Probability problem; if sum and product are not divisible by number Out of set of numbers {10,11,...,100} three are being chosen randomly. Find these probabilities:


*

*At least one out of three are divisible by 6

*Exactly two are not divisible by 4

*Their sum is not divisible by 4

*Their product is not divisible by 3


So to choose three numbers out of 91 I can do in $\binom{91}{3}$ ways. First probability I solved this way:
$\frac{\binom{15}{1}\binom{76}{2}+\binom{15}{2}\binom{76}{1}+\binom{15}{3}}{\binom{91}{3}}$
Since there are 15 numbers divisible by 6 i have 3 options that 1 is divisible two are not, 2 are divisible 1 is not and all of 3 are divisible.
Second probability:
$\frac{\binom{69}{2}22}{\binom{91}{3}}$
Since there are 22 numbers that are divisible by 4 i subtracted 22 from 91 and got numbers that are not divisible by 4 from which I chose 2 numbers and the third I chose from those 22.
I do not know how to solve third and fourth and also I am not sure if the first two probabilities are correct.
 A: 

*Divide the numbers in set of 4 groups which are of the forms 4k (23 numbers), 4k+1(22 numbers), 4k+2 (23 numbers)and 4k+3 (23 numbers). Find the cases when sum is divisible by 4.


(4k,4k,4k) which is equivalent to C(23,3);(4k,4k+2,4k+2) C(23,2)C(23,1);(4k,4k+1,4k+3);(4k+1,4k+1,4k+2);(4k+3,4k+3,4k+2) etc
Use exclusion
A: Well, I took the time to work out the hard part. Here is a table:
$$\begin{array}{cccc|c|c}0&1&2&3&\text{Number}&\text{Total}\\ \hline
23&22&23&23&&\\ \hline
3&0&0&0&\binom{23}{3}&1771\\
1&1&0&1&\binom{23}{1}\binom{22}{1}\binom{23}{1}&11638\\
1&0&2&0&\binom{23}{1}\binom{23}{2}&5819\\
0&2&1&0&\binom{22}{2}\binom{23}{1}&5313\\
0&0&1&2&\binom{23}{1}\binom{23}{2}&5819
\end{array}$$
The fist $4$ columns represent the congruence classes $\pmod4$ and we see how many from each class is in the set to form a number divisible by $4$. The fifth column counts the number of ways such a selection may be made and the sixth column is the numerical result. Adding up the results in the fifth column we see that there are $30360$ ways to sum to $4$, with probability
$$\frac{30360}{\binom{91}{3}}=0.249907$$
But we want the probability that the sum is not divisible by $4$ which is
$$1-0.249907=0.750093$$
In the third problem, since there are $30$ numbers in range divisible by $3$ we need to select $3$ out of the $61$ that are for a probability of
$$\frac{\binom{61}{3}}{\binom{91}{3}}=0.296251$$
Since there are $15$ numbers in range that are divisible by $6$, we get
$$1-\frac{\binom{76}{3}}{\binom{91}{3}}=1-0.578672=0.421328$$
for the first problem. In the second problem, we need $1$ divisible by $4$ and $2$ not, so we get
$$\frac{\binom{23}{1}\binom{68}{2}}{\binom{91}{3}}=0.43128$$
