Probability of a 5-Digit Number Divisible by 4 using Digits 1, 2, 5, 6, and 8

To find the probability of a 5-digit number using the digits 1, 2, 5, 6, and 8, which is divisible by 4, we need to consider certain conditions that must be satisfied.

Condition 1: The number must be divisible by 4
For a number to be divisible by 4, the last two digits of the number must form a multiple of 4. Thus, we need to consider the possible combinations of the last two digits using the given digits (1, 2, 5, 6, and 8) that yield multiples of 4.

Condition 2: The first three digits can be any of the given digits
Since the problem does not specify any restrictions on the first three digits, they can be any of the given digits (1, 2, 5, 6, and 8).

Solution:

Let's break down the problem into two parts: finding the possible combinations for the last two digits and calculating the probability.

Part 1: Finding the Combinations for the Last Two Digits

To find the combinations of the last two digits that form multiples of 4, we need to consider the following cases:

Case 1: The last two digits are both even (2, 6, or 8)
In this case, any combination of the last two digits can form a multiple of 4. There are 3 possibilities for each digit, so the total number of combinations is 3 * 3 = 9.

Case 2: One of the last two digits is odd (1 or 5) and the other is even (2, 6, or 8)
In this case, only specific combinations can form a multiple of 4. The possible combinations are:
- 12 (1 is odd and 2 is even)
- 16 (1 is odd and 6 is even)
- 52 (5 is odd and 2 is even)
- 56 (5 is odd and 6 is even)
- 72 (7 is odd and 2 is even)
- 76 (7 is odd and 6 is even)
- 92 (9 is odd and 2 is even)
- 96 (9 is odd and 6 is even)

So, there are 8 possible combinations for this case.

Part 2: Calculating the Probability

To calculate the probability, we need to determine the total number of possible 5-digit numbers using the given digits (1, 2, 5, 6, and 8).

Number of Possible Combinations:
Since the first three digits can be any of the given digits (1, 2, 5, 6, and 8), there are 5 choices for each digit. Therefore, the total number of possible combinations for the first three digits is 5 * 5 * 5 = 125.

Number of Combinations Divisible by 4:
Adding up the combinations from Case 1 and Case 2, we have a total of 9 + 8