Let $B_t$ denote a standard Brownian bridge, $t \in [0, 1]$.
I am interested in understanding the probability $$ f(\lambda) = P\Big(\sup_{0 \leq t \leq 1} \frac{|B_t|}{\sqrt{t(1-t)}} > \lambda\Big), \quad \lambda > 0. $$ My thinking was to rewrite $$ B_t = \sqrt{\frac{1}{t(1-t)}} \Big(W_t - t W_1 \Big), $$ where $(W_t)_{t \in [0, 1]}$ denotes a standard Brownian motion. Then, $$ f(\lambda) = P\Big(\sup_{0 \leq t \leq 1} \frac{|W_t|}{\sqrt{t(1-t)}} > \lambda \mid W_1 = 0\Big). $$ However, I was struggling to be able to calculate the righthand side above. I was thinking that if $$ g(\lambda, x) = P\Big(\sup_{0 \leq t \leq 1} \frac{|W_t|}{\sqrt{t(1-t)}} > \lambda, W_1 \leq x\Big) ,$$ Then $$ f(\lambda) = \frac{\partial_x g(\lambda, 0)}{\phi(0)}, $$ where $\phi$ denotes the standard Gaussian density, however, computing $g(\lambda, x)$ seems nontrivial to me.