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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Friday, March 2, 2012 at 6:01am.

- maths -
**David Q/R**, Saturday, March 3, 2012 at 3:51pmConcentrate on a single room for a moment. It's got two bulbs, each of which has a 1/10 chance of failing, so at the end of the time interval in question the chance of a double failure is 1/10 x 1/10 = 1/100. It's only double failures we're interested in here, so we don't need to bother to work out the chance of a single failure or no failures, even though they're easy enough to work out. What we DO need to work out is the chances of this NOT happening - and that's just (1 - 1/100) = 99/100.

Now look at all three rooms. We want to know the chances of there being AT LEAST ONE room in which both lights have failed - and that's just one minus the probability of all THREE rooms NOT having a double failure. But we know the probability of just one room not having a double failure, because we worked it out in the previous paragraph: it's 99/100. We know all failures are independent of one another, so the probability of all three rooms not having a double failure must be (99/100)^3, which is roughly 0.97. What do you think - does the logic hold together?

- maths -
**David Q/R**, Saturday, March 3, 2012 at 3:56pmOops! Actually, we want one minus that, don't we. The chances of at least one room having a double failure is one minus the chances of none of them having a double failure - and that's 1 - 0.97, which is 0.03. That's more like it.

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