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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

#1**+1 **

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%

CPhill Nov 11, 2017