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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

$$\int_{0}^2\sigma(t)^2dt$$

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  • $\begingroup$ 16 minutes. $ $ $\endgroup$ – Did Jan 17 '13 at 7:38
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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 $\left(1-e^{-2/5}\right)^2$.

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  • $\begingroup$ it does thank you very much $\endgroup$ – Cactus BAMF Jan 17 '13 at 7:18
  • $\begingroup$ @CactusBAMF: What do you mean by "it does"? Do you mean assume exponential? $\endgroup$ – André Nicolas Jan 17 '13 at 7:21
  • $\begingroup$ There is an exponential function associated with the problem. $\endgroup$ – Cactus BAMF Jan 17 '13 at 7:22

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