When a die is thrown, all possible outcomes are 1,2,3,4,5,6.
Total number of possible outcomes = 6.
(I) Let E1 be the event of getting a 3.
Then, the number of favourable outcomes = 1.
∴ P(getting a 3) =P(E1)=
1
6
.
(ii) Let E2 be the event of getting a 5.
Then, the number of favourable outcomes = 1.
∴ P(getting a 5)P(E2)=
1
6
.
(iii) Let E3) 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(E3)=
3
6
=
1
2
.
(iv) Let E4 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(E4)=
2
6
=
1
3
.