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Summer Clouds moderators

23/11/2017 at 15:27
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A number is randomly selected from the integers 1 through 25, inclusive. What is the probability that the number chosen is divisible by 2, 3, 4 or 5? Express your answer as a common fraction.




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    Phan Thanh Tinh Coordinator 23/11/2017 at 19:04

    There are 7 unsatisfied numbers : 1 ; 7 ; 11 ; 13 ; 17 ; 19 ; 23one of The number of satisfied numbers is : 25 - 7 = 18

    The answer is : \(\dfrac{18}{25}\)

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    Lê Quốc Trần Anh Coordinator 23/11/2017 at 17:48

    The numbers from 1 to 25 (total 25 numbers): 

    + Divisible by 2 are: \(2;4;6;8;10;12;14;16;18;20;22;24\left(12-numbers\right)\)

    + Divisible by 3 (but not divisible by 2) are: \(3;9;15;21\left(4-numbers\right)\)

    + Because \(4⋮2\) => The numbers divisible by 4 (but not divisible by 2 and 3) is: \(\varnothing\)

    + Divisible by 5 but not divisible by 2 and 3 is: \(5\) 

    So the probability is: \(\dfrac{12+4+1}{25}=\dfrac{17}{25}\)


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