We know that in a single throw of two dice, the total number of possible outcomes is (6×6)=36
.
Let S be the sample space. Then, n(S) = 36.
(i) Let E1
= event of getting a doublet. Then,
E1={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}.
∴n(E1)6.
∴P(E1)=
n(E1)
n(S)
=
6
36
=
1
6
.
(ii) Let E2
= event of getting an even number as the sum. Then,
E2={(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,3),(3,1),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)}
.
∴n(E2)=18
.
∴P(E2)=
n(E2)
n(S)
=
18
36
=
1
2
.
(iii) Let E3
= event of getting a prime number as the sum. Then,
E3={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}.
∴n(E3)=15.
∴P(E2)=
n(E3)
n(S)
=
15
36
=
5
12
.
(iv) Let E4
= event of getting a multiple of 3 as the sum. Then,
E4={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)}.
∴n(E4)=12
.
∴P(E4)=
n(E4)
n(S)
=
12
36
=
1
3
.
(v) Let E5=
event of getting a total of at least 10. Then,
E5=
event of getting a total of 10 or 11 or 12
={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}.
∴n(E5)=6.
∴P(E5)=
n(E5)
n(S)
=
6
36
=
1
6
.
(vi) Let E6=
event of getting a doublet of even numbers. Then,
E6={(2,2),(4,4),(6,6)}.
∴n(E6)=3.
∴P(E6)=
n(E6)
n(S)
=
3
36
=
1
12
.
(vii) Let E7=
event of getting a multiple of 2 on one die and a multiple of 3 on the other die. Then,
E7={(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(3,4),(3,6),(6,2),(6,4)}.
∴n(E7)=11.
∴P(E7)=
n(E7)
n(S)
=
11
36
.
