Question:
If a number x is chosen from the numbers 1, 2, 3, and a number y is selected from the numbers 1, 4, 9. Then, P(xy < 9)
(a) $\frac{7}{9}$
(b) $\frac{5}{9}$
(c) $\frac{2}{3}$
(d) $\frac{1}{9}$
Solution:
GIVEN: x is chosen from the numbers 1, 2, 3 and y is chosen from the numbers 1, 4, 9
TO FIND: Probability of getting 
We will make multiplication table for x and y such that
$x y=\mid \times 1=1$
$=1 \times 4=4$
$=1 \times 9=9$
$=2 \times 1=2$
$=2 \times 4=8$
$=2 \times 9=18$
$=3 \times 1=3$
$=3 \times 4=12$
$=3 \times 9=27$
So the numbers such that $(x y)<9$ is 5
We know that PROBABILITY $=\frac{\text { Number of favour able event }}{\text { Total number of event }}$
Hence $P(x y<9)=\frac{5}{9}$
Hence the correct option is $(b)$