It's a probability problem to find the odds of picking a green or red
shirt out of the 10 shirts on Thursday, Friday and Saturday since you
have randomly already know you have picked a blue shirt on the other
days. Not sure if you have this as a multiple question problem as you didn't list any possible answers (A. 7/20, B. 5/47, C. 2/5, D. 4/125) to the question. A, B, C, D being like 7 chances out of 20, 5 chances out 47, 2 chances out of chances 5 or 4 chances out of 125 (example answers only).
Probability = Number favorable outcomes / total number of outcomes
If you're asking for the probability of drawing said shirt colors on said days: Assuming there's no replacement (because clothes), the answer would be: (2/5)•(1/3)•(1/4), which equals 2/60, or 1/30. The reason: On Monday, there are 10 shirts, 4 of them blue. So, your chances are 4 out of 10, or 4/10, which reduces to 2/5. On Tuesday, there are 9 shirts left, 3 of them blue. So, 3/9, or 1/3. On Wednesday, there are 8 shirts, 2 of them blue. So, 2/8, or 1/4. You multiply all the fractions together (which is why simplifying helps A TON) Hope this helps! And sorry about not answering sooner!
You have to divide the fractions so it 3/4÷1/10 then you can do keep change flip so the new equation is 3/4×10/1 or 3/4×10 then cross simply then get 3/2×5/1 to get 15/2 then use division to get 7 1/2
