In a single throw of a die, find the probability of getting :

(i) an odd number

(ii) a number less than 5

(iii) a number greater than 5

(iv) a prime number

(v) a number less than 7

(vi) a number divisible by 3

(vii) a number between 3 and 6

(viii) a number divisible by 2 or 3.

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let E be the event of getting an odd number, then

E = {1, 3, 5}.

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

$\therefore P(E) = \dfrac{\text{No. of favourable outcomes to } E}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, the probability of getting an odd number is $\dfrac{1}{2}$.

(ii) Let E₁ be the event of getting a number less than 5, then

E₁ = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event E₁ = 4.

$\therefore P(E$1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.

Hence, the probability of getting a number less than 5 is $\dfrac{2}{3}$.

(iii) Let E₂ be the event of getting a number greater than 5, then

E₂ = {6}.

∴ The number of favourable outcomes to the event E₂ = 1.

$\therefore P(E$2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}.

Hence, the probability of getting a number greater than 5 is $\dfrac{1}{6}$.

(iv) Let E₃ be the event of getting a prime number, then

E₃ = {2, 3, 5}.

∴ 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}{6} = \dfrac{1}{2}.

Hence, the probability of getting a prime number is $\dfrac{1}{2}$.

(v) Let E₄ be the event of getting a number less than 7, then

E₄ = {1, 2, 3, 4, 5, 6}.

∴ The number of favourable outcomes to the event E₄ = 6.

$\therefore P(E$4) = \dfrac{\text{No. of favourable outcomes to } E4}{\text{Total no. of possible outcomes}} = \dfrac{6}{6} = 1.

Hence, the probability of getting a number less than 7 is 1.

(vi) Let E₅ be the event of getting a number divisible by 3, then

E₅ = {3, 6}.

∴ The number of favourable outcomes to the event E₅ = 2.

$\therefore P(E$5) = \dfrac{\text{No. of favourable outcomes to } E5}{\text{Total no. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.

Hence, the probability of getting a number divisible by 3 is $\dfrac{1}{3}$.

(vii) Let E₆ be the event of getting a number between 3 and 6.

E₆ = {4, 5}.

∴ The number of favourable outcomes to the event E₆ = 2.

$\therefore P(E$6) = \dfrac{\text{No. of favourable outcomes to } E6}{\text{Total no. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.

Hence, the probability of getting a number between 3 and 6 is $\dfrac{1}{3}$.

(viii) Let E₇ be the event of getting a number divisible by 2 or 3, then

E₇ = {2, 3, 4, 6}.

∴ The number of favourable outcomes to the event E₇ = 4.

$\therefore P(E$7) = \dfrac{\text{No. of favourable outcomes to } E7}{\text{Total no. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.

Hence, the probability of getting a number divisible by 2 or 3 is $\dfrac{2}{3}$.