寻找多项式分布的概率
一个包包含5个dices，每个袋有六个面孔，概率$ p_1 $ = $ p_2 $ = $ p_3 $ = $ 2 p_4 $ = $ 2 p_5 $ = $ 3*p_6 $。选择两个脸部4的骰子和三个用面部1选择三个骰子的概率是多少？ 有人尝试了图片中所示的R中的此问题的代码，但我不明白如何获得概率。请向我解释这个问题的答案。
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.
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首先，根据给定的平等性和以下假设查找
p_is值（替换所有p_is基于它们与p_6的关系以找到它，然后找到每个p_i）的值：要找到概率，我们需要从> 4 其概率为
2中选择5带有facep_4^2，对于其他dices，它们应该具有face1，其概率为p_1^3。现在，要了解最后一步，您应该阅读
dmultinom（x，size，prob）function的说明（来自此帖子）：因此，
dmultinom（j，n，p）表示c（5,2） * p_1^3 * p_4^2，asj =（3，0 ，0、2、0），n = 5和p =（p_1，p_1，p_2，p_3，p_4，p_4，p_5，p_5，p_6）。First, find
p_is values based on the given equalities and the following assumption (replace allp_is based on their relations withp_6to find it, then find the value of eachp_i):To find the probability, we need to select
2out of5of dices with face4which its probability isp_4^2and for other dices, they should have face1that it's probability isp_1^3.Now, to understand the last step, you should read the explanation of the
dmultinom(x, size, prob)function (from this post):Therefore,
dmultinom(j, n, p)meansC(5,2) * p_1^3 * p_4^2, asj = (3, 0, 0, 2, 0),n=5, andp = (p_1, p_2, p_3, p_4, p_5, p_6).