A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :

(i) white or blue

(ii) red or black

(iii) not white

(iv) neither white nor black?

Since, a ball is drawn at random from the bag, so all the balls are equally likely to be drawn.

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18.

So, the sample space of the experiment has 18 equally likely outcomes.

(i) Let E₁ be the event 'a white or blue ball is drawn'.

The number of white or blue balls = 5 + 2 = 7.

So, the number of favourable outcomes to the E₁ = 7.

∴ P(E₁) = P(a white or blue ball) = $\dfrac{7}{18}.$

Hence, the probability that the ball drawn is white or blue is $\dfrac{7}{18}.$

(ii) Let E₂ be the event 'a red or black ball is drawn'.

The number of red or black balls = 7 + 4 = 11.

So, the number of favourable outcomes to the E₂ = 11.

∴ P(E₂) = P(a red or black ball) = $\dfrac{11}{18}.$

Hence, the probability that the ball drawn is red or black is $\dfrac{11}{18}.$

(iii) Let E₃ be the event 'not white ball is drawn'.

Probability of not white ball drawn = Probability of any other ball drawn.

No. of balls except white balls = 18 - 5 = 13.

So, the number of favourable outcomes to the E₃ = 13.

∴ P(E₃) = P(not a white ball) = $\dfrac{13}{18}.$

Hence, the probability that the ball drawn is not white is $\dfrac{13}{18}.$

(iv) Let E₄ be the event 'neither white nor black ball is drawn'.

Probability that neither white nor black ball is drawn = Probability of a red or blue ball drawn.

No. of red or blue balls = 7 + 2 = 9.

So, the number of favourable outcomes to the E₄ = 9.

∴ P(E₄) = $\dfrac{9}{18} = \dfrac{1}{2}.$

Hence, the probability that the ball drawn is neither white nor black is $\dfrac{1}{2}.$