A three-digit number is written down at random, every three-digit number being equally likely. What is the probability that all three of its digits are odd?

The three-digit numbers run from 100 to 999, so there are 900 of them. For every digit to be odd the hundreds digit must be one of 1, 3, 5, 7, 9 and so must the tens and the units digit, giving $5\times 5\times 5=125$ numbers and a probability of 125/900 = 5/36. Taking the digits one at a time agrees: the hundreds digit is odd with probability 5/9, not 1/2, because it can never be 0, and the other two are each odd with probability 1/2, so $(5/9)(1/2)(1/2)=5/36$. Treating all three digits as odd half the time gives 1/8, which is also what counting from 000 to 999 would give; 1/2 looks only at the units digit; and 5/9 = 500/900 counts the numbers that merely BEGIN with an odd digit.