how to know when we have to add and when to multiply while finding probability? 1)A Box contain 5 green , 3 white, 4 yellow balls. three ball is drawn at random. what is the probability of get all different colored balls?
2)A Bag contain 6 white and 4 red balls. three ball is drawn. what is the probability of getting  one red ball and two white ball
In the answer we are considering n(E)=5C3+4C3+3C3  for the first question and for second n(E)=4C1 X 6C2.
why the difference how to know when we have to add and when to multiply?
 A: Think of the balls as having ID numbers to make them distinct.
There are $\binom{12}{3}$ equally likely ways to choose $3$ balls from the $12$. 
How many ways are there to choose them all of different colours?  There are $\binom{5}{1}$ ways to pick a green. For every such way there are $\binom{3}{1}$ ways to choose a white. And for every choice of green and white, there are $\binom{4}{1}$ ways to pick the yellow, for a total of $\binom{5}{1}\binom{3}{1}\binom{4}{1}$. The required probability is therefore
$$\frac{\binom{5}{1}\binom{3}{1}\binom{4}{1}}{\binom{12}{3}}.$$
The second problem uses exactly the same ideas. The solution is a little shorter.
Remark: When we are finding the probability that "$A$ or $B$" happens, where $A$ and $B$ are disjoint events, then we add probabilities. If $A$ is the event "one of the balls was green" and $B$ is the event "one of the balls was white." then $A$ and $B$ are not disjoint, in fact they can and do both happen. When for every one of the $m$ ways of doing $A$, there are $n$ ways of doing $B$, there are $mn$ ways of doing $A$ abd $B$, That is the idea we used when we multiplied $\binom{5}{1}$ and $\binom{3}{1}$. 
In general, it is best not to think immediately about what formula we should use. Analysis of the situation comes first. Sometimes addition is appropriate, sometimes multiplication. And sometimes we need to do something more complicated. Solving many different problems is the only way to acquire the background to solve new ones.
A: Answer provided by you for 1) is wrong because probability can't be >1. Even here, we would multiply the parts because these are simultaneous events. So, answer is $(5/12)*(4/11)*(3/10)=1/22$
For independent events, we will add the probabilities. Like, what if you took out the ball, checked its color and then placed it back inside the box and then again took out a ball. Here, answer to the probability of getting 2 red balls would be $(5/12)+(5/12)=10/12=5/6$
A: The probability, or more precisely number of favourable outcomes, $n(E)=\binom53+\binom43+\binom33$ you give in the question for the first problem is not that of getting three different colours (the formula gives $n(E)=15$ but there are $5\times4\times3=60$ ways to draw three different colours as André Nicholas explains. Probably it was meant to be the number of ways to get three balls of the same colour: this can happen either by choosing all three balls among the green balls (for $\binom53=10$ possibilities) or by choosing all three balls among the white balls (for $\binom33=1$ possibility) or by choosing all three balls among the yellow balls (for $\binom43=4$ possibilities). Since these are alternative, mutually exclusive, ways to get a favourable outcome, addition of the number of possibilities is called for. The final probability here will be $(\binom53+\binom33+\binom43)/\binom{12}3$.
In the second problem you want, for a favourable outcome, to get one red ball and two white balls. Although you are in fact selecting three among all ten balls here, the only favourable outcomes are those where you choose one among the $4$ red balls and two among the $6$ white balls, and you can imagine that you are doing the selection in such a way (in other words you cheat to be sure to have a favourable outcome); this will give the right number of good possibilities, but for getting the probability you must not forget to put the total number $\binom{10}3$ of all actually possible draws in the denominator. Now the "cheating" way of getting one red and two white balls consists of two independent draws (red and white) which must both be performed; the total draw is a combination of the two draws, and the number of possible combinations is obtained by multiplying the number $\binom41$ of possibilities for the red balls by the number $\binom62$ of possibilities for the white balls. The final probability here will be $\binom41\times\binom62/\binom{10}3$.
