Correct Answer -
33
100
33
Here,
S
=
{
1
,
2
,
3
,
4
,
…
,
99
,
100
}
⇒
n
(
S
)
=
100
.
S
Let
E
1
=
E
event of getting a number divisible by 4.
And,
E
2
=
E
event of getting a number divisible by 6.
Then,
E
1
∩
E
2
=
E
event of getting a number divisible by both 4 and 6, i.e., divisible by 12 (LCM of 4 and 6).
∴
E
1
=
{
4
,
8
,
12
,
...
,
100
}
⇒
n
(
E
1
)
=
25
∴
.
E
2
=
{
6
,
12
,
18
,
.
.
,
96
}
⇒
n
(
E
2
)
=
16
.
E
(
E
1
∩
E
2
)
=
{
12
,
24
,
36
,
.
.
,
96
}
⇒
n
(
E
1
∩
E
2
)
=
8
.
(
Now, use
P
(
E
1
∪
E
2
)
=
P
(
E
1
)
+
P
(
E
2
)
−
P
(
E
1
∩
E
2
)
.
P
