Given any random, 3-digit number, what is the

probability that a number will be:
a) a multiple of 5
b) divisible by 2
c) a square number

i know the answer but i don't know how to solve it

when you divide a number by 5, there are only 5 possible remainders: 0,1,2,3,4. So, since multiples of 5 have zero remainder, there's a 1/5 chance that a random number will be a multiple of 5.

In fact, there's a 1/n chance that a number will be a multiple of n.

For the squares, you need to figure out how many perfect squares there are between 100 and 999. If there are, say, 35, then there's a 1/35 chance that a random 3-digit number will be a square.

number of 3-digit numbers :

999-99 = 900

multiples of 5:
5, 10, 15, ... , 995 ---> 199 of them

prob(multiple of 5) = 199/900

divisible by 2 --- > must be even
numbers are :
100, 102, 104, ... 998 ----> 445 of them
prob(even) = 445/900 = 89/180

numbers which when squared produce a 3 digit number
10^2 = 100
11^2 = 121
..
31^2 = 961
32^2 = 1024 ... too large
so there are 22 of these 3-digit squares
prob(square) = 22/900
= 11/450

I hope you know how I am getting these counts.
e.g. how many even numbers from 100 to 998
consider them to be an arithmetic sequence
where a = 100 , d = 2 and t(n) = 998
a + (n-1)d = term(n)
100 + (n-1)(2) = 998
2(n-1) = 898
n-1 = 449
n = 449+1 = 450

Take Reiny's analysis. He paid a bit more attention to the details than did I.

but the answer is 96

"what is the probability" ...

cannot have 96 as the answer.

ok thanks

To find the probability of each event, we need to determine the total number of possible outcomes and the number of favorable outcomes for each event.

a) Probability that a number is a multiple of 5:
To find the total number of possible outcomes, let's consider any random, 3-digit number. In this case, the first digit can be any number from 1 to 9 (excluding 0 as the leading digit). The second and third digits can be any number from 0 to 9. So, the total number of possible outcomes is 9 * 10 * 10 = 900.

The favorable outcomes are the numbers that are multiples of 5. The multiples of 5 within the range of 100 to 999 are 100, 105, 110, ..., 995. To calculate how many multiples of 5 there are within this range, we can subtract the largest and smallest multiples of 5, then divide by 5 and add 1. In this case, (995 - 100) / 5 + 1 = 180.

The probability of a random, 3-digit number being a multiple of 5 is given by:
Number of favorable outcomes / Total number of possible outcomes = 180 / 900 = 1 / 5 = 0.2

Therefore, the probability of a random, 3-digit number being a multiple of 5 is 0.2 or 20%.

b) Probability that a number is divisible by 2:
In this case, we need to consider the last digit of the number since any number ending in an even digit is divisible by 2.

The total number of possible outcomes remains the same as before, which is 900.

The favorable outcomes are the numbers that end in an even digit (i.e., 0, 2, 4, 6, or 8). So, we have 5 options for the last digit.

The probability of a random, 3-digit number being divisible by 2 is given by:
Number of favorable outcomes / Total number of possible outcomes = 5 / 900

Therefore, the probability is 5 / 900 = 1 / 180 ≈ 0.0056.

c) Probability that a number is a square number:
To find the total number of possible outcomes, we still have 900 possible 3-digit numbers.

The favorable outcomes are numbers that are perfect squares between 100 and 999. The perfect squares within this range are 100, 121, 144, ..., 961. To calculate how many perfect squares there are within this range, we can find the square root of the largest and smallest perfect squares, round them to the nearest whole number, and then subtract to get the count. (31 - 10) + 1 = 22.

The probability of a random, 3-digit number being a perfect square is given by:
Number of favorable outcomes / Total number of possible outcomes = 22 / 900.

Therefore, the probability is 22 / 900 ≈ 0.0244.

Note: It's important to remember that these calculations assume a uniform distribution of random, 3-digit numbers.