a three digit number is chosen at random. what is the probability that the sum of the digits is less than 6?
i got 3.78% but im not sure if that is correct. I assumed that there were three possible outcomes because 012 would just be considered 12
From the way you have stated the problem,
012 = 0 + 1 + 2 = 3 = 120 = 102 etc.
Does that change your mind?
To determine the probability, we need to find the number of three-digit numbers whose digits sum up to less than 6, and divide it by the total number of three-digit numbers.
Let's break it down step by step:
Step 1: Find the three-digit numbers whose digits sum up to less than 6:
- The maximum sum of three digits is 9+9+9 = 27.
- We need to find the numbers where the sum is less than 6.
- We can break it down into cases based on the smallest digit:
- If the smallest digit is 1, the other two digits can be 1, 2, or 3.
- If the smallest digit is 2, the other two digits can be 1, 2, 3, or 4.
- If the smallest digit is 3, the other two digits can be 1, 2, 3, 4, or 5.
- Counting the possibilities in each case:
- Case 1: 1, 11, 12, or 13 (4 possibilities)
- Case 2: 2, 21, 22, 23, 24 (5 possibilities)
- Case 3: 3, 31, 32, 33, 34, 35 (6 possibilities)
Step 2: Calculate the total number of three-digit numbers:
- The smallest three-digit number is 100.
- The largest three-digit number is 999.
- To find the total count, subtract the smallest possible number from the largest possible number and add 1:
Total count = 999 - 100 + 1 = 900
Step 3: Calculate the probability:
- Divide the count of three-digit numbers with digits summing up to less than 6 by the total count:
Probability = (4 + 5 + 6) / 900
Now, let's calculate it:
Probability = (4 + 5 + 6) / 900 = 15 / 900 = 1/60 ≈ 0.0167 (16.67%)
So, the correct probability that the sum of the digits is less than 6 is approximately 1/60, which is approximately 0.0167 or 16.67%. Therefore, your answer of 3.78% is not correct.