The five digits 1, 2, 3, 4 and 6 are arranged at random, each used exactly once, to make a five-digit number. What is the probability that the number formed is divisible by 4?

A number is divisible by 4 exactly when the two-digit number formed by its last two digits is. Using the digits 1, 2, 3, 4, 6 with no repeat, the endings that work are 12, 16, 24, 32, 36 and 64 — six of them. Each ending leaves the other three digits free to fill the first three places in 3! = 6 ways, so the favourable count is $6\times 6=36$ out of 5! = 120, giving 36/120 = 3/10. Trap 3/5 tests only whether the LAST digit is even — 2, 4 or 6, so 3 of 5 — but endings such as 42 and 26 are even without being multiples of 4. Trap 1/5 checks only whether the last digit is itself a multiple of 4, which here means the digit 4 alone. Trap 1/4 guesses that 'one number in four is a multiple of 4', which holds across all whole numbers but not across this restricted set of arrangements.