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

No. of red ball = 10

No. of blue ball = 5

No. of green balls = 7

If a ball is drawn out of 22 balls (5 blue

+ 7 green + 10 red), then the total number of outcomes are

n(S) = 22

(i) Let E_{1} = Event of getting a red ball

∴ n(E_{1}) = 10

(ii) Let E_{2} = Event of getting a green ball

n(E_{2}) = 7

(iii) Let E_{3} = Event getting a red ball or a green ball i.e., not a blue ball.

n(E_{3}) = (10 + 7) = 17

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