Solve a combinatorial problem using a probabilistic approach. Consider two arrays of length 100, among those 10 entries are 1 and rest are 0. 
I want to add these two arrays co-ordinate wise such that there is no 2 in the sum. 
Number of such arrays will be $N_1= \binom{100}{10} \times \binom{90}{10}$. Instead of combinatorial approach, I want to take a probabilistic approach. Any co-ordinate of any array is zero with probability 0.9. Thus probability that sum of two co-ordinates is 0 or 1 is $p=(0.9)^2+2\times 0.9\times 0.1$. So I calculate ﻿number of arrays for which sum is 0/1 in each co-ordinate is  
$N_2= (\binom{100}{10})^2 \times p^{100}$. But it does not match with $N_1$. I understand it is not correct to assume independence among the co-ordinate sum. But can we find good approximation using a probabilistic approach? Actually, instead of 0 or 1, I want to include -1 also in the arrays. Also I start with 3 arrays instead of 2 arrays and want to calculutale number of arrays with give binary sum in the co-ordinate wise. It seems combinatorially problem becomes dificult. So I want probabilistic approach, which can give good approximation. Please help me.
 A: If you want it in a probabilistic way, you can't take $0.9$ as the probability. The correct value will be
Pr $= \frac{90}{100}\frac{89}{99}\frac{88}{98} ...\frac{81}{91}$
which will make the answers tally, but I don't know if this will help you.
PS:
The first array could have the $1's$ in any $10$ places, and the probability that the second array is aligned appropriately is given by the expression, indicating a probability of $\approx 0.33$ for "second arrays" that are aligned in the desired manner for each of the possible first arrays
PPS
I have worked it out for your three arrays. Denoting $-1, 0,1$ as $N,O,P$ respectively, there can be various favorable alignments looking vertically that sum to $0, \;or\; 1$, viz
$NOP\; 6$ permutations
$NPP\; 3$ permutations
$OOO\; 1$ permutation, and
$OOP\;3$ permutations
P(a random vertical slice of the array sums to $0$ or $1$)
= $\frac{1}{100^3}(6\cdot5\cdot85\cdot15 + 3\cdot5\cdot15^2+ 85^3+ 3\cdot85^2\cdot15) = \frac{7847}{8000}, \approx 0.98$
Thus by linearity of expectation, expected # of co-ordinates summing to $0$ or $1 = 100*0.98 = 98$
This is happening as the overwhelming majority  are $O's$
I checked that the probabilities taking all possible configurations does sum up to $1$
