Let A denote the event that Lily goes to practice, and let B denote the event that rains. Then

\(\displaystyle{P}{\left({A}\right)}={P}{\left({A}{\mid}{B}\right)}{P}{\left({B}\right)}+{P}{\left({A}{\mid}{B}^{{c}}\right)}{P}{\left({B}^{{c}}\right)}\)

(B* is the complementary event of B — here it is the event that it does not rain). Also,

\(\displaystyle{P}{\left({B}\right)}={0.3}\Rightarrow{P}{\left({B}^{{c}}\right)}={1}-{P}{\left({B}\right)}={0.7}\)

Also, we are given that

\(\displaystyle{P}{\left({A}{\mid}{B}\right)}={0.1}\)

\(\displaystyle{P}{\left({A}{\mid}{B}^{{c}}\right)}={0.9}\)

Therefore,

\(\displaystyle{P}{\left({A}\right)}={0.1}\cdot{0.3}+{0.9}\cdot{0.7}={0.66}\)

\(\displaystyle{P}{\left({A}\right)}={P}{\left({A}{\mid}{B}\right)}{P}{\left({B}\right)}+{P}{\left({A}{\mid}{B}^{{c}}\right)}{P}{\left({B}^{{c}}\right)}\)

(B* is the complementary event of B — here it is the event that it does not rain). Also,

\(\displaystyle{P}{\left({B}\right)}={0.3}\Rightarrow{P}{\left({B}^{{c}}\right)}={1}-{P}{\left({B}\right)}={0.7}\)

Also, we are given that

\(\displaystyle{P}{\left({A}{\mid}{B}\right)}={0.1}\)

\(\displaystyle{P}{\left({A}{\mid}{B}^{{c}}\right)}={0.9}\)

Therefore,

\(\displaystyle{P}{\left({A}\right)}={0.1}\cdot{0.3}+{0.9}\cdot{0.7}={0.66}\)