A bag contains 16 coloured balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is :

(i) red

(ii) not red

(iii) white

(iv) not white

(v) green or red

(vi) white or green

(vii) green or red or white

There are 16 coloured balls.

∴ No. of possible outcomes = 16.

(i) There are 7 red balls.

∴ No. of favourable outcomes = 7

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{16}$.

Hence, the probability of drawing a red ball = $\dfrac{7}{16}$.

(ii) There are 9 non red balls (6 green + 3 white).

∴ No. of favourable outcomes = 9

P(not drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{9}{16}$.

Hence, the probability of not drawing a red ball = $\dfrac{9}{16}$.

(iii) There are 3 white balls.

∴ No. of favourable outcomes = 3

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{16}$.

Hence, the probability of drawing a white ball = $\dfrac{3}{16}$.

(iv) There are 13 non white balls (6 green + 7 red).

∴ No. of favourable outcomes = 13

P(not drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{16}$.

Hence, the probability of not drawing a white ball = $\dfrac{13}{16}$.

(v) Since, there are only 3 coloured (red, green and white) balls.

We can say that,

P(drawing a green or red ball) = P(not drawing a white ball) = $\dfrac{13}{16}$.

Hence, the probability of drawing a green or red ball = $\dfrac{13}{16}$.

(vi) Since, there are only 3 coloured (red, green and white) balls.

We can say that,

P(drawing a white or green ball) = P(not drawing a red ball).

From part (ii),

P(not drawing a red ball) = $\dfrac{9}{16}$.

∴ P(drawing a white or green ball) = $\dfrac{9}{16}$.

Hence, the probability of drawing a white or green ball = $\dfrac{9}{16}$.

(vii) There are 6 green, 7 red and 3 white balls.

∴ No. of favourable outcomes = 16.

P(drawing a green or red or white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{16}$ = 1.

Hence, the probability of drawing a green or red or white ball = 1.