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Each outcome on the spinner below has equal probability. If you spin the spinner three times and form a three-digit number from the three outcomes, such that the first outcome is the hundreds digit, the second outcome is the tens digit and the third outcome is the units digit, what is the probability that you will end up with a three-digit number that is divisible by 4? Express your answer as a common fraction.

 
Logic  Oct 12, 2018
 #1
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There's no particular magic to this.  You have equiprobable numbers in the range 111-333.

 

Count how many of those are divisible by 4.

 

112 is divisible by 4

332 is divisible by 4

 

this is (332-112)/4 + 1 = 56 numbers.

 

There are a total of (333-111)+1 = 223 possible numbers.

 

P[spin choice is divisible by 4] = 56/223

 
Rom  Oct 12, 2018

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