Show that \({\text{P}}(X \leqslant n) = 1 - {(1 - p)^n},{\text{ }}n \in {\mathbb{Z}^ + }\) .

Deduce an expression for \({\text{P}}(m < X \leqslant n)\,,{\text{ }}m\,,{\text{ }}n \in {\mathbb{Z}^ + }\) and m < n .

Given that p = 0.2, find the least value of n for which \({\text{P}}(1 < X \leqslant n) > 0.5\,,{\text{ }}n \in {\mathbb{Z}^ + }\) .

\({\text{P}}(X \leqslant n) = \sum\limits_{{\text{i}} = 1}^n {{\text{P}}(X = {\text{i}}) = \sum\limits_{{\text{i}} = 1}^n {p{q^{{\text{i}} - 1}}} } \)     M1A1

\( = p\frac{{1 - {q^n}}}{{1 - q}}\)     A1

\( = 1 - {(1 - p)^n}\)     AG

[3 marks]

\({(1 - p)^m} - {(1 - p)^n}\)     A1

[1 mark]

attempt to solve \(0.8 - {(0.8)^n} > 0.5\)     M1

obtain n = 6     A1

[2 marks]

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.