Find the probability that he wins exactly four games.

Write down an expression, in terms of p , for the probability that he wins exactly four games.

Hence, find the values of p such that the probability that he wins exactly four games is 0.15.

evidence of recognizing binomial probability (may be seen in (b) or (c))     (M1)

e.g. probability \( = \left( {\begin{array}{*{20}{c}}
7\\
4
\end{array}} \right){(0.9)^4}{(0.1)^3}\) , \(X \sim {\rm{B}}(7,0.9)\) , complementary probabilities

probability \(= 0.0230\)     A1    N2

[2 marks]

correct expression     A1A1     N2

e.g. \(\left( {\begin{array}{*{20}{c}}
7\\
4
\end{array}} \right){p^4}{(1 - p)^3}\) , \(35{p^4}{(1 - p)^3}\)

Note: Award A1 for binomial coefficient (accept \(\left( {\begin{array}{*{20}{c}}
7\\
3
\end{array}} \right)\) ) , A1 for \({p^4}{(1 - p)^3}\) .

[2 marks]

evidence of attempting to solve their equation     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
7\\
4
\end{array}} \right){p^4}{(1 - p)^3} = 0.15\) , sketch

\(p = 0.356\), \(0.770\)     A1A1     N3

[3 marks]

Parts of this question were handled very well by a great many candidates. Most were able to recognize the binomial condition and had little difficulty with part (a). However, more than a few reported the answer as 0.23, thus incurring the accuracy penalty.

Those candidates that were successful in part (a) could easily write the required expression for part (b).

In part (c), many candidates set up the question correctly or set their expression from (b) equal to 0.15, however few candidates considered the GDC as a method to solve the equation. Rather, those who attempted usually tried to expand the polynomial, and still did not use the GDC to solve this equation. A graphical approach to the solution would reveal that there are two solutions for p, but few caught this subtlety.