For N = 4, all those "at least"s should be "exactly".
My 7-grade daughter brought this problem home today and having trouble figuring it out.
Each box contains a pen with one of four colors, equally likely. Each box is independent.
After opening N (N=4) boxes, what is probability you have a full set (all 4 colors)?
My daughter came up with the following solution which I verified with a quick Matlab run.
P(set)=P(any color in box 1)*P(different color in box 2)*P(different color in box 3)*P(different color in box 4)
P(set)=1*.75*.5*.25
P(set)=.094
I tried a more formal method (because the next problem requires a higher value of N) and got a different answer.
P(set)=P(at least one blue)*P(at least one red|at least one blue)*P(at least one yellow|at least one red & blue)*P(at least one green|at least one red & blue & yellow)
P(at least one blue)=[1-P(no blue in any box)]=[1-.75^4]=.684
P(at least one red|at least one blue)=[1-P(no red in 3 boxes)]=[1-.75^3]=.578
P(at least one yellow|at least one red & blue)=[1-P(no yellow in 2 boxes)]=[1-.75^2]=.438
P(at least one green|at least one red & blue & yellow)=[1-P(no green in last box)]=[1-.75]=.25
So, P(set)=.684*.578*.438*.25=.043
I know the first answer is right. I don't know why the second method doesn't work. Can someone please point out my mistake?
For N = 4, all those "at least"s should be "exactly".
But for N>4, it wouldn't be "exactly", it would be "at least". I'm trying to find the general solution and then apply it to N=4 to verify it.
The problem is making sure that you don't count anything twice. Here is a solution if you want one -
http://math.stackexchange.com/questions/1814316/
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