Find the probability that

(i)     he hits the target exactly 4 times in his first 8 shots;

(ii)     he hits the target for the \({4^{{\text{th}}}}\) time with his \({8^{{\text{th}}}}\) shot.

Ben hits the target for the \({10^{{\text{th}}}}\) time with his \({X^{{\text{th}}}}\) shot.

(i)     Determine the expected value of the random variable X.

(ii)     Write down an expression for \({\text{P}}(X = x)\) and show that

\[\frac{{{\text{P}}(X = x)}}{{{\text{P}}(X = x - 1)}} = \frac{{3(x - 1)}}{{5(x - 10)}}.\]

(iii)     Hence, or otherwise, find the most likely value of X.

(i)     the number of hits, \(X \sim {\text{B(8, 0.4)}}\)     (A1)

\(P(X = 4) = \left( {\begin{array}{*{20}{c}}
  8 \\
  4
\end{array}} \right) \times {0.4^4} \times {0.6^4}\)     (M1)

= 0.232     A1

Note: Accept any answer that rounds to 0.23.

(ii)     let the \({4^{{\text{th}}}}\) hit occur on the \({Y^{{\text{th}}}}\) shot so that \(Y \sim {\text{NB(4, 0.4)}}\)     (A1)

\(P(Y = 8) = \left( {\begin{array}{*{20}{c}}
  7 \\
  3
\end{array}} \right) \times {0.4^4} \times {0.6^4}\)     (M1)

= 0.116     A1

Note: Accept any answer that rounds to 0.12.

[6 marks]

(i)     \(X \sim {\text{NB(10, 0.4)}}\)     (M1)

\({\text{E}}(X) = \frac{{10}}{{0.4}} = 25\)     A1

(ii)     let \({{\text{P}}_x}\) denote \({\text{P}}(X = x)\)

\({P_x} = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  9
\end{array}} \right) \times {0.4^{10}} \times {0.6^{x - 10}}\)     A1

\(\frac{{{P_x}}}{{{P_{x - 1}}}} = \frac{{\left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  9
\end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x - 10}}}}{{\left( {\begin{array}{*{20}{c}}
  {x - 2} \\
  9
\end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x - 11}}}}\)     M1A1

\( = \frac{{(x - 1)!}}{{9!(x - 10)!}} \times \frac{{9!(x - 11)! \times 0.6}}{{(x - 2)!}}\)     A1

Note: Award A1 for correct evaluation of combinatorial terms.

\( = \frac{{3(x - 1)}}{{5(x - 10)}}\)     AG

(iii)     \({{\text{P}}_x} > {{\text{P}}_{x - 1}}\) as long as

\(3x - 3 > 5x - 50\)     (M1)

i.e. \(x < 23.5\)     (A1)

the most likely value is 23     A1

Note: Allow solutions based on creating a table of values of \({{\text{P}}_x}\).

[9 marks]

Part (a) was well answered in general although some candidates were unable to distinguish between the binomial and negative binomial distributions.

In (b)(ii), most candidates knew what to do but algebraic errors were not uncommon. Candidates often used equal instead of inequality signs and this was accepted if it led to \(x = 23.5\). The difficulty for these candidates was whether to choose \(23\) or \(24\) for the final answer and some made the wrong choice. Some candidates failed to see the relevance of the result in (b)(ii) to finding the most likely value of \(X\) and chose an ‘otherwise’ method, usually by creating a table of probabilities and selecting the largest.