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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
Jan plays a game where she tosses two fair six-sided dice. She wins a prize if the sum of her scores is 5.
Jan tosses the two dice once. Find the probability that she wins a prize.
Jan tosses the two dice 8 times. Find the probability that she wins 3 prizes.
Answer/Explanation
Markscheme
36 outcomes (seen anywhere, even in denominator) (A1)
valid approach of listing ways to get sum of 5, showing at least two pairs (M1)
e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram
\({\rm{P(prize)}} = \frac{4}{{36}}\) \(\left( { = \frac{1}{9}} \right)\) A1 N3
[3 marks]
recognizing binomial probability (M1)
e.g. \({\rm{B}}\left( {8,\frac{1}{9}} \right)\) , binomial pdf, \(\left( {\begin{array}{*{20}{c}}
8\\
3
\end{array}} \right){\left( {\frac{1}{9}} \right)^3}{\left( {\frac{8}{9}} \right)^5}\)
\({\text{P(3 prizes)}} = 0.0426\) A1 N2
[2 marks]
Question
A company uses two machines, A and B, to make boxes. Machine A makes \(60\% \) of the boxes.
\(80\% \) of the boxes made by machine A pass inspection.
\(90\% \) of the boxes made by machine B pass inspection.
A box is selected at random.
Find the probability that it passes inspection.
The company would like the probability that a box passes inspection to be 0.87.
Find the percentage of boxes that should be made by machine B to achieve this.
Answer/Explanation
Markscheme
evidence of valid approach involving A and B (M1)
e.g. \({\rm{P}}(A \cap {\rm{pass}}) + {\rm{P}}(B \cap {\rm{pass}})\) , tree diagram
correct expression (A1)
e.g. \({\rm{P}}({\rm{pass}}) = 0.6 \times 0.8 + 0.4 \times 0.9\)
\({\rm{P}}({\rm{pass}}) = 0.84\) A1 N2
[3 marks]
evidence of recognizing complement (seen anywhere) (M1)
e.g. \({\rm{P}}(B) = x\) , \({\rm{P}}(A) = 1 – x\) , \(1 – {\rm{P}}(B)\) , \(100 – x\) , \(x + y = 1\)
evidence of valid approach (M1)
e.g. \(0.8(1 – x) + 0.9x\) , \(0.8x + 0.9y\)
correct expression A1
e.g. \(0.87 = 0.8(1 – x) + 0.9x\) , \(0.8 \times 0.3 + 0.9 \times 0.7 = 0.87\) , \(0.8x + 0.9y = 0.87\)
\(70\% \) from B A1 N2
[4 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
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
The following table shows a probability distribution for the random variable \(X\), where \({\text{E}}(X) = 1.2\).
A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable \(X\).
A game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.
Find \(q\).
Find \(p\).
Write down the probability of drawing three blue marbles.
Explain why the probability of drawing three white marbles is \(\frac{1}{6}\).
The bag contains a total of ten marbles of which \(w\) are white. Find \(w\).
Grant plays the game until he wins two prizes. Find the probability that he wins his second prize on his eighth attempt.
Answer/Explanation
Markscheme
correct substitution into \({\text{E}}(X)\) formula (A1)
eg\(\,\,\,\,\,\)\(0(p) + 1(0.5) + 2(0.3) + 3(q) = 1.2\)
\(q = \frac{1}{{30}}\), 0.0333 A1 N2
[2 marks]
evidence of summing probabilities to 1 (M1)
eg\(\,\,\,\,\,\)\(p + 0.5 + 0.3 + q = 1\)
\(p = \frac{1}{6},{\text{ }}0.167\) A1 N2
[2 marks]
\({\text{P (3 blue)}} = \frac{1}{{30}},{\text{ }}0.0333\) A1 N1
[1 mark]
valid reasoning R1
eg\(\,\,\,\,\,\)\({\text{P (3 white)}} = {\text{P(0 blue)}}\)
\({\text{P(3 white)}} = \frac{1}{6}\) AG N0
[1 mark]
valid method (M1)
eg\(\,\,\,\,\,\)\({\text{P(3 white)}} = \frac{w}{{10}} \times \frac{{w – 1}}{9} \times \frac{{w – 2}}{8},{\text{ }}\frac{{_w{C_3}}}{{_{10}{C_3}}}\)
correct equation A1
eg\(\,\,\,\,\,\)\(\frac{w}{{10}} \times \frac{{w – 1}}{9} \times \frac{{w – 2}}{8} = \frac{1}{6},{\text{ }}\frac{{_w{C_3}}}{{_{10}{C_3}}} = 0.167\)
\(w = 6\) A1 N2
[3 marks]
recognizing one prize in first seven attempts (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right),{\text{ }}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6}\)
correct working (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right){\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6},{\text{ }}0.390714\)
correct approach (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right){\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6} \times \frac{1}{6}\)
0.065119
0.0651 A1 N2
[4 marks]
Question
The following table shows a probability distribution for the random variable \(X\), where \({\text{E}}(X) = 1.2\).
A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable \(X\).
A game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.
Jill plays the game nine times. Find the probability that she wins exactly two prizes.
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\({\text{B}}(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n – r}},{\text{ }}{(0.167)^2}{(0.833)^7},{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ 2 \end{array}} \right)\)
0.279081
0.279 A1 N2
[2 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]