Rolling a biased dice, Multinomial probability What is the easiest way to answer this question.
Lets say you had a biased 6-sided die
P(rolling '1') = P(rolling '3') = 0.1
P(rolling '2') = P(rolling '4') = P(rolling '6') = 0.25
P(rolling '5') = 0.05
The sum is equal to 1.
By rolling 6 dice what is the probability of getting:
P(exactly two 4's and three '1')
$(P(4))^2 * (P(1))^3 * P(x)$ 
where $P(x)  = 1 - 0.1 - 0.25 = 0.65,$
for $x \ne 4$ and $x \ne 1.$
$(0.25)^2 * (0.1)^3 * 0.65 = 13/320000$
how do you find the possible arrangements?
 A: Comment: This is a straightforward problem using a multinomial distribution.
It seems with a combination of what you have done and the Comment by @calculus, you are well on the way. 
As a check on your answer, here is a simulation of a million
performances of the experiment. Some related binomial probabilities
are included to show that the simulation is working as it should.
Simulation approximations are accurate to about two or three places, maybe a little more for very small probabilities.
[Note: $ .00244 \pm 1.96\sqrt{.00244(1-.00244)/1000000}$ amounts to $( 0.002343, 0.002537).$]
Intuitively, why can't you multiply two binomial probabilities to get $your$ answer?
 m = 10^6;  ones.3 = fours.2 = fours.2p = numeric(m)
 pr = c(.1, .25, .1, .25, .05, .25)
 for (i in 1:m) {
    faces = sample(1:6, 6, rep=T, prob=pr)
    ones.3[i]  = (sum(faces==1) == 3)
    fours.2[i] = (sum(faces==4) == 2) }
 mean(ones.3 & fours.2)  
 ## 0.002363     # Approx P(three 1's & two 4's)
 (0.25)^2 * (0.1)^3 * 0.65 * 60 
 ## 0.0024375    # Exact multinomial

 mean(ones.3)    
 ## 0.014494     # Approx P(three 1's)
 dbinom(3,6,.1)
 ## 0.01458      # Exact binomial

 mean(fours.2)
 ## 0.295885     # Approx P(two 4's)
 dbinom(2,6,.25)
 ## 0.2966309    # Exact binomial

 dbinom(3,6,.1)*dbinom(2,6,.25)
 ## 0.004324878  # Events {three 1's} and {two 4's} NOT independent

