Asked by naomi
                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
            
        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
Answers
                    Answered by
            Steve
            
    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.
    
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.
                    Answered by
            Reiny
            
    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
    
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
                    Answered by
            Steve
            
    Take Reiny's analysis. He paid a bit more attention to the details than did I.
    
                    Answered by
            da
            
    but the answer is 96
    
                    Answered by
            Steve
            
    "what is the probability" ...
cannot have 96 as the answer.
    
cannot have 96 as the answer.
                    Answered by
            da
            
    ok thanks
    
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