Let S be the sample space. Then, clearly n(S) = 500.
Let E1
= event of getting a number divisible by 3,
and E2
= event of getting a number divisible by 5. Then,
(E1∩E2)=
event of getting a number divisible by both 3 and 5
= event of getting a number divisible by 15.
∴E1={3,6,9,..,495,498},E2={5,10,15,..,495,500}
and (E1∩E2)={15,30,45,..,495}.
∴n(E1)=(
498
3
)=166,n(E2)=(
500
5
)=100
and n(E1∩E2)=(
495
15
)=33.
∴P(E1)=
n(E1)
n(S)
=
166
500
=
83
250
,P(E2)=
n(E2)
n(S)
=
100
500
=
1
5
and P(E1∩E2)=
n(E1∩E2)
n(S)
=
33
500
.
∴
P(the chosen number is divisible by 3 or 5)
=P(E1orE2)=P(E1∪E2)
=P(E1)+P(E2)-P(E1∩E2)
=(
83
250
+
1
5
-
33
500
)=
233
500
.
