Given $ X ~ Beta(4,2) $ , How can I calculate $P(X<0.5)$?

Question

Given $ X ~ Beta(4,2) $ , How can I calculate $P(X<0.5)$ (the probability that X is lesser than $0.5$) ?

Answer 1

In general if $X\sim \text{Beta}(\alpha, \beta)$, the density is given by $$ f_X(x)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}\quad (0\leq x\leq 1). $$ In particular, using the fact that $\Gamma(n)= (n-1)!$ for integers $n\geq 1$, we have that in your case the density is given by $$ f(x)=\frac{5!}{3!1!}x^3(1-x)I_{[0, 1]}(x). $$ Thus you need to compute $$ P(X< 0.5)=\int_{0}^{0.5} f(x)\, dx. $$