Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is :

(i) a prime number

(ii) a number divisible by 4

(iii) a number that is a multiple of 6

(iv) an odd number.

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. Hence, total no. of cards = 10.

(i) Let E₁ be the event of choosing a prime number card.

E₁ = {2}.

∴ The number of favourable outcomes to the event E₁ = 1.

$\therefore P(E$1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{1}{10}.

Hence, the probability of choosing a prime number card is $\dfrac{1}{10}$.

(ii) Let E₂ be the event of choosing a card with number that is divisible by 4.

E₂ = {4, 8, 12, 16, 20}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E$2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}.

Hence, the probability of choosing a card with number that is divisible by 4 is $\dfrac{1}{2}$.

(iii) Let E₃ be the event of choosing card with number that is a multiple of 6.

E₃ = {6, 12, 18}.

∴ The number of favourable outcomes to the event E₃ = 3.

$\therefore P(E$3) = \dfrac{\text{No. of favourable outcomes to } E3}{\text{Total no. of possible outcomes}} = \dfrac{3}{10}.

Hence, the probability of choosing a card with number that is multiple of 6 is $\dfrac{3}{10}$.

(iv) Let E₄ be the event of choosing card with odd number.

E₄ = {}.

∴ The number of favourable outcomes to the event E₄ = 0.

$\therefore P(E$4) = \dfrac{\text{No. of favourable outcomes to } E4}{\text{Total no. of possible outcomes}} = \dfrac{0}{10} = 0.

Hence, the probability of choosing a card with odd number is 0.