in reply to Testing by Contract
Let T be the total unknown number of possible bugs associated with all combinations. Let A be the number of bugs found by Tester One. Let B be the number of bugs found by Tester Two. Let C be the number of bugs found by both Tester One and Two.Hence (let P(X) be probability of X)
P(A and B) = P(C) (by definition)That means, the less bugs both Tester One and Two found at the same time, the more likely there're still a large number of unknown bugs yet to be found.
P(A)P(B) = P(C) (independence assumption) A B C --- * --- = --- T T T A*B ----- = T C
+----------------------------------+ | | | +------------+ | | | | T | | | A | | | | | | | | +------|-------+ | | | | C | | | | +-----|------+ | | | | B | | | | | | | +--------------+ | | | +----------------------------------+
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Re: Re: Testing by Contract
by BrowserUk (Patriarch) on Jul 01, 2003 at 11:51 UTC | |
by chunlou (Curate) on Jul 01, 2003 at 12:52 UTC |