probability question involving numbers 100 to 999 inclusive A whole number between 100 and 999 inclusive is chosen at random. Find the probability that it is exactly divisible by 3. If it is exactly divisible by 3, what is the probability that it is exactly divisible by 9?
 A: Please do not count...
For each $ab$ in $\{0,1,\ldots,9\}\times\{0,1,\ldots,9\}$ there are $9$ numbers in the set which end by $ab$. Among them, exactly $3$ are multiples of $3$ and exactly $1$ is a multiple of $9$. Thus the probability that a number chosen unformly at random in the set $\{100,101,\ldots,999\}$ is some multiple of $3$ is exactly $\frac13$ and the probability that it is some multiple of $9$ is exactly $\frac19$.
This applies to every set $\{9k+i+1,9k+i+2,\ldots,9\ell+i\}$ with $0\leqslant k\lt\ell$ and $i\geqslant0$.
A: You can count total no of numbers divisible by 3 in given range and then divide it with the range, fraction is the answer. 
For conditional probability count no of numbers divisible by 9 in given range and then divide it with no of numbers divisble by 3.
Use 3*3=9 expression while calculating conditional probability. It will simplify calculation. 
If the total no of numbers divisible by 3 is even then conditional probability will be 1/2.
A: For a $3$-digit number $abc$ (with $a\not=0$) to be divisible by $3$, the sum of the digits, $a+b+c$, must be divisible by $3$.  No matter what $b+c$ is, there will be $3$ choices of $a$ that make $a+b+c$ divisible by $3$, out of $9$ choices altogether.  So the probability is $3/9=1/3$.
For divisibility by $9$, the same trick works, except now there is only one choice of $a$, so the probability is $1/9$.  (And this make the conditional probability, which is what the OP asked about, equal to $(1/9)/(1/3)=1/3$.)
Note, this works for $n$-digit numbers in general, not just $3$-digit numbers.  
