(a)     Ahmed is typing Section A of a mathematics examination paper. The number of mistakes that he makes, X , can be modelled by a Poisson distribution with mean 3.2 . Find the probability that Ahmed makes exactly four mistakes.

(b)     His colleague, Levi, is typing Section B of this paper. The number of mistakes that he makes, Y , can be modelled by a Poisson distribution with mean m.

(i)     If \({\text{E}}({Y^2}) = 5.5\) , find the value of m.

(ii)     Find the probability that Levi makes exactly three mistakes.

(c)     Given that X and Y are independent, find the probability that Ahmed makes exactly four mistakes and Levi makes exactly three mistakes.

(a)     \(X \sim {\text{Po(3.2)}}\)

\({\text{P}}(X = 4) = \frac{{{{\text{e}}^{ - 3.2}}{{3.2}^4}}}{{4!}}\)

= 0.178     A1

(b)     (i)     \({\text{Var}}(Y) = {\text{E}}({Y^2}) - {{\text{E}}^2}(Y)\)     (M1)

\(m = 5.5 - {m^2}\)     A1

m = 1.90 (or m = –2.90 which is valid)     A1

(ii)     \(Y \sim {\text{Po(1.90)}}\)

\({\text{P}}(Y = 3) = \frac{{{{\text{e}}^{ - 1.90}}{{1.90}^4}}}{{3!}}\)     (M1)

= 0.171     A1

(c)     Required probability \( = 0.171 \times 0.178 = 0.0304\) (accept 0.0305)     (M1)A1

[8 marks]

Part (a) was correctly solved by most candidates, either using the formula or directly from their GDC. Solutions to (b), however, were extremely disappointing with the majority of candidates giving \(\sqrt 5 \), incorrectly, as their value of m . It was possible to apply follow through in (b) (ii) and (c) which were well done in general.