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algebra 1

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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

  • algebra 1 -

    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.

  • algebra 1 -

    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

  • algebra 1 - go with Reiny -

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

  • algebra 1 -

    but the answer is 96

  • algebra 1 -

    "what is the probability" ...
    cannot have 96 as the answer.

  • algebra 1 -

    ok thanks

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