# Geometric brownian motion and probabilities

A stock's price movement is described by the equations $$dS_t=0.02S_tdt+0.25S_tdW_t$$ and $$S_0=100$$. An investor buys a call option on said stock with a strike price $$K=95$$ which expires in $$T=2$$ years. What is the probability that the investor makes a profit greater than $$20$$ at expiry? The risk-free rate is $$2\%$$ (continuously compounded).

I'm also given $$N(0.08144)=0.532454, N(0.434993)=0.668216,N(0.8907)=0.808041$$ so no calculator will be needed.

My attempt: We know that (skipping Ito's lemma) $$S_t=S_0e^{-0.01125t+0.25W_t}$$ The probability I need to find is $$P(S_2-K>20)=P(S_2>115)=P(e^{-0.01125*2+0.25W_2}>1.15)=P(\frac{W_2}{\sqrt{2}}>\frac{ln(1.15)+0.0225}{0.25\sqrt{2}})=1-N(0.458946)$$

which must be wrong.

I've gone through this countless times without success, can anyone tell what I'm doing wrong? If it helps the correct answer is given as $$19.2\%$$ Thanks in advance.

• You paid the call option premium so if you just factor in the future value in 2 years of the call option premium, you should get the answer of 19.2% – Magic is in the chain May 12 '20 at 21:29
• I thought it meant just the exercise payoff, not the net profit. Including the premium yields the correct result, thanks a lot. – actuarialboi9 May 12 '20 at 22:30