A bag contains $5$ red balls , $15$ black balls, and $10$ green balls. Three balls are randomly drawn, one-by-one, with replacement, the probability of the event that "all are different coloured balls" is $P$, then $36P$ is
My solution:
$P(R)=\dfrac{5}{30}, P(B)=\dfrac{15}{30}, P(G)=\dfrac{10}{30}$
Hence probability that all balls are present is $\dfrac{1}{36}$
But order of colour can be anything so according to me we should multiply above answer by $6$ because we have $6$ permutation of three color as $RBG, RGB, BGR, BRG, GRB, GBR$
So final answer must be $36\cdot 6\cdot \dfrac{1}{36}=6$ but given answer is $1$. Am I doing anything wrong?
Best Answer
Your approach is correct. We can verify it by looking at a smaller example which can be manually listed. Let's assume we have $1$ red ball, $1$ black ball and $1$ green ball and we are looking for three balls randomly drawn one-by-one with replacement.
Here we have a total of $3^3=27$ outcomes, namely \begin{align*} \begin{array}{ccc} BBB&\qquad GBB&\qquad RBB\\ BBG&\qquad GBG&\qquad \color{blue}{RBG}\\ BBR&\qquad \color{blue}{GBR}&\qquad RBR\\ BGB&\qquad GGB&\qquad \color{blue}{RGB}\\ BGG&\qquad GGG&\qquad RGG\\ \color{blue}{BGR}&\qquad GGR&\qquad RGR\\ BRB&\qquad \color{blue}{GRB}&\qquad RRB\\ \color{blue}{BRG}&\qquad GRG&\qquad RRG\\ BRR&\qquad GRR&\qquad RRR \end{array} \end{align*} To count all valid outcomes, we have to count all $6$ blue marked results and the probability to get three different coloured balls is \begin{align*} \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}\color{blue}{\cdot 6}=\frac{2}{9} \end{align*}