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 Spotty editing (Posted on 2006-02-14)
Anne, Edward, and Isaac are the three editors of the Perplexus Weekly newspaper. This week, Anne spotted 300 errors, Edward spotted 100 errors, and Isaac spotted 200 errors. Altogether, they spotted 404 errors. Assuming that each error is equally easy to spot, about how many errors did they miss?

 See The Solution Submitted by Tristan Rating: 4.3333 (3 votes)

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 Assuming that ... | Comment 3 of 13 |

Assuming that the numbers observed are proportional to the probabilities that the individuals would spot errors, and that the error spotting is independent among the editors, and x is the number of errors altogether, the probability that an error is found (404/x) is the sum of the probabilities of each editor finding the error (100/x + 200/x + 300/x), minus the sum of the probabilities of pairwise finding (100*200/x^2 + 200*300/x^2 + 100*300/x^2), plus the probability that all would find the given error (100*200*300/x^3). So we need to solve:

404/x = 100/x + 200/x + 300/x - (100*200/x^2 + 200*300/x^2 + 100*300/x^2) + 100*200*300/x^3

The above equation is satisfied by x = 500. Since 404 errors were found, most likely 96 were not found. BTW, in this scenario 24 errors were found by all 3 editors. 40 were found by Edward and Isaac (including the 24 found by all); 60 were found by Anne and Edward and 120 by Anne and Isaac.

x = 555
FOR i = 1 TO 40
x2 = x * x: x3 = x2 * x
v = 100 / x + 200 / x + 300 / x - 20000 / x2 - 30000 / x2 - 60000 / x2 + 6000000 / x3
x = 404 / v
PRINT x
NEXT

which converges to 500:

532.2342
518.7455
510.8466
506.2567
503.6024
502.0718
501.1908
500.6841
500.393
500.2257
500.1296
500.0745
500.0428
500.0246
500.0142
500.0081
500.0047
500.0027
500.0016
500.0009
500.0005
500.0003
500.0002
500.0001
500.0001
500
500
500
500

 Posted by Charlie on 2006-02-14 14:44:47

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