A probability problem by Anubhav Sinha

The 3 part is $\frac{1}{6}$ of the circle. The probability of landing on it is also $\frac{1}{6}$ , so the chance of not landing on 3 part will be same as the chance of landing anywhere else - that means, that there is (1 - $\frac{1}{6}$ ) chance of landing anywhere else and that is $\frac{5}{6}$ .