Brownian Motion maximum principle

First some notation.

B(t) is a standard Brownian Motion.

M(t) = max(B(u): 0<=u<=t)

T(x) = min(u=>0: B(u)=x)

phi(x) = standard normal cdf

I know that P(B(t) > x) = 1-phi(x/sqrt(t)) and P(M(t)=>x) = P(T(x)<=t) = 2*(1-phi(x/sqrt(t))).

I have a problem that asks for P(M(4)<=2). Wouldn't this be equal to 2*(phi(x/sqrt(t))) = 2*(phi(2/sqrt(4))) = 2*phi(1)?? However, the book says the answer is 0.6826, which is not equal to 2*phi(1). What am I doing wrong?

Re: Brownian Motion maximum principle

Hey BrownianMan.

What did you get for your distribution of the maximum of B(u) for 0<=u<=t?