For independent <exponential distributions>, the numerator from part a(ii) is
$$
\int_0^c\lambda_1e^{-(\lambda_1+\lambda_2)t}\,dt
=\frac{\lambda_1}{\lambda_1+\lambda_2}
\{1-e^{-(\lambda_1+\lambda_2)c}\},
$$
while the denominator is the expression in braces. Their ratio is
$$
\frac{\lambda_1}{\lambda_1+\lambda_2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 5 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-207/5/b/i/solution