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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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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):
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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)]
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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.