Given any random, 3-digit number, what is the probability that a number will be:
a) a multiple of 5
b) a multiple of 2
c) a square number
How?
Can the numbers include 0?
Every fifth number will be a multiple of 5 (005, 010... 105, 110...), and every other number will be a multiple of 2.
I don't know about the squares.
To calculate the probability of a random 3-digit number having a particular property, we need to determine the number of favorable outcomes (numbers that satisfy the property) and the total number of possible outcomes (all the 3-digit numbers).
a) Probability of being a multiple of 5:
A number will be a multiple of 5 if its units digit is either 0 or 5. Since there are 10 digits (0-9), the units digit can take on 10 possibilities. Therefore, the probability of a random 3-digit number being a multiple of 5 is 2/10 = 1/5.
b) Probability of being a multiple of 2:
A number will be a multiple of 2 if its units digit is an even number (0, 2, 4, 6, or 8). Again, since there are 10 digits, the units digit has 5 possibilities. Hence, the probability of a random 3-digit number being a multiple of 2 is 5/10 = 1/2.
c) Probability of being a square number:
A square number is the result of multiplying an integer by itself (e.g., 9 is the square of 3). Since we are looking for 3-digit numbers, we need to find the integer values between 10 and 31 (since the square root of 999 is approximately 31.6). There are 6 such integers (10, 11, 12, 13, 14, 15). Thus, the probability of a random 3-digit number being a square number is 6/100 = 3/50.
To calculate the probabilities, we divide the number of favorable outcomes by the total number of possible outcomes. In this case, the total number of 3-digit numbers is 900 (from 100 to 999).