t (to avoid starting with 0): 4 choices (1, 2, 3, or 4).

Choose the second digit: 4 remaining choices.

Choose the third digit: 3 remaining choices.

Choose the fourth digit: 2 remaining choices.

The last digit is the final remaining digit.

Thus, the total number of possible numbers is:

$ 4 \times 4 \times 3 \times 2 \times 1 = 96 $

For a number to be divisible by 4, its last two digits must form a number divisible by 4. From the set of digits {0, 1, 2, 3, 4}, the pairs that are divisible by 4 are: 12, 20, 24, 32, and 40.

Let's consider the formation of numbers with each of these pairs as the last two digits:

Pairs 20, 40:

The first digit (non-zero) has 3 choices.

The second digit has 2 remaining choices.

The third digit is the last remaining digit.

$ 3 \times 2 \times 1 = 6 \text{ arrangements per pair} $

Pairs 12, 24, 32:

The first digit (non-zero) has 2 choices.

The second digit has 2 remaining choices.

The third digit is the last remaining digit.

$ 2 \times 2 \times 1 = 4 \text{ arrangements per pair} $

Now, combine these results:

Two pairs (20, 40) with 6 arrangements each contribute:

$ 2 \times 6 = 12 $

Three pairs (12, 24, 32) with 4 arrangements each contribute:

$ 3 \times 4 = 12 $

Hence, the total number of numbers divisible by 4 is:

$ 18 + 12 = 30 $

Therefore, the probability that a randomly chosen number is divisible by 4 is:

$ \frac{30}{96} = \frac{5}{16} $

Probability

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