probability, divisibility

Dec 2015
1,084
169
Earth
Find the probability such that a 3-digit number is divisible by 3.
 

skipjack

Forum Staff
Dec 2006
21,480
2,470
1/3 (assuming numbers beginning with a zero are excluded and each 3-digit number is equally likely)
 
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Dec 2015
1,084
169
Earth
My approach :
Number of 3-digit numbers is 999-99=900 . Number of 3-digit numbers divisible by 3 in interval [1,999] is 999/3=333.
Number of 3-digit numbers divisible by 3 in interval [1,99] is 99/3=33. Number of 3-digit numbers divisible by 3 in interval [100,999] is 333-33=300.
P=300/900=3/9=1/3.
 

mathman

Forum Staff
May 2007
6,932
774
My approach :
Number of 3-digit numbers is 999-99=900 . Number of 3-digit numbers divisible by 3 in interval [1,999] is 999/3=333.
Number of 3-digit numbers divisible by 3 in interval [1,99] is 99/3=33. Number of 3-digit numbers divisible by 3 in interval [100,999] is 333-33=300.
P=300/900=3/9=1/3.
You seem to be overly complicating a trivial problem. Every third number is divisible by 3.
 
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