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Meyer rolls two fair, ordinary dice with the numbers  on their sides. What is the probability that at least one of the dice shows a square number?

 

 

 

this is what i found

 

 

(square numbers for this are 1,4,9... and 9 is too big so just 1 and 4)

 Jan 17, 2019
 #1
avatar+16487 
+1

36 possible rolls

 

20 out of 36 will display al least one '1'  or at least one '4'     20/36 = 5/9

 Jan 17, 2019
 #2
avatar+96439 
+1

Here's another way to see this

 

P(at least one square)  =  1 - P( no square on either die)  =

 

1 - (4/6) (4/6)  =

 

1 -  16/36  =

 

[ 36 - 16 ] / 36  =

 

20 /36  =

 

5 / 9

 

 

cool  cool cool

 Jan 17, 2019

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