Science:Math Exam Resources/Courses/MATH103/April 2011/Question 05 (c)

MATH103 April 2011

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Question 05 (c)
In a room two different lightbulbs $A$ and $B$ are installed. According to the packaging, lightbulb $A$ fails before time $t$ with probability $F_{A}(t)=1-e^{-t}$ ($t$ in months). Lightbulb $B$ has a probability density function for the failure time given by $p_{B}(t)=Ce^{-2t}.$ What is the probability that both lightbulbs are still working after $t$ months?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A light bulb is still working after $t$ months if and only if it has not failed before $t$ months.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The probability that lightbulb $A$ is still working after $t$ months is equal to the probability that it has not failed before $t$ months, which is $1-F_{A}(t)=e^{-t}.$ The probability that lightbulb $B$ is still working after $t$ months is equal to the probability that it fails at a time greater than $t$: ${\begin{aligned}\int _{t}^{\infty }p_{B}(x)\,dx=2\int _{t}^{\infty }e^{-2x}\,dx=2\left[{\frac {e^{-2x}}{-2}}\right]_{t}^{\infty }=e^{-2t}.\end{aligned}}$ The probability that both lightbulbs are still working after $t$ months is equal to the product of the two probabilities found above, which is ${\begin{aligned}e^{-t}e^{-2t}=e^{-3t}.\end{aligned}}$

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