]
Question
The weights, in grams, of oranges grown in an orchard, are normally distributed with a mean of 297 g. It is known that 79 % of the oranges weigh more than 289 g and 9.5 % of the oranges weigh more than 310 g.
The weights of the oranges have a standard deviation of σ.
The grocer at a local grocery store will buy the oranges whose weights exceed the 35th percentile.
The orchard packs oranges in boxes of 36.
Find the probability that an orange weighs between 289 g and 310 g.
Find the standardized value for 289 g.
Hence, find the value of σ.
To the nearest gram, find the minimum weight of an orange that the grocer will buy.
Find the probability that the grocer buys more than half the oranges in a box selected at random.
The grocer selects two boxes at random.
Find the probability that the grocer buys more than half the oranges in each box.
Answer/Explanation
Markscheme
correct approach indicating subtraction (A1)
eg 0.79 − 0.095, appropriate shading in diagram
P(289 < w < 310) = 0.695 (exact), 69.5 % A1 N2
[2 marks]
METHOD 1
valid approach (M1)
eg 1 − p, 21
−0.806421
z = −0.806 A1 N2
METHOD 2
(i) & (ii)
correct expression for z (seen anywhere) (A1)
eg \(\frac{{289 – u}}{\sigma }\)
valid approach (M1)
eg 1 − p, 21
−0.806421
z = −0.806 (seen anywhere) A1 N2
[2 marks]
METHOD 1
attempt to standardize (M1)
eg \(\sigma = \frac{{289 – 297}}{z},\,\,\frac{{289 – 297}}{\sigma }\)
correct substitution with their z (do not accept a probability) A1
eg \( – 0.806 = \frac{{289 – 297}}{\sigma },\,\,\frac{{289 – 297}}{{ – 0.806}}\)
9.92037
σ = 9.92 A1 N2
METHOD 2
(i) & (ii)
correct expression for z (seen anywhere) (A1)
eg \(\frac{{289 – u}}{\sigma }\)
valid approach (M1)
eg 1 − p, 21
−0.806421
z = −0.806 (seen anywhere) A1 N2
valid attempt to set up an equation with their z (do not accept a probability) (M1)
eg \( – 0.806 = \frac{{289 – 297}}{\sigma },\,\,\frac{{289 – 297}}{{ – 0.806}}\)
9.92037
σ = 9.92 A1 N2
[3 marks]
valid approach (M1)
eg P(W < w) = 0.35, −0.338520 (accept 0.385320), diagram showing values in a standard normal distribution
correct score at the 35th percentile (A1)
eg 293.177
294 (g) A1 N2
Note: If working shown, award (M1)(A1)A0 for 293.
If no working shown, award N1 for 293.177, N1 for 293.
Exception to the FT rule: If the score is incorrect, and working shown, award A1FT for correctly finding their minimum weight (by rounding up)
[3 marks]
evidence of recognizing binomial (seen anywhere) (M1)
eg \(X \sim {\text{B}}\left( {36,\,\,p} \right),\,\,{}_n{C_a} \times {p^a} \times {q^{n – a}}\)
correct probability (seen anywhere) (A1)
eg 0.65
EITHER
finding P(X ≤ 18) from GDC (A1)
eg 0.045720
evidence of using complement (M1)
eg 1−P(X ≤ 18)
0.954279
P(X > 18) = 0.954 A1 N2
OR
recognizing P(X > 18) = P(X ≥ 19) (M1)
summing terms from 19 to 36 (A1)
eg P(X = 19) + P(X = 20) + … + P(X = 36)
0.954279
P(X > 18) = 0.954 A1 N2
[5 marks]
correct calculation (A1)
\({0.954^2},\,\,\left( \begin{gathered}
2 \hfill \\
2 \hfill \\
\end{gathered} \right){0.954^2}{\left( {1 – 0.954} \right)^0}\)
0.910650
0.911 A1 N2
[2 marks]