probabilitymultinomialprobability-distribution

Finding Probability of Multinomial distribution


A bag contain 5 dices and each have six faces with probability $p_1$=$p_2$=$p_3$=$2p_4$=$2p_5$=$3*p_6$. What is the probability of selecting two dice with face 4, and three dice with face 1? Someone have try the codes for this problem in r shown in picture but I not understand how the probability is obtained. Kindly explain me the answer for this problem.

enter image description here


Solution

  • First, find p_is values based on the given equalities and the following assumption (replace all p_is based on their relations with p_6 to find it, then find the value of each p_i):

    p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 1
    

    To find the probability, we need to select 2 out of 5 of dices with face 4 which its probability is p_4^2 and for other dices, they should have face 1 that it's probability is p_1^3.

    Now, to understand the last step, you should read the explanation of the dmultinom(x, size, prob) function (from this post):

    Generate multinomially distributed random number vectors and compute multinomial probabilities. If x is a K-component vector, dmultinom(x, prob) is the probability P(X[1]=x[1], … , X[K]=x[k]) = C * prod(j=1 , …, K) p[j]^x[j]. where C is the ‘multinomial coefficient’ C = N! / (x[1]! * … * x[K]!) and N = sum(j=1, …, K) x[j]. By definition, each component X[j] is binomially distributed as Bin(size, prob[j]) for j = 1, …, K.

    Therefore, dmultinom(j, n, p) means C(5,2) * p_1^3 * p_4^2, as j = (3, 0, 0, 2, 0), n=5, and p = (p_1, p_2, p_3, p_4, p_5, p_6).