A bag contains 10 red 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a
(i) red ball (ii) green ball (iii) not a blue ball

Explanation

If a ball drawn out 22 balls (5 blue+7green+10red), then the total number of outcomes are
n(S)=22
(i) Let `E_(1)` =Event of getting of a red ball `n(E_ (1))=20`
`therefore "Required probability"=(n(E_(1)))/(n(S))=(10)/(22)=(5)/(11)`
Let `E_(2)`
`N(E_(2))=7`
`therefore "Required probability"=(n(E_(1)))/(n(S))=(7)/(22)`
(iii) Let `E_(3)`=Event gettin a red ball or a green ball i.e. not a blue ball
`n(E_(3))=(10+7)=17`
`therefore "Required probability"=(n(E_(3)))/(n(S))=(17)/(22)`