In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Explanation:-

$$\eqalign{ & {\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {8 + 7 + 6} \right) = 21. \cr & {\text{Let}}\,{\text{E}} \cr & {\text{ = event}}\,{\text{that}}\,{\text{the}}\,{\text{ball}}\,{\text{drawn}}\,{\text{is}} \cr & {\text{ }}\,\,\,\,\,\,{\text{neither}}\,{\text{red}}\,{\text{nor}}\,{\text{green}} \cr & {\text{ = event}}\,{\text{that}}\,{\text{the}}\,{\text{ball}}\,{\text{drawn}}\,{\text{is}}\,{\text{blue}}{\text{.}} \cr & \therefore n\left( E \right) = 7. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{{21}} = \frac{1}{3} \cr}$$