The first part of the question is also here Probability/Combinatorics Question
At a picnic, there was a bowl of chocolate candy that had 10 pieces of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.
(a) What is the probability that she got one of each variety?
My approach was- let the variable $x_1,x_2,\cdots ,x_6$ denote the number of candies of first type, second type and so on..
Thus total number of ways in which Jen could take out the candies would be equal to the number of positive integral solutions of
$$x_1+x_2+\cdots +x_6=6$$
which is ${6+5}\choose {5}$
And only one configuration ($x_1=x_2=\cdots =x_6=1$) is favourable.
Thus the probability is $$\frac{1}{{6+5}\choose {5}}$$ But this answer doesn't match with the answer given in the linked question .
Where am I going wrong ?
Edit: I understand most answers are claiming that all integer solutions are not 'equally likely', this somehow presupposes that one choice of the probability measure is correct over the other, eg: I could have defined the probability measure such that each integer solution is equally likely. Is there any reason to chose one measure over the other?
Best Answer
Addendum just added to respond to questions from the Original Poster.
As other answers have indicated, while it is true that there are $\binom{11}{5}$ non-negative integer solutions to
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6$,
these solutions are not equally likely.
Let's consider $3$ such solutions: I will use the syntax $(a,b,c,d,e,f)$ to indicate that
$x_1 = a, x_2 = b, x_3 = c, x_4 = d, x_5 = e, x_6 = f.$
Solution 1: (3,2,1,0,0,0).
There are $\binom{10}{3}$ ways that $3$ of the 1st type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 2nd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 3rd type of candy could be selected.
Computation equals $\binom{10}{3} \times \binom{10}{2} \times \binom{10}{1}.$
Solution 2: (2,2,2,0,0,0).
There are $\binom{10}{2}$ ways that $2$ of the 1st type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 2nd type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 3rd type of candy could be selected.
Computation equals $\binom{10}{2} \times \binom{10}{2} \times \binom{10}{2}.$
Solution 3: (1,1,1,1,1,1).
There are $\binom{10}{1}$ ways that $1$ of the 1st type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 2nd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 3rd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 4th type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 5th type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 6th type of candy could be selected.
Computation equals $\binom{10}{1}^6.$
My solution:
Probability equals $$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$
Where $D = \binom{60}{6} = $ the number of ways of selecting $6$ candies out of $60$. Here:
Because of the second point above, the numerator, $N$ must be enumerated in a consistent manner. That means that when enumerating $N$, you must presume that the order that the candies where selected is irrelevant.
Suppose that the candies were numbered $1,2,\cdots,60$, where :
There are $\binom{10}{1} = 10$ choices for which Milky Way candy is selected.
Independently, regardless of which Milky Way candy is selected, there are $\binom{10}{1} = 10$ choices for which Almond Joy candy is selected.
The same type of reasoning indicates that you will have $6$ independent factors to consider: that is, for each type of candy, there are $(10)$ choices for which candy is selected.
In other words, you have $10$ choices for which Milky Way is selected, $10$ choices for which Almond Joy is selected, $10$ choices for which Butterfinger is selected, and so forth.
Therefore, $N = \binom{10}{1}^6 = \left(10\right)^6.$
Final computation:
$$\frac{N}{D} = \frac{(10)^6}{\binom{60}{6}}.$$
Addendum
Responding to the OP's comment and the latest editing of his question.
In order to explore these issues, I will first try to examine in greater detail, each of the $\binom{11}{5} = 462$ solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 \tag1 $$ $$~~: ~x_1, x_2, x_3, x_4, x_5, x_6 \in \Bbb{Z_{\geq 0}}. $$
I will partition the solutions into all of the possible formats. For each individual solution that corresponds to a specific format, I will determine how many distinct distributions are represented by this solution. Then the probability will be computed as
$$\frac{\text{# of satisfying distributions}}{\text{total # of possible distributions}}.$$
I will use the syntax $(a,b,c,d,e,f)$ to indicate that
$x_1 = a, x_2 = b, x_3 = c, x_4 = d, x_5 = e, x_6 = f.$
For a solution that follows format $k$, as detailed below, I will use the variable $t_k$ to denote how many of the $(462)$ solutions have format $k$.
I will then use the variable $w_k$ to denote how many different distributions are associated with each individual solution to (1) above that follows format $k$.
In effect, $w_k$, which denotes a number of distributions, will be used to represent a weight that is given to each of the solutions represented by format $k$. In effect, $w_k$ will represent the relative weight to be given to each of the $462$ solutions, based on how many distributions are associated with the specific solution.
This implies that the product $(t_k \times w_k)$ will represent all of the distributions possible that are represented by format $k$.
The following table examines each of the possible formats:
\begin{array}{| r | r | r | r |r |} \hline \text{#} & \text{Pattern} & t_k & w_k & t_k \times w_k \\ \hline 1 & (6,0,0,0,0,0) & 6 & \binom{10}{6} & 1,260 \\ \hline 2 & (5,1,0,0,0,0) & 30 & \binom{10}{5} \times \binom{10}{1} & 75,600 \\ \hline 3 & (4,2,0,0,0,0) & 30 & \binom{10}{4} \times \binom{10}{2} & 283,500 \\ \hline 4 & (4,1,1,0,0,0) & 60 & \binom{10}{4} \times \binom{10}{1}^2 & 1,260,000 \\ \hline 5 & (3,3,0,0,0,0) & 15 & \binom{10}{3}^2 & 216,000 \\ \hline 6 & (3,2,1,0,0,0) & 120 & \binom{10}{3} \times \binom{10}{2} \times \binom{10}{1} & 6,480,000 \\ \hline 7 & (3,1,1,1,0,0) & 60 & \binom{10}{3} \times \binom{10}{1}^3 & 7,200,000 \\ \hline 8 & (2,2,2,0,0,0) & 20 & \binom{10}{2}^3 & 1,822,500 \\ \hline 9 & (2,2,1,1,0,0) & 90 & \binom{10}{2}^2 \times \binom{10}{1}^2 & 18,225,000 \\ \hline 10 & (2,1,1,1,1,0) & 30 & \binom{10}{2} \times \binom{10}{1}^4 & 13,500,000 \\ \hline 11 & (1,1,1,1,1,1) & 1 & \binom{10}{1}^6 & 1,000,000 \\ \hline \text{sum} & & 462 & & 50,063,860 \\ \hline \end{array}
Notice that $\sum_{k=1}^{11} t_k = 462 = \binom{11}{5}.$
Notice that $\sum_{k=1}^{11} (t_k \times w_k) = 50,063,860 = \binom{60}{6}.$
This means that in the above chart, all of the $\binom{11}{5}$ possible solutions are accounted for and all of the $\binom{60}{6}$ possible distributions are accounted for.
Then, the probability of a satisfying distribution is
$$\frac{t_{11} \times w_{11}}{\sum_{k=1}^{11} (t_k \times w_k)} = \frac{1,000,000}{50,063,860} = \frac{10^6}{\binom{60}{6}}.$$
For illustrative purposes, I will discuss Format # $7$ in greater detail.
There are $6$ choices for which of the $6$ variables will be set to the value $(3)$. Once this selection is made, there are $\binom{5}{3} = 10$ ways of selecting $3$ of the remaining $5$ variables, and then setting these variables to the value $(1)$.
Therefore, $t_7 = 6 \times 10 = 60.$
For each of the $60$ out of a possible $462$ solutions to the equation $x_1 + \cdots + x_6 = 6$, how many distributions are represented by each solution? To answer that, simply determine how many distributions are represented by the specific solution $(3,1,1,1,0,0)$.
This solution calls for $3$ candies to be selected from the $10$ candies represented by the candy type associated with the variable $x_1$. This can be done in $\binom{10}{3}$ ways.
Similarly, there are $\binom{10}{1}$ possible ways of selecting $1$ candy out of $10$, for each of the candy types represented by the variables $x_2, x_3, x_4.$
This explains the value of $w_7$, shown in the chart, and explains the product $(t_7 \times w_7)$ shown in the chart.