Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:

(i) an even number

(ii) a multiple of 3

(iii) an even number and a multiple of 3

(iv) an even number or a multiple of 3

We know that, there are totally 9 cards from which one card is drawn.

Total number of possible outcomes = 9

(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10

∴ Favorable number of outcomes = 5

P(selecting a card with an even number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{9}$.

Hence, the probability that card selected will be an even number = $\dfrac{5}{9}$.

(ii) From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e. 3, 6, 9.

∴ Favorable number of outcomes = 3

P(selecting a card with multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{9} = \dfrac{1}{3}$.

Hence, the probability that card selected will be a multiple of 3 = $\dfrac{3}{9} = \dfrac{1}{3}$.

(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6

∴ Number of favourable outcomes = 1

P(selecting a card with even number and multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{9}$.

Hence, the probability of selecting a card with even number and multiple of 3 = $\dfrac{1}{9}$.

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10

∴ Number of favourable outcomes = 7.

P(selecting a card with even number or multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{9}$.

Hence, the probability of selecting a card with even number or multiple of 3 = $\dfrac{7}{9}$.