mmeyer@m2.csc.ti.com (Mark Meyer) (04/04/90)
In article <1609@nvuxr.UUCP> deej@nvuxr.UUCP (David Lewis) writes: >A better approach would be to calculate the chance of losing at least >one orbiter in N launches, because the above approach discounts the >chance of two (or more) catastrophic failures on different flights. >Although in reality the first would probably put a long hold, if not a >stop, on the shuttle program, this calculation understates the >probability. I'm not up to derive the calculation for this in real >time, though. Maybe later. That's easy. All you have to do is compute the probability of no failures at all and subtract from 1. This gives the probability of at least one failure. If p is the reliability, then the probability of no failures in N launches is simply p^N. Therefore, the probability of losing at least one orbiter in N launches is 1 - p^N. By the way, if p=0.98, then the N such that 1-p^N = 0.9 turns out to be 114, the same number David derived. -- Mark Meyer USENET: {ut-sally!im4u,convex!smu,sun!texsun}!ti-csl!mmeyer Texas Instruments, Inc. CSNET : mmeyer@TI-CSL Every day, Jerry Junkins is grateful that I don't speak for TI. "He founded an Origami magazine, but it quickly folded."