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Fri 20 Jan, 2017 11:23 am
My 7th-grade daughter brought this problem home today and I'm 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?
@bekmonster,
Bek you might find that many of us resent others' homework
However no offense and good lukt'ya
@dalehileman,
The homework was due this morning. At this point I'm just frustrated I couldn't work out the answer. I was fairly certain my method would work but clearly didn't because I couldn't verify it for N=4.
Whatever. I was in the pursuit of knowledge. I'll look elsewhere.
@bekmonster,
Bek hang in there! Some of our guys are experts at nearly everything and if interest aroused will respond anyhow
There's another group, however, we csll the TAATANE, who can find a reason to crit nearly anything so you must learn to ignor'em
@dalehileman,
sorry. its just been 25 yrs since my last statistics class but its still embarrassing that I can't answer my 7th grader's questions. I thought for sure I had this one but it clearly doesn't verify for N=4. I clearly put some work into this so I'm not just asking for the answer. I'm looking for where I may have gone wrong in my thinking. I was hoping this was a no-brainer for a probability geek and could get a quick answer.
@bekmonster,
Well, Kmon, it's only one day. I'd be willing to bet you'll hear from Eng or one of those other fellsa who are real math wizards
Oft thought ever mem oughta register in a link describing his specialty
@dalehileman,
I've figured out why my method won't work. I was calculating the P(at least one green|at least one red & blue & yellow) incorrectly. For N=4, I assumed that the color of the last pen was equally likely - its not. Given the permutations, you actually have a 40% chance of having the last color if you already have at least one of the other three. So I need to start over anyway.
@bekmonster,
Aha! Bek (as if I understood perfectly)
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