Solution:

There are total 900 three digit integers.

Now, to find an integer whose product is not even, neither of the 3 digits in the integer should be even as e*o*o=e
Thus, all the three digits has to be even and this can be done in 5* 5* 5=125 ways

Therefore, probability that the product of the digit is even=\(\frac{900-125}{900}\)=\(\frac{31}{35}\)

IMO B

Hope this helps!

Total Even numbers can be found by taking even digit in any of the three places. So there will be lot of combinations and it will be difficult to list down everything.
This is why an approach to find odd number is easier and will save you time.
If you want to find total number of even numbers then they this.
First digit even (4 ways) * anything in second and third. So 400 ways.
Second digit even and first digit odd and anything in third place.
So, 5*5*10 = 250 ways
Third digit even and first and second are odd = 5*5*5 = 125 ways
Total = 400+250+125 = 775.

Hope it helps!

_________________

is it option A?

For a three digit number
- First digit can be any number from 1 to 9 => 9 possibilities => 5 odd, 4 even
[since it is a three digit number so first digit cannot be zero]
- Second digit can be any number from 0 to 9 => 10 possibilities => 5 odd, 5 even
- Third digit can be any number from 0 to 9 => 10 possibilities => 5 odd, 5 even

Total number of three digit numbers possible are 999-100 + 1 = 899 + 1 = 900
[This can also be achieved by multiplying the possibilities of first digit * second digit * third digit = 9*10*10 = 900]

The three digit number will be odd only when all digits are odd
=> First digit has 5 possibilities * second digit has 5 possibilities * third digit has 5 possibilities = 125 numbers
Remaining Numbers out of 900 will be even then
=> Even numbers = 900 - 125 = 775

=> Probability of getting even number = \(\frac{775}{900}\) = \(\frac{31}{36}\)

So, answer will be B
Hope it helps!
_________________

why total even no. can be 9*10*5
9= from 1 to 9 for hundred position
10=for 10 choice tenth digit position
5= 0,2,4,6,8 possible option unit place

Hi!

This can be one of the possibility. But, it misses the possibility of considering the units place to be odd.

For a number to have an even product, it can be EOO;OEO;OOE;EEO;EEE;OEE;EOE etc.

Thus, the above possibilities are not total number of possibilities.

Hope this helps!

thank, i was narrow sighted while considering possibilities....

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