A bag contains 5 red balls, 4 white balls, 2 black balls and 4 green balls. A ball is drawn at random from the big. Find the probability that it is (i) black, (ii) not green, (iii) red or white, (iv) neither red nor green. [CBSE 2012]

Total numbers of elementary events are: 5 + 4 + 2 + 4 = 15

(i) Let E be the event of getting a black ball at the random draw

Then, numbers of favourable outcomes are: 2

∴ P (getting a black ball) = P (E) = 2/15

(ii) Let E be the event of getting non green ball at the random draw

Then, the numbers of unfavourable outcomes are: 4

Probability of getting a green ball = P (green ball) = 4/15

Then, the number of favourable outcome P (not green ball) = 1- P (green ball)

∴ (P non green ball)= P (E) = 1- 4/15 =11/15

(iii) Let E be the event of getting a red or white ball

Let A be the event of getting a red ball

Then, favourable outcomes are: 5

Probability (getting a red ball) = P (A) = 5/15

Let B be the event of getting a white ball

Then, the numbers of favourable outcomes are: 4

Probability (getting white ball) = P (B) = 4/15

P (E) = P (A) + P (B)

∴ P(red ball or white ball) = P (E) = 5/15 + 4/15 = 9/15 = 3/5

(iv) Let E be the event of getting neither red nor green

Let A be the probability of getting a red ball

Then, the favourable outcomes are: 5

∴ P (getting red ball) =P (A) = 5/15

Let B be the event of getting a green ball

Then, the favourable outcomes are: 4

P (getting green ball) = 4/15

Let C be the getting red or green ball

P (getting red or green ball) = P(C) = 5/15 + 4/15 = 9/15 = 3/5

P (getting neither Red nor green ball) = P (E) = 1- P (C) = 1-3/5 = 2/5