The mean condition gives $a/(a+b)=0.1$, hence $b=9a$. The <Beta distribution> variance is then
$$
\frac{ab}{(a+b)^2(a+b+1)}
=\frac{0.09}{10a+1}.
$$
Since $0.02985^2=9/10100$, equating variances gives $10a+1=101$. Therefore
$$
\boxed{a=10,\qquad b=90.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 207 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-207/1/b/solution