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

-- All of the non-zero digits are: 1, 2, 3, 4, 5, 6, 7, 8, and 9 .
that's nine digits all together.

-- The odd ones are 1, 3, 5, 7, and 9 .
There are five odd ones.

-- The even ones are 2, 4, 6, and 8.
There are four of them.

-- Two odd digits are already used, and we can't repeat them.
So we have 7 choices left for the last digit ... 3 odds and 4 evens.
 
-- Probability = (number of successful choices) / (total possibilities)

Successful choices . . . an odd digit, 3 available
Total possibilities . . . 7 unused digits

Probability = 3/7  (choice C)