Scramble five digits, is the number even?

Question

You write the digits 1, 2, 3, 4, 5 in a random order to form a five-digit number (each digit used exactly once). What is the probability the number is even?

Accepted Answer

A number is even exactly when its last digit is even. Among the digits 1, 2, 3, 4, 5, the even ones are 2 and 4, that's 2 of the 5 digits. By symmetry, in a random arrangement every digit is equally likely to land in the last position. So the last digit is even with probability P(	ext{even}) = \frac{2}{5} = 0.4. The other four positions don't matter for evenness, so no further counting is needed. Takeaway. Zoom in on the one feature that decides the event, here only the final digit matters. By symmetry each digit is equally likely to occupy any given slot, turning the problem into a simple c…

Scramble five digits, is the number even?

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