Students sign up at a desk for an activity during the course of an afternoon. The arrival of each student is independent of the arrival of any other student and the number of students arriving per hour can be modelled as a Poisson distribution with a mean of \(\lambda \).

The desk is open for 4 hours. If exactly 5 people arrive to sign up for the activity during that time find the probability that exactly 3 of them arrived during the first hour.

P(3 in the first hour) \( = \frac{{{\lambda ^3}{e^{ - \lambda }}}}{{3!}}\)     A1

number to arrive in the four hours follows \(Po(4\lambda )\)     M1

P(5 arrive in total) \( = \frac{{{{(4\lambda )}^5}{e^{ - 4\lambda }}}}{{5!}}\)     A1

attempt to find P(2 arrive in the next three hours)     M1

\( = \frac{{{{(3\lambda )}^2}{e^{ - 3\lambda }}}}{{2!}}\)     A1

use of conditional probability formula     M1

P(3 in the first hour given 5 in total) \( = \frac{{\frac{{{\lambda ^3}{e^{ - \lambda }}}}{{3!}} \times \frac{{{{(3\lambda )}^2}{e^{ - 3\lambda }}}}{{2!}}}}{{\frac{{{{(4\lambda )}^5}{e^{ - 4\lambda }}}}{{5!}}}}\)     A1

\(\frac{{\left( {\frac{9}{{2!3!}}} \right)}}{{\left( {\frac{{{4^5}}}{{5!}}} \right)}} = \frac{{45}}{{512}} = 0.0879\)     A1

[8 marks]

A more difficult question, but it was still surprising how many candidates were unable to make a good start with it. Many were using \(\lambda  = \frac{5}{4}\) and consequently unable to progress very far. Many students failed to recognise that a conditional probability should be used.