Q. 15 4.0( 3 Votes )

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.

Answer :

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


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