binomial distribution - colors In a package of M&Ms there are 20 pieces. The pieces come in 6 different colors: red, blue, green, yellow, orange and brown.  Assuming that the M&M colors occur with equal probability, what is the probability of getting 5 red M&Ms in a package. 
Solution:
Would I be doing:
$$\dfrac{20!}{15!\cdot 5!}  \cdot  \left(\dfrac{1}{6}\right)^5   \left(\dfrac{5}{6}\right)^{15}  = 0.1294$$
Thx.
 A: That's correct. If you have a probability $p$ of success (getting a red M&M), and probability $q = 1 - p$ of failure (getting a blue or green or yellow or orange or brown M&M), then the probability of getting exactly $k$ successes in $n$ random trials (a package of $n$ randomly picked M&Ms) is
$${\large ^nC_k}\, p^k q^{n - k}$$
In this case, that's ${\large ^{20}C_5}\left(\dfrac{1}{6}\right)^5\left(\dfrac{5}{6}\right)^{15}$, as you've calculated.
A: That is assuming the M&M's are ordered inside the bag, which doesn't make a lot of sense, if we don't care about the order within the bag:
Total number of packages can be obtained using stars and bars. We get $\binom{25}{5}$.Since we have to divide the $20$ sweets into $6$ colors.
There are $\binom{19}{4}$ bags containing exactly $5$ red M&M's.
There are $\binom{20}{5}$ bags containing at least $5$ red M&M's. So the probabilites for having exactly $5$ reds or at least $5$ reds are $7\%$ and $29\%$ respectively
A: $$\dfrac{1}{\sqrt{npq}}\dfrac{1}{\sqrt{2\pi}}{\Large e}^{-\dfrac{(k-np)^2}{2npq}}$$
In addition to the binomial distribution, I attempted to solve this using the above formula (approximation) since n is rather large and p is > 0.1.  However, my results between the 2 approaches are not spot on (not the same) (.1452 and .1294)
