Posted by Anonymous on .
A building has three rooms. Each room has two separate electric lights. There are thus six Electric lights altogether. After a certain time there is a probability of 0.1 that a given light will have failed and all light are independent of all other lights. Find the probability that, after this time, there is at least one room in which both lights have failed

maths 
David Q/R,
Concentrate 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,
Oops! 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.