If one three-digit number ( 0 cannot be a left digit) is chosen at random from all those that can be made from the following set of digits, find the probability that the one chose a multiple of 2 .
\[
\{0,1,2,3,4,5,6,7,8\}
\]
Answer
Final Answer: The probability that a randomly chosen three-digit number from the given set is a multiple of 2 is approximately \(\boxed{0.635}\).
Step 1 ：We are given a set of digits \(\{0,1,2,3,4,5,6,7,8\}\) and we are asked to find the probability that a randomly chosen three-digit number from this set is a multiple of 2.
Step 2 ：A three-digit number is a multiple of 2 if its last digit is even. In the given set, the even digits are \(\{0,2,4,6,8\}\). However, 0 cannot be the leftmost digit.
Step 3 ：First, let's calculate the total number of three-digit numbers that can be formed from the given set of digits. The first digit can be any of the 8 non-zero digits, the second digit can be any of the 9 digits (including 0), and the third digit can also be any of the 9 digits. So, the total number of three-digit numbers is \(8 \times 9 \times 9 = 504\).
Step 4 ：Next, let's calculate the number of three-digit numbers that end with an even digit. The first digit can be any of the 8 non-zero digits, the second digit can be any of the 9 digits, and the third digit must be one of the 5 even digits. So, the total number of three-digit numbers that are multiples of 2 is \(8 \times 9 \times 5 = 320\).
Step 5 ：Finally, the probability that a randomly chosen three-digit number from the given set is a multiple of 2 is the ratio of the number of favorable outcomes to the total number of outcomes, which is \(\frac{320}{504} \approx 0.635\).
Step 6 ：Final Answer: The probability that a randomly chosen three-digit number from the given set is a multiple of 2 is approximately \(\boxed{0.635}\).
