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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