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?

up vote 0 down vote accepted

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
  • 1
    Its a little hard to follow but I believe by simply multiplying my answer with 60 gives 39/16000 which is between the interval you gave. – TaZlyy Mar 20 '16 at 2:23
  • Yes, that's 9th and 10th line of code. Suggest your text or Wikipedia on 'multinomial distribution' to pull the use of the distribution together in a memorable way. Generalization of binomial. R code uses 'logical' vectors with elements TRUE and FALSE. When 'coerced' into arithmetic TRUE=1, FALSE=0. The 'mean' of a logical vector is its proportion of 'TRUEs'. – BruceET Mar 20 '16 at 2:34

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