You just need one!

Those numbers whose first digit is even will result in even number. There are 40 of them (20s, 40s, 60s, 80s). For rest of them (50 nos), the product is determined by the 2nd digit. 25 will give odd product and 25 even. So there will be a total of 65 numbers out of 90, which will give even product

First we notice, that the only way a two digit number can be odd, is if both the digits are odd. We are given a two digit number. The first digit can be between $1$ and $9$ , containing $\frac {5}{9}$ odd numbers. The second digit can be between $0$ and $9$ , containing $\frac {5}{10}$ odd numbers. The chance of both the numbers being odd is $\frac {5}{9} \cdot \frac {5}{10} = \frac {25}{90}$ This means the chances of the number being even is the remaining $1 - \frac {25}{90}= \frac {65}{90}$ .

There are 90 whole numbers on the interval [10, 99]. Let's count those numbers whose digits are both odd.

11, 13, 15, 17, 19,
31, 33, 35, 37, 39,
51, 53, 55, 57, 59,
71, 73, 75, 77, 79,
91, 93, 95, 97, 99.

With five numbers like this in each of five rows, there are a total of 25 numbers for which the product of the digits is odd. These are the only numbers with this property, so we subtract it from 90 to figure out the number of wholes whose digit product is even.

90 - 25 = 65, so that means the probability that a whole number selected on this interval will have an even digit product is $\dfrac{65}{90}$ .

Remember: 0 is even.

$1-\frac { 5 }{ 9 } \times \frac { 5 }{ 10 } =\frac { 65 }{ 90 }$

The complement is : to get 2 digits number which each digit is odd.

for the first digit, there are 5 choices, and for the second digit, there are 5 choices as well. $\Rightarrow$ 25 numbers.

the probability that we're looking for: $1- \frac{25}{90} = \boxed{\frac{65}{90}}$

10 to 20;11,13,15,17,19, ; 30 to 4031,33,35,37,39.......Thus 5x5=25, Numbers give odd product. Tota numbers between 10 to 99 = 99 - 10 + 1 = 90; Numbers given even product = 99 - 25 = 65. There fore Probability = 65/90 = 13/18.