Show that the probability that Alfred wins exactly 4 of the games is \(\frac{{80}}{{243}}\).

(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.

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 \(\left( {\begin{array}{*{20}{c}}
  {12} \\
  6
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
  6 \\
  0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
  6 \\
  6
\end{array}} \right)^2}\).

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 \(\frac{{2n}}{3}\).

\(B\left( {6,\frac{2}{3}} \right)\)     (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]

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.