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?
Best Answer
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.