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
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?
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.
To find the probability that at least one room has both lights failed, we can calculate the probability that all six lights are not failed and subtract it from 1.
First, let's calculate the probability that a given light has not failed after the certain time. The probability that a given light fails is 0.1, so the probability that it does not fail is 1 - 0.1 = 0.9.
Since all lights are independent of each other, the probability that all six lights have not failed is (0.9)^6. We raise 0.9 to the power of 6 because each light has a 0.9 probability of not failing.
So, the probability that all six lights have not failed is (0.9)^6 = 0.531441.
Finally, we subtract this probability from 1 to find the probability that at least one room has both lights failed:
1 - 0.531441 = 0.468559.
Therefore, the probability that after this time there is at least one room in which both lights have failed is approximately 0.4686, or 46.86%.