At this time, the third pile of sunspots accounts for two-fifths of all sunspots, and the first and second piles account for three-fifths of all sunspots. That is, X+Y sunspots account for 3/5 of all sunspots, that is, the number of each pile is 3/5 of all sunspots, so the total number of three piles is 9/5 times of all sunspots, and the number of each pile is X+Y, accounting for 3/5 of all sunspots.
The third pile of sunspots accounts for 2/5 of all sunspots, and the total amount accounts for 3/5 of all sunspots, so the whitening amount of the third pile accounts for 1/5 of all sunspots.
The Bai Zi number of the first and second piles is the same as the number of sunspots, accounting for 3/5 of all sunspots.
So (1/5+3/5)/(9/5) = 4/9.
Problem solving process: Let all sunspots be x, then the number of sunspots in the third pile is 2/5X.
The sum of sunspot numbers in the first and second piles is 3/5X.
Because there are as many sunspots in the first pile as there are Bai Zi in the second pile, if the first pile and the second pile are mixed together, the black and white pieces will be half, and because there are as many pieces in the three piles, the number of pieces in each pile is 3/5X.
After the first and second piles are mixed, the black and Bai Zi numbers are both 3/5X.
In the third pile, Bai Zi number is 3/5x-2/5x = 1/5x.
So the total Bai Zi number is the sum of three piles of mixed Bai Zi divided by the total number of fragments.
There is (3/5x+ 1/5x)/[(9/5) x] = 4/9.