Calculating coffee bags distribution probability A barista has $20$ coffee bags. Then, those Coffee Bags are given randomly and equally to $5$ customers ($1$ customer gets $4$ bags). $5$ out of $20$ coffee bags are raw and the rest are roasted.
A. Calculate the probability of each customer getting at least $1$ raw coffee bag!
B. Calculate the probability of $2$ customers each receiving $1$ raw coffee bag and the other $3$ raw coffee bags are distributed to the same customer!
Here are my attempts to answer the question
A.
$$\frac{5P1 × 15P3}{20P4} = \frac{5×2730}{116280} = \frac{13650}{116280} =0,117$$
B. $$\frac{5P2 × 8P2}{20P4} × \frac{3P1 × 9P3}{12P3} = \frac{1120}{116280} × \frac{1512}{1320} = 0.01$$
I know the second answer is wrong. However, I just can't find the correct solution. Any ideas?
 A: Well for the first part there is no way that if each customer gets one bag, that another one can get more than one bag, so the words "at least" seem misleading.
There are $20\choose5$ ways to distribute the five "raw" bags into 20 slots. Of those choices, there are ${4\choose1}$ different ways to give one bag to each customer, so the number of ways to give out raw bags is
$$
{{4\choose1}^5\over {20\choose5}}={64\over 969}\approx0.066
$$
by my calculations, which disagrees with your number.
There are ${4\choose1}^2\times{4\choose3}\times{5\choose2}\times{3\choose1}$ ways to distribute one bag $4\choose1$ to two customers $5\choose2$ and three bags $4\choose3$ to one of the other three customers, $3\choose1$, which gives us $40/323\approx0.123$ probability by my calculations.
A: A. There are 4 different bags that a person has to collect. There are 5 coffee bags. The chances of picking a coffee bag is 1/4 and it is 1/5 of those bags that the person has to pick. I'd say that it is 1/5 divided by 1/4 which will equal to 4/5 chance picking a raw coffee bag. I might be wrong idk. Just giving a go cus its my first time answering a question. Ok good luck have a nice day.
