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?
$ \dfrac{10}{21} $
$ \dfrac{11}{21} $
$ \dfrac{2}{7} $
$ \dfrac{5}{7} $
Explanation:
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, $ n\left(S\right)$ = Number of ways of drawing 2 balls out of 7
= 7C2 `
=$ \dfrac{(7 \times 6)}{(2 \times 1)} $
= 21.
Let E = Event of drawing 2 balls, none of which is blue.
$\therefore n \left(E\right)$ = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
=$ \dfrac{(5 \times 4)}{(2 \times 1)} $
= 10.
$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{10}{21} $.