Good question! IMO D as well.
We know that 60% of n shirts are white in colour. So 3n/5 shirts are white. And 2n/5 shirts are non-white.
p(non white shirt with tailoring defect) = 3/4
So total non white shirts with tailoring defects are 3/4*2n/5 = 3n/10.
Statement 1: If one shirt is selected at random from the box, the probability that it will have a tailoring defect is 0.5
So total defect shirts(white+non white) = n/2. Out of these n/2 shirts defect shirts, 3n/10 shirts are non white defects. So shirts with white defects = n/2-3n/10 = 0.2n shirts.
% of white defect shirts: (0.2n/n)*100=20%
Statement 2: The probability that a shirt with tailoring defect will be white in color is 2/5
White shirts=3n/5, Non white shirts=2n/5
Non white shirts with defects= 3/4*2n/5=3n/10
Given: p(white shirt with tailoring defect)=2/5
# white shirts with tailoring defect/All shirts with tailoring defect = 2/5
Let # white shirts with tailoring defect=a
a/(3n/10+a)=2/5.....(Non white shirts with defects=3n/10)
5a=6n/10+2a
a=2n/10
% of white defect shirts: (0.2n/n)*100=20%
Thus, option D. Hope this helps!