A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (i) of red colour (ii) not of green colour.

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Solution

Total possible outcomes = Total number of balls
= ( 5 + 4 + 3) = 12
(i) Number of red balls = 5
∴ P (getting a red ball ) = $\frac{5}{12}$

(ii) Total number of balls which are not green = (5 + 4) = 9
∴ P (getting a ball which is not green) = $\frac{9}{12} = \frac{3}{4}$

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A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (i) of red colour (ii) not of green colour.

Total possible outcomes = Total number of balls = ( 5 + 4 + 3) = 12 (i) Number of red balls = 5 ∴ P (getting a red ball ) = 512 (ii) Total number of balls which are not green = (5 + 4) = 9 ∴ P (getting a ball which is not green) = 912=34

Total possible outcomes = Total number of balls
= ( 5 + 4 + 3) = 12
(i) Number of red balls = 5
∴ P (getting a red ball ) = $\frac{5}{12}$

(ii) Total number of balls which are not green = (5 + 4) = 9
∴ P (getting a ball which is not green) = $\frac{9}{12} = \frac{3}{4}$