The prices of a certain security followa geometric brownian motion with parameters μ = .12 and σ = .24. if the security’s price is presently 40, what is the probability that a call option, having four months until its expiration time and with a strike price of k = 42, will be exercised? (a security whose price at the time of expiration of a call option is above the strike price is said to finish in the money.)

Probability = 45%

Explanation:

The stock is above 42 after 4 months. Well the distribution of the stock after 4 months (1/3 of a year) satisfies.

ln(S/40) = a normal distribution with mean μt - 0.5σ^{2t} and

std dev σsqrt(t) => a normal distribution with mean 0.03 and std dev 0.1385

ln(S/40) => ln( 42 / 40 ) => 0.049.

Convert to standard normal => (0.049-0.03)/0.1385 => 0.133

The answer is the area of the normal distribution above 0.133 which is about 45%

Hope this helps!