What is the probability of picking $3$ $A$'s, $4$ $B$'s, $5$ $C$'s and $8$ $D$'s when you pick $20$ balls?

Question

Q:

Choose $20$ balls from an urn with infinite numbers of balls.
The balls are labeled with $A, B, C, D$ and each has $25$% of chances getting picked.
What is the probability of picking $3$ $A$'s, $4$ $B$'s, $5$ $C$'s and $8$ $D$'s when you pick $20$ balls?

My approach to this question is:

total # of permutations = $4^{20}$
ways of having $8$ $D$'s = ${20 \choose 8}$
ways of having $5$ $C$'s = ${12 \choose 5}$
ways of having $4$ $B$'s = ${7 \choose 4}$
permutation of these = $4!$

Probability: $(4! \cdot {20 \choose 8} \cdot {12 \choose 5} \cdot {7 \choose 4}) / 4^{20}$

I am not sure if I am doing this correctly or not.
Can someone please give me some pointers.
Also, how to deal with it if the probability distribution is not uniform?

Think a little bit more about the $4!$ term. What do you think it's counting? — Mike F, Oct 23 '13 at 16:56
^that's very insightful. Also, (+1) for showing some work, OP! — The Chaz 2.0, Oct 23 '13 at 17:10
Ah, I realized the 4! is not necessary at all. $20 \choose 8$$12 \choose 5$$7 \choose 4$ has already covered all the cases. — jamesclyeh, Oct 23 '13 at 18:11
You may find that that expression makes a lot more sense to you if you don't pack it up like that but just say $\frac{20!}{{8!}{5!}{4!}{3!}}$. — Martin Kochanski, Aug 15 '16 at 10:54

Ross Millikan · Accepted Answer · 2013-11-07T16:41:33.210

1

The factor $\frac 1{4^{20}}$ is the chance of a given configuration. For the non-uniform case, if the chance of drawing (in one draw) an $A$ is $a$ and so on, the factor becomes $d^8c^5b^4a^3$. And, like Mike said.....

edited Nov 07 '13 at 16:41

answered Oct 23 '13 at 18:04

Ross Millikan

374,822

What is the probability of picking $3$ $A$'s, $4$ $B$'s, $5$ $C$'s and $8$ $D$'s when you pick $20$ balls?

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