I am stuck on a question. I know how to attempt it but I was wondering if there is an easier way to do it.
Question
Suppose you have one red ball, one green ball, one blue ball, and three white balls. Three balls are drawn at random without replacement. What is the probability that the drawn balls are of different colours
How I think of it: To do this, I would try to find the probability of all the different cases. Such as case where the balls are drawn in this pattern: RGB, RBG, RBW…etc but as you can see this will get too long and time consuming.
Is there an easier way to do this?
Best Answer
There are $\binom{6}{3}$ equally likely ways to pick $3$ balls from $6$.
To count the ways in which the colours are all different, there is RGB ($1$ way) plus the ways that involve white.
To count these, the non-whites can be chosen in $\binom{3}{2}$ ways, and for each choice the whites can be chosen in $\binom{3}{1}$ ways.
Thus our probability is $\dfrac{1+\binom{3}{2}\binom{3}{1}}{\binom{6}{3}}$. This simplifies nicely.