Probability and permutation A triangular spinner has one red side, one blue side and one green side. The red side is weighted so that the spinner is four times more likely to land on the red side than on the blue side. The green side is weighted so that the spinner is three times more likely to land on the green side than on the blue side. The spinner is spun 3 times. Find the probability that it lands on a different coloured side each time. I get it that we have p(RBG) = 1/8*4/8*3/8 But then we multiply it with 3p3 why do we consider arrangements in probability question? Please I need an explanation.

From your conditions we can deduce $P(B) = \frac{1}{8}, P(R) = \frac{4}{8}=\frac{1}{2}, P(G) = \frac{3}{8}$.

We are seeking the probability that a sequence of spins happens where exactly one of each of the colours is landed on.

These possibilities are $RGB$ in any order, or, explicitly: $RGB, RBG, GRB, GBR, BRG, BGR$.

We can calculate the probability of getting the sequence $RGB$ as $P(R) \cdot P(G) \cdot P(B) = \frac{3}{128}$. But we want the probability of _any_ arrangement of this sequence occurring, so we multiply by ${}^3\\!P_3 = 6$ (as this event is $6$ times more likely) to give $\frac{9}{64}$.