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Question
A bag contains four gold balls and six silver balls.
Two balls are drawn at random from the bag, with replacement. Let \(X\) be the number of gold balls drawn from the bag.
Fourteen balls are drawn from the bag, with replacement.
(i) Find \({\rm{P}}(X = 0)\) .
(ii) Find \({\rm{P}}(X = 1)\) .
(iii) Hence, find \({\rm{E}}(X)\) .
Hence, find \({\rm{E}}(X)\) .
Find the probability that exactly five of the balls are gold.
Find the probability that at most five of the balls are gold.
Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correct to two decimal places.
Answer/Explanation
Markscheme
METHOD 1
(i) appropriate approach (M1)
eg \(\frac{6}{{10}} \times \frac{6}{{10}}\) , \(\frac{6}{{10}} \times \frac{5}{9}\) , \(\frac{6}{{10}} \times \frac{5}{{10}}\)
\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\) A1 N2
(ii) multiplying one pair of gold and silver probabilities (M1)
eg \(\frac{6}{{10}} \times \frac{4}{{10}}\) , \(\frac{6}{{10}} \times \frac{4}{9}\) , 0.24
adding the product of both pairs of gold and silver probabilities (M1)
eg \(\frac{6}{{10}} \times \frac{4}{{10}} \times 2\) , \(\frac{6}{{10}} \times \frac{4}{9} + \frac{4}{{10}} \times \frac{6}{9}\)
\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\) A1 N3
(iii)
\({\rm{P}}(X = 2) = 0.16\) (seen anywhere) (A1)
correct substitution into formula for \({\rm{E}}(X)\) (A1)
eg \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)
\({\rm{E}}(X) = \frac{4}{5} = 0.8\) A1 N3
METHOD 2
(i) evidence of recognizing binomial (may be seen in part (ii)) (M1)
eg \(X \sim {\rm{B}}(2,0.6)\) , \(\left( \begin{array}{l}
2\\
0
\end{array} \right){(0.4)^2}{(0.6)^0}\)
correct probability for use in binomial (A1)
eg \(p = 0.4\) , \(X \sim {\rm{B}}(2,0.4)\) , \(^2{C_0}{(0.4)^0}{(0.6)^2}\)
\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\) A1 N3
(ii) correct set up (A1)
eg \(_2{C_1}{(0.4)^1}{(0.6)^1}\)
\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\) A1 N2
(iii)
attempt to substitute into \(np\) (M1)
eg \(2 \times 0.6\)
correct substitution into \(np\) (A1)
eg \(2 \times 0.4\)
\({\rm{E}}(X) = \frac{4}{5} = 0.8\) A1 N3
[8 marks]
METHOD 1
\({\rm{P}}(X = 2) = 0.16\) (seen anywhere) (A1)
correct substitution into formula for \({\rm{E}}(X)\) (A1)
eg \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)
\({\rm{E}}(X) = \frac{4}{5} = 0.8\) A1 N3
METHOD 2
attempt to substitute into \(np\) (M1)
eg \(2 \times 0.6\)
correct substitution into \(np\) (A1)
eg \(2 \times 0.4\)
\({\rm{E}}(X) = \frac{4}{5} = 0.8\) A1 N3
[3 marks]
Let \(Y\) be the number of gold balls drawn from the bag.
evidence of recognizing binomial (seen anywhere) (M1)
eg \(_{14}{C_5}{(0.4)^5}{(0.6)^9}\) , \({\rm{B}}(14,0.4)\)
\({\rm{P}}(Y = 5) = 0.207\) A1 N2
[2 marks]
recognize need to find \({\rm{P}}(Y \le 5)\) (M1)
\({\rm{P}}(Y \le 5) = 0.486\) A1 N2
[2 marks]
Let \(Y\) be the number of gold balls drawn from the bag.
recognizing conditional probability (M1)
eg \({\rm{P}}(A|B)\) , \({\rm{P}}(Y = 5|Y \le 5)\) , \(\frac{{{\rm{P}}(Y = 5)}}{{{\rm{P}}(Y \le 5)}}\) , \(\frac{{0.207}}{{0.486}}\)
\({\rm{P}}(Y = 5|Y \le 5) = 0.42522518\) (A1)
\({\rm{P}}(Y = 5|Y \le 5) = 0.43\) (to \(2\) dp) A1 N2
[3 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
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]