Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.
Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Solution:
We have integers 1,2,3, \ldots 1000
We have integers 1,2,3, \ldots 1000
(S)=1000
No. of integers that are multiple of 2=500
Let’s say the no. of integers that are multiple of 9 be n.
The nth term =999=9+(n-1) 9=999
\Rightarrow \quad 9+9 n-9=999 \Rightarrow \quad n=111
From 1 to 1000 , the no. of multiples of 9 is 111 .
The multiple of 2 and 9 both are 18,36, \ldots, 990.
Let’s say that \mathrm{m} be the no. of terms in above series.
\begin{array}{ll} \therefore & & \text { mth term }=990 \\ \Rightarrow & & 18+(\mathrm{m}-1) 18=990 \\ \Rightarrow & & 18+18 \mathrm{~m}-18=990 \\ \Rightarrow & \mathrm{m} & =55 \end{array}
No. of multiples of 2 or 9=500+111-55=556=n(E)
\therefore \quad The required probability =\frac{n(E)}{n(S)}=\frac{556}{1000}=0.556