(a) the probability that he catches at least one fish in the first hour;

(b) the probability that he catches exactly three fish if he fishes for four hours;

(c) the number of complete hours that Mr Lee needs to fish so that the probability of catching more than two fish exceeds 80 %.

(a)     \(X{\text{ ~ Po(0}}{\text{.6)}}\)

\({\text{P}}(X \geqslant 1) = 1 - {\text{P}}(X = 0)\)     M1

\( = 0.451\)     A1     N1

(b)     \(Y{\text{ \~ Po(2}}{\text{.4)}}\)     (M1)

\({\text{P}}(Y = 3) = 0.209\)     A1

(c)     \(Z{\text{ \~ Po(}}0.6n{\text{)}}\)     (M1)

\({\text{P}}(Z \geqslant 3) = 1 - {\text{P}}(Z \leqslant 2) > 0.8\)     (M1)

Note: Only one of these M1 marks may be implied.

\(n \geqslant 7.132...\) (hours)

so, Mr Lee needs to fish for at least \(8\) complete hours     A1     N2

Note: Accept a shown trial and error method that leads to a correct solution.

[7 marks]

It was clear that many students had not been taught the topic and were consequently unable to make an attempt at the question. Of those students who were able to start, common errors were in a misunderstanding of the language. Many had difficulties in part (c) and “at least” in part (a) was sometimes misinterpreted.