Ball Probability

First, we find that the total number of possibilities to pick the balls are $12 \times 11 \times 10 \times 9 = 11880$

Let us consider one configuration first. (The subscript denotes which ball it is since there are three balls)

Let $\text{b}_1$ denote blue, $\text{y}_1$ yellow, $\text{r}_1$ red and $\text{bl}_1$ black.

b_1 r_1 y_1 bl_1

We want these balls to NOT be in the same original position. This can be done in $\mathsf{Der}(4) = !4 = 9$ ways where $\mathsf{Der}$ is the derangement function.

Now, how many such configurations are there? Well, for the first position we have $3$ choices, $3$ for the second and so on. Thus, $3^{4}$ choices. For example, we have to derange

b_2 r_3 y_1 bl_3

and so on...

Thus, the total number of ways are $3^4 \times 9$ .

Our answer is thus $\dfrac{ 3^4 \times 9}{ 11880} = \dfrac{27}{440}$ giving us $27 + 440 = \boxed{467}$

This problem based on derangement. Assume that position 1st is blue , 2 - red and so on...

P(E)= no of derangement of 4 balls* no of ways of choosing ball of each color/ total combinations.

$= 9 *3C1*3C1*3C1*3C1/ (12C4 * 4!)$ = 27/440

This approach works for discussing more complicated patterns of balls too.

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from itertools import combinations,permutations
from re import compile
from fractions import Fraction

blue,red,yellow,black='1','2','3','4'
bag=3*[blue,red,yellow,black]

patt=compile('[234][134][124][123]')

count=0
total=0
for pick in combinations(bag,4):
    for order in permutations(''.join(pick)):
        total+=1
        if patt.match(''.join(order)) and len(set(order))==4:
            count+=1
print count,total,Fraction(count,total)
print Fraction(count,total).numerator+Fraction(count,total).denominator

Hey Daniel, would you please post this solution about this problem because my approach is really a mess. It's kind like 'brute force' method because I used Excel to tackle this problem. I believe your solution is more elegant than mine. Thank you... :)

UPDATE :

Let $P$ be the probability that the first is not blue, the second is not red, the third is not yellow, the fourth is not black and all four balls are different-colored, then $$

$P=\;!4\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}\cdot\frac{3}{9}=9\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}\cdot\frac{3}{9}=\frac{27}{440}.$