brian bovril

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17 years, 288 days

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These are replies submitted by brian bovril

@Carl Love edited. It seems to me there are two probabilities. The probability of the particular configuration occurring, which is 1/1001. Then testing whether that configuration was a result of proper randomisation. You applied Fishers test and got probability 6/1001. Because that is a lot less than 0.05 it is significant. i.e there is evidence the pick wasn’t done randomly. I didn’t really think about a test for significance when I originally posted the question.

I’m not sure if you are tearing your hair out at my interpretation but I’m going to bed now.

@Carl Love I don't have Maple 2019. But thanks anyway. I put it through your 2018 code and it works....

So my old maths prof was correct for the probability...1 / 1001 = 0.000999

@mmcdara 

Thankyou for your effort. Just to clarify, each contestant was given an opaque bag containing a buff.

Once they were dispensed, they were told to reveal their buffs (simultaneously). So that would be Scenario 2.

 

@acer On yours it worked, but I tried this in my worksheet, but label 9 still displays USD....

Feel free to peruse the problem i'm trying to solve, its a work in progress..

Solar_Battery_Payoff.mw

 

Yuri, how can I use you code to solve a 2 bucket problem with unlimited water.

https://mindyourdecisions.com/blog/2013/02/04/the-water-jug-riddle/

Happy Xmas!

@Joe Riel Sorry for being thick, but whats the instructors number prediction? Would it be an envelope containing one (or more?) of the 8 most frequent sums 1755,1638,1656,...?

is our friend ?

Markiyan Hirnyk 

http://www.mapleprimes.com/posts/200573-Putnam-Done-With-Maple

@Rouben Rostamian  for your effort!

As the error says, you have a symbolic upper limit z. You need to change it to a number

@Rouben Rostamian  I suppose its too "human crafted" to make a general procedure out of it.

@acer thanks

@tomleslie Tom and Joe and Carl

I assume you're using 2D input. Try it with 1D input

@Carl Love Thanks. So for LambertW(0.4) 

first iteration: --> x=y (keep x and y the same so Ln(1)=0). For Order 3: W(0.4,0.4) = 0.297061159

next iteration: W (0.297061159, 0.4) = 0.2971677 .

good!

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