Let S be the sample space. Then, n(S) = 36.
Let E1
= event that a doublet appears,
and E2=
event of getting a total of 10.
Then, E1={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)},
and E2={(4,6),(5,5,),(6,4)}.
∴(E1∩E2)={(5,5)}.
Thus, n(E1)=6,m(E2)=3andn(E1∩E2)=1.
∴P(E1)=
n(E1)
n(S)
=
6
36
=
1
6
,P(E2)=
n(E2)
n(S)
=
3
36
=
1
12
and P(E1∩E2)=
n(E1∩E2)
n(S)
=
1
36
.
∴
P(getting a doublet or a total of 10)
=P(E1orE2)=P(E1∪E2)
=P(E1)+P(E2)-P(E1∩E2)
=(
1
6
+
1
12
-
1
36
)=
8
36
=
2
9
.
∴
P(getting neither a doublet nor a total of 10)
=P(
¯
E1
and
¯
E2
)=P(
¯
E1
∩
¯
E2
)
=P(
¯
E1∪E2
)=1-P(E1∪E2)=(1-
2
9
)=
7
9
.
Hence, the required probability is
7
9
.
