Probability

We consider all distinct rational numbers $\;\frac{p}{q}$ such that $p, q \in \{1,2,3,4,5,6\}$ and GCD $(p, q)=1$Now, we will consider each possible $q$ from 1 to 6 and for each, count the $p \in \{1,2,3,4,5,6\}$ such that g.c.d. $(p, q)=1$ So, there are 22 distinct reduced rational numbers of the form $\;\frac{p}{q}$.As we know that, a proper fraction is one where pAmong the 22 reduced fractions, we will now count how many satisfy pWhen, $q=1;\; p=2,3,4,5,6 \longrightarrow$ all $p>q \longrightarrow$ None proper fraction.when $q=2,\; p=1,3,5 \longrightarrow$ only $p=1 \longrightarrow 1$ proper fraction.When $q=3;\; p=1,2,4,5 \longrightarrow p=1,2 \longrightarrow 2$ proper fraction.when $q=4;\; p=1,3,5 \longrightarrow p=1,3 \longrightarrow 2$ proper fraction when$q=5;\; p=1,2,3,4,6 \longrightarrow p=1,2,3,4 \longrightarrow 4$ proper fraction when$q=6,\; p=1,5 \longrightarrow p=1,5 \longrightarrow$ both $< 6 \longrightarrow 2$ proper fraction ∴ Total proper fraction$=1+2+2+4+2=11$So, required probability$\;=\;\frac{\;\text{Number of proper fraction}\;}{\;\text{Total distinct reduced fractions}\;}$$\;=\;\frac{11}{22}=\;\frac{1}{2}$