Question 3. [Maximum mark: 8]
At a school, 70% of the students play a sport and 20% of the students are involved in
theatre. 18% of the students do neither activity.
A student is selected at random.
(a) Find the probability that the student plays a sport and is involved in theatre. [2]
(b) Find the probability that the student is involved in theatre, but does not play a sport. [2]
At the school 48% of the students are girls, and 25% of the girls are involved in theatre.
A student is selected at random. Let G be the event “the student is a girl” and let T be the
event “the student is involved in theatre”.
(c) Find P(G ∩ T). [2]
(d) Determine if the events G and T are independent. Justify your answer.
▶️Answer/Explanation
(a) EITHER
\(P\left ( S \right )+P\left ( T \right )+P\left ( S{}’\cap T{}’ \right )-P\left ( S\cap T \right )=1\) OR \( P\left ( S\cup T \right )=P\left ( \left ( S{}’\cap T{}’ \right ) {}’\right )\)
\(0.7+0.2+0.18-P\left ( S\cap T \right )=1\) OR \(P\left ( S\cup T \right )=1-0.18\)
OR
a clearly labelled Venn diagram
THEN
\(P\left ( S\cap T\right )=0.08\) (accept 8%)
(b) EITHER
\(P\left ( T\cap S{}’ \right )=P\left ( T \right )-P\left ( T\cap S \right )\left ( =0.2-0.08 \right )\) OR \(P\left ( T\cap S{}’ \right )=P\left ( T\cup S \right )-P\left ( S \right )\left ( =0.82-0.7 \right )\)
OR
a clearly labelled Venn diagram including \(P\left ( S \right ),P\left ( T \right )\) and \(P\left ( S \cap T\right )\)
THEN
= 0.12 (accept 12%)
(c)\( P\left ( G \cap T\right )=P\left ( T/G \right )P\left ( G \right )\left ( 0.25\times 0.48 \right )=0.12\)
(d) METHOD 1
\(P\left ( G \right )\times P\left ( T \right )\left ( =0.48\times 0.2\right )=0.096\)
\(P\left ( G \right )\times P\left ( T \right )\neq P\left ( G\cap T \right )\Rightarrow\) G and T arer not independent
METHOD 2
\(P\left ( T|G \right )=0.25\)
\(P\left ( T|G \right )\neq P\left ( T \right )\Rightarrow\) G and T are not independent
Question: [Maximum mark: 6]
Events A and B are independent and P(A) = 3P(B).
Given that P(A ∪ B) = 0.68, find P(B).
▶️Answer/Explanation
Ans:
P(A∪B) = P(A) + P(B) – P(A∩B) = 0.68
substitution of P(A). P(B) for P(A∩B) in P(A∪B)
P(A) + P(B) – P(A) P(B) (=0.68)
substitution of 3P(B) for P(A)
3P(B) +P(B)-3P(B)P(B) = 0.68 (or equivalent)
Note: The first two marks are independent of each other.
attempts to solve their quadratic equation
\(P(B)= 0.2, 1.133…\left ( \frac{1}{5},\frac{17}{15} \right )\)
\(P(B)= 0.2\left ( =\frac{1}{5} \right )\)
Note: Award A1 if both answers are given as final answers for P(B) .
Question
At a school, 70 % of the students play a sport and 20 % of the students are involved in theatre. 18 % of the students do neither activity.
A student is selected at random.
(a) Find the probability that the student plays a sport and is involved in theatre. [2]
(b) Find the probability that the student is involved in theatre, but does not play a sport. [2]
At the school 48 % of the students are girls, and 25 % of the girls are involved in theatre.
A student is selected at random. Let G be the event “the student is a girl” and let T be the event “the student is involved in theatre”.
(c) Find P (G ∩ T ) . [2]
(d) Determine if the events G and T are independent. Justify your answer. [2]
▶️Answer/Explanation
Ans:
(a) By using the combined events formula, we have $\begin{eqnarray} \text{P}\left(S\cup T\right) = \text{P}\left(S\right)+\text{P}\left(T\right)-\text{P}\left(S\cap T\right) \nonumber \\ 0.82 = 0.7+0.2-\text{P}\left(S\cap T\right) \nonumber \\ \text{P}\left(S\cap T\right) = 0.08. \end{eqnarray}$ (b) $\begin{eqnarray} \text{P}\left(T\cap S’\right) &=& \text{P}\left(t\right)-\text{P}\left(S\cap T\right) \nonumber \\ &=& 0.2-0.08 \nonumber \\ &=& 0.12. \end{eqnarray}$ (c) $\begin{eqnarray} \text{P}\left(G\cap T\right) &=& 0.48\times 0.25 \nonumber \\ &=& 0.12. \end{eqnarray}$ (d) Since $\begin{eqnarray} \text{P}\left(G\right)\times\text{P}\left(T\right) &=& 0.48\times 0.2 \nonumber \\ &=& 0.096 \nonumber \\ &\neq & 0.12 \nonumber \\ &=& \text{P}\left(G\cap T\right), \end{eqnarray}$ $G$ and $T$ are not independent.
Question
A Chocolate Shop advertises free gifts to customers that collect three vouchers. The vouchers are placed at random into 10% of all chocolate bars sold at this shop. Kati buys some of these bars and she opens them one at a time to see if they contain a voucher. Let \({\text{P}}(X = n)\) be the probability that Kati obtains her third voucher on the \(n{\text{th}}\) bar opened.
(It is assumed that the probability that a chocolate bar contains a voucher stays at 10% throughout the question.)
It is given that \({\text{P}}(X = n) = \frac{{{n^2} + an + b}}{{2000}} \times {0.9^{n – 3}}\) for \(n \geqslant 3,{\text{ }}n \in \mathbb{N}\).
Kati’s mother goes to the shop and buys \(x\) chocolate bars. She takes the bars home for Kati to open.
a.Show that \({\text{P}}(X = 3) = 0.001\) and \({\text{P}}(X = 4) = 0.0027\).[3]
b.Find the values of the constants \(a\) and \(b\).[5]
c.Deduce that \(\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n – 1)}} = \frac{{0.9(n – 1)}}{{n – 3}}\) for \(n > 3\).[4]
d.(i) Hence show that \(X\) has two modes \({m_1}\) and \({m_2}\).
(ii) State the values of \({m_1}\) and \({m_2}\).[5]
e.Determine the minimum value of \(x\) such that the probability Kati receives at least one free gift is greater than 0.5.[3]
▶️Answer/Explanation
Markscheme
\({\text{P}}(X = 3) = {(0.1)^3}\) A1
\( = 0.001\) AG
\({\text{P}}(X = 4) = {\text{P}}(VV\bar VV) + {\text{P}}(V\bar VVV) + {\text{P}}(\bar VVVV)\) (M1)
\( = 3 \times {(0.1)^3} \times 0.9\) (or equivalent) A1
\( = 0.0027\) AG
[3 marks]
METHOD 1
attempting to form equations in \(a\) and \(b\) M1
\(\frac{{9 + 3a + b}}{{2000}} = \frac{1}{{1000}}{\text{ }}(3a + b = – 7)\) A1
\(\frac{{16 + 4a + b}}{{2000}} \times \frac{9}{{10}} = \frac{{27}}{{10\,000}}{\text{ }}(4a + b = – 10)\) A1
attempting to solve simultaneously (M1)
\(a = – 3,{\text{ }}b = 2\) A1
METHOD 2
\({\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n – 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n – 3}}\) M1
\( = \frac{{(n – 1)(n – 2)}}{{2000}} \times {0.9^{n – 3}}\) (M1)A1
\( = \frac{{{n^2} – 3n + 2}}{{2000}} \times {0.9^{n – 3}}\) A1
\(a = – 3,b = 2\) A1
Note: Condone the absence of \({0.9^{n – 3}}\) in the determination of the values of \(a\) and \(b\).
[5 marks]
METHOD 1
EITHER
\({\text{P}}(X = n) = \frac{{{n^2} – 3n + 2}}{{2000}} \times {0.9^{n – 3}}\) (M1)
OR
\({\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n – 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n – 3}}\) (M1)
THEN
\( = \frac{{(n – 1)(n – 2)}}{{2000}} \times {0.9^{n – 3}}\) A1
\({\text{P}}(X = n – 1) = \frac{{(n – 2)(n – 3)}}{{2000}} \times {0.9^{n – 4}}\) A1
\(\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n – 1)}} = \frac{{(n – 1)(n – 2)}}{{(n – 2)(n – 3)}} \times 0.9\) A1
\( = \frac{{0.9(n – 1)}}{{n – 3}}\) AG
METHOD 2
\(\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n – 1)}} = \frac{{\frac{{{n^2} – 3n + 2}}{{2000}} \times {{0.9}^{n – 3}}}}{{\frac{{{{(n – 1)}^2} – 3(n – 1) + 2}}{{2000}} \times {{0.9}^{n – 4}}}}\) (M1)
\( = \frac{{0.9({n^2} – 3n + 2)}}{{({n^2} – 5n + 6)}}\) A1A1
Note: Award A1 for a correct numerator and A1 for a correct denominator.
\( = \frac{{0.9(n – 1)(n – 2)}}{{(n – 2)(n – 3)}}\) A1
\( = \frac{{0.9(n – 1)}}{{n – 3}}\) AG
[4 marks]
(i) attempting to solve \(\frac{{0.9(n – 1)}}{{n – 3}} = 1\) for \(n\) M1
\(n = 21\) A1
\(\frac{{0.9(n – 1)}}{{n – 3}} < 1 \Rightarrow n > 21\) R1
\(\frac{{0.9(n – 1)}}{{n – 3}} > 1 \Rightarrow n < 21\) R1
\(X\) has two modes AG
Note: Award R1R1 for a clearly labelled graphical representation of the two inequalities (using \(\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n – 1)}}\)).
(ii) the modes are 20 and 21 A1
[5 marks]
METHOD 1
\(Y \sim {\text{B}}(x,{\text{ }}0.1)\) (A1)
attempting to solve \({\text{P}}(Y \geqslant 3) > 0.5\) (or equivalent eg \(1 – {\text{P}}(Y \leqslant 2) > 0.5\)) for \(x\) (M1)
Note: Award (M1) for attempting to solve an equality (obtaining \(x = 26.4\)).
\(x = 27\) A1
METHOD 2
\(\sum\limits_{n = 0}^x {{\text{P}}(X = n) > 0.5} \) (A1)
attempting to solve for \(x\) (M1)
\(x = 27\) A1
[3 marks]
Question
Each of the 25 students in a class are asked how many pets they own. Two students own three pets and no students own more than three pets. The mean and standard deviation of the number of pets owned by students in the class are \(\frac{{18}}{{25}}\) and \(\frac{{24}}{{25}}\) respectively.
Find the number of students in the class who do not own a pet.
▶️Answer/Explanation
Markscheme
METHOD 1
let p have no pets, q have one pet and r have two pets (M1)
p + q + r + 2 = 25 (A1)
0p + 1q + 2r + 6 = 18 A1
Note: Accept a statement that there are a total of 12 pets.
attempt to use variance equation, or evidence of trial and error (M1)
\(\frac{{0p + 1q + 4r + 18}}{{25}} – {\left( {\frac{{18}}{{25}}} \right)^2} = {\left( {\frac{{24}}{{25}}} \right)^2}\) (A1)
attempt to solve a system of linear equations (M1)
p = 14 A1
METHOD 2
(M1)
\(p + q + r + \frac{2}{{25}} = 1\) (A1)
\(q + 2r + \frac{6}{{25}} = \frac{{18}}{{25}}\left( { \Rightarrow q + 2r = \frac{{12}}{{25}}} \right)\) A1
\(q + 4r + \frac{{18}}{{25}} – {\left( {\frac{{18}}{{25}}} \right)^2} = \frac{{576}}{{625}}\left( { \Rightarrow q + 4r = \frac{{18}}{{25}}} \right)\) (M1)(A1)
\(q = \frac{6}{{25}},\,\,r = \frac{3}{{25}}\) (M1)
\(p = \frac{{14}}{{25}}\) A1
so 14 have no pets
[7 marks]
Examiners report
Question
The mean number of squirrels in a certain area is known to be 3.2 squirrels per hectare of woodland. Within this area, there is a 56 hectare woodland nature reserve. It is known that there are currently at least 168 squirrels in this reserve.
Assuming the population of squirrels follow a Poisson distribution, calculate the probability that there are more than 190 squirrels in the reserve.
▶️Answer/Explanation
Markscheme
X is number of squirrels in reserve
X ∼ Po(179.2) A1
Note: Award A1 if 179.2 or 56 × 3.2 seen or implicit in future calculations.
recognising conditional probability M1
P(X > 190 | X ≥ 168)
\( = \frac{{{\text{P}}\left( {X > 190} \right)}}{{{\text{P}}\left( {X \geqslant 168} \right)}} = \left( {\frac{{0.19827 \ldots }}{{0.80817 \ldots }}} \right)\) (A1)(A1)
= 0.245 A1
[5 marks]