A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting

(i) white ball or a green ball.
(ii) neither a green ball not a red ball.

Solution

Total number of balls in the bag = 5 + 8 + 7 = 20
i.e. n( S) = 20
(i) Let A be the favourable outcomes of getting a white ball or a green ball. Then n( A) = 15
Therefore,
P(getting a white ball or a green ball)
$equals fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 15 over 20 equals 3 over 4$

(ii) Let B be the favourable outcomes of getting neither of a green ball nor a red ball. Then h(B) = 7
Therefore, P(getting neither a green ball nor a red ball)
$equals space fraction numerator straight n left parenthesis straight B right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 7 over 20.$

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Real Numbers

A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting

(i) white ball or a green ball.
(ii) neither a green ball not a red ball.

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For Daily Free Study Material Join wiredfaculty

Real Numbers

A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting (i) white ball or a green ball.(ii) neither a green ball not a red ball.

Some More Questions From Real Numbers Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting

(i) white ball or a green ball.
(ii) neither a green ball not a red ball.