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Ben twice chooses a random integer between 1 and 50, inclusive (and he may choose the same integer both times). What is the probability that at least one of the numbers Ben chooses is a multiple of 3?

Guest Nov 11, 2017
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We can count sets, here

 

No. of sets possible  by choosing  any 2 numbers from 50  = C(50,2) = 1225

And we need to add to this 50 sets that have the same integer selected twice  = 50

And there are 16 integers that are multiples of 3  between 1 - 50

So.......the number of sets that don't contain a multiple of 3 is :

C (  50 - 16 , 2)  = C( 33 , 2) =   528  plus 33 additional sets of the same integer selected twice =

561

 

So......the probablity of selecting a set that contains one or two multiples of three is

 

1 -  [ 528 + 33 ] / [ 1225 + 50 ]  =  

 

1 - [ 561 ] / [1275 ]  =   

 

[1275 - 561 ] / 1275  =    14 / 25  =   56 / 100  =   56%    

 

 

cool cool cool

CPhill  Nov 11, 2017

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