Question

A locker combination has three nonzero digits, with no digit repeated. If the first two digits are odd, what is the probability that the third digit is also odd?

Answer 1

If you can get this using evens,
The digits are probablly  1 to 9, then the even is 2,4,6,8 and the odd are 1,3,5,7,9. 

Prodd(ways/choices)=5/(2+5)

So, Using odd numbers, The digits are likely to be 1-9, so the odds are 1,3,5,7,9. Look at the middle number, 5⇒6+3 (because of odds not evens that use 2). 6,7,8,9. 5/9 is your answer.


Maybe this helped and if it did then yw. :D