A lamp has two bulbs, each of a type with an average lifetime of 5 hours. The probability density function for the lifetime of a bulb is $\sigma(t)$,
What is the probability that both of the bulbs will fail within 2 hours?
If the probability of an event occurring is given by:
$$\int_{0}^\infty \sigma(t)dt$$
Then is it true that for two independent events, the probability is given by
$$\int_{0}^2\sigma(t)^2dt$$
Best Answer
The probability that the first bulb fails within $2$ hours is $\int_0^2 \sigma(t)\,dt$. For the probability that both bulbs fail, assuming independence, just square the answer for one bulb. This gives $\left(\int_0^2 \sigma(t)\,dt\right)^2$, which is not the same as the answer proposed in the OP.
Without knowing more about $\sigma(t)$, we cannot produce an explicit numerical answer. The fact that the mean is $5$ is not enough to determine the probability a single bulb has lifetime $\le 2$.
Possibly (but unreasonably) you are expected to assume the lifetime has exponential distribution.
Added: It looks as if one is expected to assume exponential distribution. This is not a good model. The exponential distribution is an appropriate model for the lifetime of things that die but do not age, like atoms of a radioactive substance. Lightbulbs age!
But if we assume exponential distribution with mean $5$, then the density function is $\frac{1}{5}e^{-t/5}$ (for $t\ge 0$). Integration shows that the probability the lifetime is $\le x$ is $1-e^{-x/5}$. So the probability both our bulbs are dead by time $2$ is $\left(1-e^{-2/5}\right)^2$.