# probability, divisibility

#### idontknow

Find the probability such that a 3-digit number is divisible by 3.

#### skipjack

Forum Staff
1/3 (assuming numbers beginning with a zero are excluded and each 3-digit number is equally likely)

• idontknow

#### idontknow

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
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.

• idontknow