In tossing a die once, the sample space is given by
S={1,2,3,4,5,6}
and, therefore, n(S) = 6.
(i) Let E1
= event of getting the number 4.
Then, E1={4}
and, therefore, n(E1)=1
.
∴
P(getting the number 4) =P(E1)=
n(E1)
n(S)
=
1
6
.
(ii) Let E2
= event of getting an even number. Then,
E2={2,4,6}
and, therefore, n(E2)=3
.
∴
P(getting the number 4) =P(E2)=
n(E2)
n(S)
=
3
6
=
1
2
.
(iii) Let E3
= event of getting a number less than 5. Then,
E3={1,2,3,4}
and, therefore, n(E3)=4
.
∴
P(getting a number less than 5) =P(E3)=
n(E3)
n(S)
=
4
6
=
2
3
.
(iv) Let E4
= event of getting a number greater than 4. Then,
E4={5,6}
and, therefore, n(E4)=2
.
∴
P(getting a number greater than 4) =P(E4)=
n(E4)
n(S)
=
2
6
=
1
3
.
(v) Let E5
= event of getting the number 8.
Since no face of the die is marked with the number 8,
we have E5=ϕ
and, therefore, n(E5)=0
.
∴
P(getting the number 8) =P(E5)=
n(E5)
n(S)
=
0
8
=0.
(vi) Let E6
= event of getting a number less than 8.
Then, E6={1,2,3,4,5,6}
and, therefore, n(E6)=6
.
∴
P(getting a number less than 8) =P(E6)=
n(E6)
n(S)
=
6
6
=1.
