probability難題XDD

將抽到的可能性全部相加便行，抽到的次序有6個可能性，括號內為抽到的機會率：
1) BBWW (2/5 x 1/4)
2) BWBW (2/5 x 3/4 x 1/3)
3) BWWB (2/5 x 3/4 x 2/3 x 1/2)
4) WBBW (3/5 x 2/4 x 1/3) (與第2個情況有相同機會率)
5) WBWB (3/5 x 2/4 x 2/3 x 1/2) (與第3個情況有相同機會率)
6) WWBB (3/5 x 2/4 x 2/3 x 1/2) (與第3個情況有相同機會率)
將以上情況相加：
(1) + (2) + (3) + (4) + (5) + (6)
= (2/5 x 1/4) + 2(2/5 x 3/4 x 1/3) + 3((3/5 x 2/4 x 2/3 x 1/2)
= 1/10 + 1/5 + 3/10
= 6/10
= 3/5 (或0.6)
抽到兩個黑波及兩個白波的機會率是3/5。

This is a question about hypergeometic distribution
The total combinations to draw 4 balls from 5 balls are
5C4=5
The total combinations to draw 2 black balls and 2 white balls from 2 balck balls and 3 white balls are (2C2)(3C2)=3
So the required probability is
3/5