Given that \(2{\text{P}}(X = 4) = {\text{P}}(X = 5)\), show that \(m = 10\).

Given that \(X \le 11\), find the probability that \(X = 6\).

\(2\frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}} = \frac{{{{\text{e}}^{ - m}}{m^5}}}{{5!}}\)     M1A1

\(\frac{2}{{4!}} = \frac{m}{{5!}}\;\;\;\)or other simplification     M1

Note:     accept a labelled graph showing clearly the solution to the equation. Do not accept simple verification that \(m = 10\) is a solution.

\( \Rightarrow m = 10\)     AG

[3 marks]

\({\text{P}}(X = 6|X \le 11) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(X \le 11)}}\)     (M1) (A1)

\( = \frac{{0.063055 \ldots }}{{0.696776 \ldots }}\)     (A1)

\( = 0.0905\)     A1

[4 marks]

Total [7 marks]

Most candidates successfully finished part (a) with two fundamental errors occurring regularly. Either e was granted powers of 4 and 5 or an attempt to show that the value of \(m\) was 10 was made by evaluation.

Part (b) was challenging for many candidates that showed that the idea of conditional probability was poorly understood. There were many incorrect solutions where often candidates only found \({\text{P}}(X = 6)\).