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  #21  
Old 10-28-2005, 04:18 PM
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Default Re: Riddle -- Probability of Expectation

I got a slightly lower chance for the lines to have all of the characters, and my methods may have been off for it.

That said; Bruce, there is a sign error in the inclusion/exclusion expression as posted.
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  #22  
Old 10-28-2005, 10:42 PM
AlphaWice AlphaWice is offline
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Default Re: Riddle -- Probability of Expectation

This is such an easy problem.

Work with one line only: if we know the probability for one line, we are done.

Let A, B be the probability of "hues", "hews" occuring in that line. P(A union B) = P(A) + P(B) - P(A intersection B)
where all of them are easy to calculate.

Here I haven't actually calculated anything, but it is very easy to do, and requires only grade 11 math.
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  #23  
Old 10-28-2005, 11:19 PM
BruceZ BruceZ is offline
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Default Re: Riddle -- Probability of Expectation

[ QUOTE ]
I got a slightly lower chance for the lines to have all of the characters, and my methods may have been off for it.

That said; Bruce, there is a sign error in the inclusion/exclusion expression as posted.

[/ QUOTE ]

The only sign error I found wasn't in the numerical calculation. It was this one at the end of the 4th line of the set up (corrected in red):

[ QUOTE ]
P = P(no "h-u-e-s" AND no "h-e-w-s") =

3*P(no h) + P(no u AND no w) -

3*P(no h AND no e) - 3*P[no h AND (no u AND no w)] +

P(no h AND no e AND no s) + 3*P[no h AND no e AND (no u AND no w)] <font color="red"> -</font>

P[no h AND no e AND no s AND (no u AND no w)]

[/ QUOTE ]

Is that the one you meant?

Each term in the 3rd line is meant to be subtracted, each term in the 4th line is meant to be added, and the final term is subtracted. There are no implied parentheses around each line.

I don't think there is any question that this calculation is correct is there? I'm not sure why people are still working on it. I think some were scared off by the complicated looking expression above, but the terms themselves were very simple to compute. All I'm doing is noting that the event of NOT getting "h-u-e-s" or "h-e-w-s" in one line is the the union of

(no h) U (no e) U (no s) U (no u AND no w).

Then I'm computing the probability of this union as the sum of the probabilities minus the probabilities of the intersections.
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  #24  
Old 12-19-2005, 01:39 AM
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Default Re: Riddle -- Probability of Expectation

Answer to come tomorrow morning. My error was that I didn't think it out thoroughly.
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  #25  
Old 12-19-2005, 04:52 AM
bigt2k4 bigt2k4 is offline
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Default Re: Riddle -- Probability of Expectation

I am getting 92% in 2nd year stats and I could do this queston most likely, but I read this 2 days after my final for the course, so I do not want to, and screw you for reminding me of the library(and not the fun 5th floor aka the sex floor)
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  #26  
Old 12-19-2005, 05:12 AM
ohnonotthat ohnonotthat is offline
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Default There\'s a problem here

If we have the monkeys spend all this time typing will they be too tired to play 3-6 ?

I'd miss them - although of late even they have been beating me like a red-headed step child.

I need to check my Chineese calender; I think this was the year of the FISH.

[img]/images/graemlins/confused.gif[/img]
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