A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Answer: Option A

Total number of balls
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
$$\begin{array}{rl}& \text{n}\left(\text{S}\right)={}^7{C}_2\\ & \text{n}\left(\text{S}\right)=\frac{\left(7\times 6\right)}{\left(2\times 1\right)}\\ & \text{n}\left(\text{S}\right)=21\end{array}$$
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
$$\begin{array}{rl}& \text{n}\left(\text{E}\right)={}^5{C}_2\\ & \text{n}\left(\text{E}\right)=\frac{\left(5\times 4\right)}{\left(2\times 1\right)}\\ & \text{n}\left(\text{E}\right)=10\\ & \therefore \text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{10}{21}\end{array}$$