When a die is thrown, all possible outcomes are 1,2,3,4,5,6.
Total number of possible outcomes = 6.
(I) Let `E_(1)` be the event of getting a 3.
Then, the number of favourable outcomes = 1.
`:. " P(getting a 3) " = P(E_(1)) = 1/6.`
(ii) Let `E_(2)` be the event of getting a 5.
Then, the number of favourable outcomes = 1.
`:. " P(getting a 5)" P(E_(2)) = 1/6.`
(iii) Let `E_(3))` be the event of getting an odd number.
Then, the favourable outcomes are 1, 3, 5.
Number of favourable outcomes = 3.
`:. ` P(getting an old number ) ` =P(E_(3)) = 3/6 = 1/2.`
(iv) Let `E_(4)` be the event of getting a number greater than 4.
Then, the favourable outcomes are 5, 6.
Number of favourable outcomes = 2.
`:. ` P(getting a number greater than 4) `P(E_(4)) = 2/6 = 1/3.`