## Typing test

Posted by Chris on February 24, 2011 – 5:14 am

A document was independently checked for errors by two people. One of them found 32 errors, the other found 24 errors. It turned out that 8 of those errors were noticed by both people.

Estimate how many errors there actually were in the document.

February 24th, 2011 at 5:57 am

It says that 8 of those were noticed by both but tells nothing about the remaining ones.

Moreover, we know of errors which were found and not those which weren’t.

So let’s first see what we can deduce about errors which were found :

There are at least 8 errors both people found, and in the case these are the only such ones, we would have this number of found errors :

found_by_A + found_by_B – found_by_both = 32 + 24 – 8 = 48 errors.

In fact if we imagine that Chris did not think of possible additional errors found by both which the text does not tell us of, this number, 48, would be the minimum number of errors in all this text.

However, if there are more errors which were found twice, the minimum number of errors drops down to 32 i.e. one of them did a bad job and only found errors the other also already found and not even all of them.

Now for a probabilistic analysis. Let’s say there are X errors in this text, let respectively P and Q be the probabilities that the first one finds a given error and that the second one does. We will call A the first one (who found 32) and B the second one.

We will call a and b the numbers of errors A and B respectively found and c the number of those both found. (here a=32, b=24, c=8, but letters will be more practical to generalize)

If the probabilities match exactly with the numbers, we have :

p*X = a, q*X = b,

and

p*q*X = c.

because we know the document was independently checked by 2 ppl.

finally we get from substitution in the third equation :

p*(q*X)=c, and q*X=b, then p=c/b

same thing on the other side : q=c/a.

it seems strange that the probability that A found an error only depends on c and b ; we may think “where is the a ???” but in fact the number c also depends on a…

then we just have to solve it all :

p=c/b=1/3

q=c/a=1/4

X=a*b/c=96.

Those two are in fact very bad at checking errors…

We have an approximate value of 96 errors in the document (i.e. if the probabilities were not exactly those we calculated from the text)

Error check : Our formulas for p, q and X seem coherent with extreme cases : if A was so competent he found all errors (p=1) we clearly have b=c (that is every error B finds has been found by A), and X=a… same thing with B. We could also check the case in which A and B were absolutely incompetent…

February 24th, 2011 at 6:35 am

Hi Karys. That was quick and correct, well done. I had thought this one was going to last a fair bit longer.

I probably should have made the commonly discovered errors a bit bigger. As you say, both of the checkers are dreadful. So this analysis provides a nice way of checking the checkers.

Recap, let E be the total number of errors. Let the first checker have a probability a of detecting each error and let A = 32 be the number of errors he found. Let checker B have a probability b of detecting each error and let B = 24 be the number of errors he found. Let C = 8 be the commonly discovered errors.

Then A = aE, B = bE and C = abE. The last can be thought of as a(bE) or b(aE). So AB = abE² = CE => E = AB/C = 32*24/8 = 96.

We also see that a = 1/3 and b = 1/4. Sack both of them.

February 27th, 2011 at 2:23 am

48

February 27th, 2011 at 7:53 am

Hi bibek. They actually found 32+24-8 = 48. But there were more like 96 errors altogether. They missed half of them.

In fact the number of errors they missed is

E – (A+B-C) = (A-C)(B-C)/C

The fraction they missed is (A-C)(B-C)/(AB)

February 27th, 2011 at 12:41 pm

I think there’s eight errors in the document.

February 27th, 2011 at 1:15 pm

Hi laura. I’d love to know how you worked that out.

February 27th, 2011 at 2:40 pm

40

February 28th, 2011 at 3:48 am

48 errors

February 28th, 2011 at 1:32 pm

Their efficiency is considered to be a fixed number so, if total number of errors is x, then eff of A a=p/x; n B is b=q/x; out of p errors A found, B found r common errors. so b=r/p. so r/p=q/x; so x=pq/r; x=96. Its the same way we can count th number of fish in a small pond.

March 3rd, 2011 at 8:34 pm

48 errors

March 3rd, 2011 at 9:09 pm

Hi is it right and anyone else who doesn’t say 96. 96 is about the best estimate that you can make. 48 = 32+24-8 is the minimum number of errors. By the way, that 96 is 2*48 is a coincidence.

If I’d said that checker A had found 32 errors, and that checker B had found 30 errors and that 29 were common, then there would definitely be 32+30-29 = 33 errors. But the best estimate of the real number of errors is 32*30/29 = 33.1, so between them they’d found 99.7% of all the errors.

Then checker A’s quality is about 29/30 = 96.7%

and checker B’s quality is about 29/32 = 90.6%

March 7th, 2011 at 3:59 am

It is easy . A found 32 errors and B found 24 . 8 errors which they found were common . So ,

X = (A+B)-C

X = (32+24)-8

X = 56-8

X = 48

So , the total no. of errors were 48

March 7th, 2011 at 5:33 am

Hi Tanvi. You’ve not answered the question.

March 9th, 2011 at 3:17 pm

there would be 48

March 10th, 2011 at 6:57 am

LOL. What a lot of write only posts. Several of you have failed to learn anything from this problem..

48 is how many errors have definitely been found. It is not an estimate of 96 or so likely errors.

March 10th, 2011 at 10:11 am

There are 48 errors.

They found 8 errors in common, so we need to add 8 to the number of errors they found in common. How do we do that?

If you take 8 away from both errors totals(since 8 was found by both, you need to subtract 8 each from both. The number of errors they found “not in common”:

(32 – + (24 – = 24 + 16 = 40 –> errors not in common

Now add errors in common with errors not in common:

40 + 8 = 48

March 10th, 2011 at 10:45 am

Hi Dr 96. That’s right, they actually found 48 errors; that’s 32 + 24 – 8 = 48. But the question was to estimate the total number of errors, not to say how many errors had actually been found.

March 14th, 2011 at 11:41 am

8

April 5th, 2011 at 5:51 am

24+32-8 = 48

April 5th, 2011 at 5:53 am

“That’s right, they actually found 48 errors; that’s 32 + 24 – 8 = 48. But the question was to estimate the total number of errors, not to say how many errors had actually been found.”

then we can treat it as insufficient detail..right Chris?

April 5th, 2011 at 6:39 am

Hi srinu. The answer is 96. See posts 1 and 2 (and others).