Let $B_{t}$ a brownian motion (stochastic process) then I know $B_{t} -B_{s}$ has a normal distribution with mean$=0$ and variance $=t-s$
I want to calculate the following probability: $P(3B_{2}>4)$
I know that if $X \sim N(\mu,\sigma^2$) then if $k$ is an constant $kX \sim N(k \mu,k^2 \sigma^2)$
therefore:
$3B_{2} \sim 3N(0,2)=N(0,2*3^2)=N(0,18)$ then $P(3B_{2}>4)=0.41$
but in $P(B_{2}>4/3)$,
$ B_{2} \sim N(0,2)$ and $P(B_{2}>4/3)=.25$
I don't know what happens, should not give the same probability? on the other hand if $3B_{2} \sim N(0,6)$ the probability of $P(3B_{2}>4)=0.25$
Can andybody helpme?