* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

$\Rightarrow 1-{\text{e}}^{-m}=\frac{3}{4}$ or equivalent A1

$\Rightarrow m=\ln 4$ A1

[3 marks]

$\text{P}(Y>1)=1-\text{P}(Y=0)-\text{P}(Y=1)$ (M1)

$=1-{\text{e}}^{-2\ln 4}-{\text{e}}^{-2\ln 4}\times 2\ln 4$ A1

recognition that $2\ln 4=\ln 16$ (A1)

$\text{P}(Y>1)=\frac{15-\ln 16}{16}$ A1

[4 marks]