Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is :

(i) a multiple of 4 or 6.

(ii) a multiple of 3 and 5.

(iii) a multiple of 3 or 5.

Since, there are 30 identical cards.

No. of possible outcomes = 30.

(i) Cards numbered 4, 6, 8, 12, 16, 18, 20, 24, 28, 30 are multiple of either 4 or 6.

∴ No. of favourable outcomes = 10.

P(getting a card which is a multiple of 4 or 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{30} = \dfrac{1}{3}$.

Hence, the probability of drawing a card which is a multiple of 4 or 6 = $\dfrac{1}{3}$.

(ii) Cards numbered 15, 30 are multiple of 3 and 5.

∴ No. of favourable outcomes = 2.

P(getting a card which is a multiple of 3 and 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{30} = \dfrac{1}{15}$.

Hence, the probability of drawing a card which is a multiple of 3 and 5 = $\dfrac{1}{15}$.

(iii) Cards numbered 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 are multiple of either 3 or 5.

∴ No. of favourable outcomes = 14.

P(getting a card which is a multiple of 3 or 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{14}{30} = \dfrac{7}{15}$.

Hence, the probability of drawing a card which is a multiple of 3 or 5 = $\dfrac{7}{15}$.