Question
The aircraft for a particular flight has 72 seats. The airline’s records show that historically
for this flight only 90% of the people who purchase a ticket arrive to board the flight.
They assume this trend will continue and decide to sell extra tickets and hope that no
more than 72 passengers will arrive.
The number of passengers that arrive to board this flight is assumed to follow a binomial
distribution with a probability of 0.9.
(a) The airline sells 74 tickets for this flight. Find the probability that more than 72
passengers arrive to board the flight.
(b) (i) Write down the expected number of passengers who will arrive to board the flight if 72 tickets are sold.
(ii) Find the maximum number of tickets that could be sold if the expected number
of passengers who arrive to board the flight must be less than or equal to 72.
Each passenger pays $150 for a ticket. If too many passengers arrive, then the airline will
give $300 in compensation to each passenger that cannot board.
(c) Find, to the nearest integer, the expected increase or decrease in the money made
by the airline if they decide to sell 74 tickets rather than 72.
▶️Answer/Explanation
Ans:
Question
A Principal would like to compare the students in his school with a national standard.
He decides to give a test to eight students made up of four boys and four girls. One of
the teachers offers to find the volunteers from his class.
(a) Name the type of sampling that best describes the method used by the Principal.
The marks out of 40, for the students who took the test, are:
25, 29, 38, 37, 12, 18, 27, 31.
(b) For the eight students find
(i) the mean mark.
(ii) the standard deviation of the marks.
The national standard mark is 25.2 out of 40.
(c) Perform an appropriate test at the 5% significance level to see if the mean marks
achieved by the students in the school are higher than the national standard. It can
be assumed that the marks come from a normal population.
(d) State one reason why the test might not be valid.
Two additional students take the test at a later date and the mean mark for all ten students
is 28.1 and the standard deviation is 8.4.
For further analysis, a standardized score out of 100 for the ten students is obtained by
multiplying the scores by 2 and adding 20.
(e) For the ten students, find
(i) their mean standardized score.
(ii) the standard deviation of their standardized score.
▶️Answer/Explanation
Ans:
(a) quote
(b)(i) \(27.125 \approx 27.1\)
(ii) \(8.29815… \approx 8.30\)
(c) (let \(\mu be the national mean)
\(H_0: \mu = 25.2\)
\(H_1: \mu > 25.2\)
recognizing t-test
p-value = 0.279391…
0.279391…> 0.05
insufficient evidence to reject the null hypothesis (that the mean for the school is 25.2)
(d) EITHER
the sampling process is not random
For example:
the school asked for volunteers
the students were selected from a single class
OR
the quota might not be representative of the student population
For example:
the school may have only 4 boys and 400 girls.
(e) (i) \((28.1 \times 2 + 20 = ) 76.2
(ii) \(8.4 \times 2\)
= 16.8
Question
A biased coin is weighted such that the probability of obtaining a head is \(\frac{4}{7}\). The coin is tossed 6 times and X denotes the number of heads observed. Find the value of the ratio \(\frac{{{\text{P}}(X = 3)}}{{{\text{P}}(X = 2)}}\).
▶️Answer/Explanation
Markscheme
recognition of \(X \sim {\text{B}}\left( {6,\frac{4}{7}} \right)\) (M1)
\({\text{P}}(X = 3) = \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right){\left( {\frac{4}{7}} \right)^3}{\left( {\frac{3}{7}} \right)^3}\left( { = 20 \times \frac{{{4^3} \times {3^3}}}{{{7^6}}}} \right)\) A1
\({\text{P}}(X = 2) = \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right){\left( {\frac{4}{7}} \right)^2}{\left( {\frac{3}{7}} \right)^4}\left( { = 15 \times \frac{{{4^2} \times 34}}{{{7^6}}}} \right)\) A1
\(\frac{{{\text{P}}(X = 3)}}{{{\text{P}}(X = 2)}} = \frac{{80}}{{45}}\left( { = \frac{{16}}{9}} \right)\) A1
[4 marks]
Question
a.On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any one of these games is \(\frac{2}{3}\).
Show that the probability that Alfred wins exactly 4 of the games is \(\frac{{80}}{{243}}\).[3]
b.(i) Explain why the total number of possible outcomes for the results of the 6 games is 64.
(ii) By expanding \({(1 + x)^6}\) and choosing a suitable value for x, prove
\[64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\]
(iii) State the meaning of this equality in the context of the 6 games played.[4]
c.The following day Alfred and Beatrice play the 6 games again. Assume that the probability that Alfred wins any one of these games is still \(\frac{2}{3}\).
(i) Find an expression for the probability Alfred wins 4 games on the first day and 2 on the second day. Give your answer in the form \({\left( {\begin{array}{*{20}{c}}
6 \\
r
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^s}{\left( {\frac{1}{3}} \right)^t}\) where the values of r, s and t are to be found.
(ii) Using your answer to (c) (i) and 6 similar expressions write down the probability that Alfred wins a total of 6 games over the two days as the sum of 7 probabilities.
(iii) Hence prove that \(\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}\).[9]
d.Alfred and Beatrice play n games. Let A denote the number of games Alfred wins. The expected value of A can be written as \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{a^r}}}{{{b^n}}}\).
(i) Find the values of a and b.
(ii) By differentiating the expansion of \({(1 + x)^n}\), prove that the expected number of games Alfred wins is \(\frac{{2n}}{3}\).[6]
▶️Answer/Explanation
Markscheme
\(B\left( {6,\frac{2}{3}} \right)\) (M1)
\(p(4) = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2}\) A1
\(\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) = 15\) A1
\( = 15 \times \frac{{{2^4}}}{{{3^6}}} = \frac{{80}}{{243}}\) AG
[3 marks]
(i) 2 outcomes for each of the 6 games or \({2^6} = 64\) R1
(ii) \({(1 + x)^6} = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)x + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right){x^2} + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right){x^3} + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){x^4} + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right){x^5} + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right){x^6}\) A1
Note: Accept \(^n{C_r}\) notation or \(1 + 6x + 15{x^2} + 20{x^3} + 15{x^4} + 6{x^5} + {x^6}\)
setting x = 1 in both sides of the expression R1
Note: Do not award R1 if the right hand side is not in the correct form.
\(64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\) AG
(iii) the total number of outcomes = number of ways Alfred can win no games, plus the number of ways he can win one game etc. R1
[4 marks]
(i) Let \({\text{P}}(x,{\text{ }}y)\) be the probability that Alfred wins x games on the first day and y on the second.
\({\text{P(4, 2)}} = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^4} \times {\left( {\frac{1}{3}} \right)^2} \times \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^2} \times {\left( {\frac{1}{3}} \right)^4}\) M1A1
\({\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) or \({\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A1
r = 2 or 4, s = t = 6
(ii) P(Total = 6) =
P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0) (M1)
\( = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + … + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A2
\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)}^2}} \right)\)
Note: Accept any valid sum of 7 probabilities.
(iii) use of \(\left( {\begin{array}{*{20}{c}}
6 \\
i
\end{array}} \right) = \left( {\begin{array}{*{20}{l}}
6 \\
{6 – i}
\end{array}} \right)\) (M1)
(can be used either here or in (c)(ii))
P(wins 6 out of 12) \( = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^6} \times {\left( {\frac{1}{3}} \right)^6} = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) A1
\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)}^2}} \right) = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) A1
therefore \({\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2} = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) AG
[9 marks]
(i) \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} {\left( {\frac{2}{3}} \right)^r}{\left( {\frac{1}{3}} \right)^{n – r}} = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}\)
(a = 2, b = 3) M1A1
Note: M0A0 for a = 2, b = 3 without any method.
(ii) \(n{(1 + x)^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{x^{r – 1}}\) A1A1
(sigma notation not necessary)
(if sigma notation used also allow lower limit to be r = 0)
let x = 2 M1
\(n{3^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{2^{r – 1}}\)
multiply by 2 and divide by \({3^n}\) (M1)
\(\frac{{2n}}{3} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r\frac{{{2^r}}}{{{3^n}}}\left( { = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}} \right)\) AG
[6 marks]
Examiners report
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(a) Candidates need to be aware how to work out binomial coefficients without a calculator
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(b) (ii) A surprising number of candidates chose to work out the values of all the binomial coefficients (or use Pascal’s triangle) to make a total of 64 rather than simply putting 1 into the left hand side of the expression.
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(d) This was poorly done. Candidates were not able to manipulate expressions given using sigma notation.
Question
A biased coin is tossed five times. The probability of obtaining a head in any one throw is \(p\).
Let \(X\) be the number of heads obtained.
a.Find, in terms of \(p\), an expression for \({\text{P}}(X = 4)\).[2]
(ii) For this value of \(p\), determine the expected number of heads.[6]
▶️Answer/Explanation
Markscheme
\(X \sim {\text{B}}(5,{\text{ }}p)\) (M1)
\({\text{P}}(X = 4) = \left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right){p^4}(1 – p)\) (or equivalent) A1
[2 marks]
(i) \(\frac{{\text{d}}}{{{\text{d}}p}}(5{p^4} – 5{p^5}) = 20{p^3} – 25{p^4}\) M1A1
\(5{p^3}(4 – 5p) = 0 \Rightarrow p = \frac{4}{5}\) M1A1
Note: Do not award the final A1 if \(p = 0\) is included in the answer.
(ii) \({\text{E}}(X) = np = 5\left( {\frac{4}{5}} \right)\) (M1)
\( = 4\) A1
[6 marks]
Examiners report
This question was generally very well done and posed few problems except for the weakest candidates.
This question was generally very well done and posed few problems except for the weakest candidates.