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Question statement is like this: You brought a box of lightbulbs from a shop. You know that the bulbs are all either short-life bulbs with mean life of 500 hours or long-life bulbs with a mean life of 2500 hours, but you can’t tell which, because there was no label on the box.

As you have not shopped from this shop before, you initially have no opinion as to whether you have been sold long-life bulbs or the cheaper alternative.

After approximately 300 hours you find that 5 bulbs are alive. Assuming that life of an individual light bulb has an exponential distribution, how would you now assess the probability that you bought the long life bulbs?

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I dont understand the approach to solve it. I tried testing of hypothesis, but neither 'n' nor significance level is given. Tested using MP test, H0:2500 vs H1:500.

Another method I thought to try failure rate. but instead of failure rate, survival rate is given. I don't know if survival rate exist in stats ( 5 bulbs survived in 300 hrs).

Please help

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  • $\begingroup$ You will get a more enthusiastic reception on this site if you replace "Please help" by saying what you have studied recently and showing something related that you have tried. The hypothesis testing approach doesn't seem a good start. My 'Answer' gives a clue that may help. $\endgroup$
    – BruceET
    Aug 26, 2017 at 5:59
  • $\begingroup$ If the mean time to failure of an exponential distribution is $\mu$ hours, then the failure rate is $\lambda = 1/\mu$ per hour. $\endgroup$
    – BruceET
    Aug 26, 2017 at 6:19

1 Answer 1

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The solution uses Bayes' Theorem. I'll try to get you started.

(1) Either you bought short-lived bulbs $S$ or long-lived bulbs $L$. Your prior probabilties (before testing the bulbs at home) is $P(S)=P(L) = 1/2.$

(2) Let $E$ be the event that five bulbs tested (presumably out of five) are still alive after 300 hours.

(3) You seek $P(L|E).$

Bayes' Theorem says $$P(L|E) = \frac{P(L\cap E)}{P(E)} = \frac{P(L)P(E|L)}{P(L)P(E|L)+P(S)P(E|S)}.$$

Knowing what you do about the exponential distribution, you should be able to find $P(E|L)$ and $P(E|S)$ in order to finish the problem.

What is the probability one 'long-lived' bulb lasts for at least 300 hours? What is the probability all of five long-lived bulbs last for at least 300 hours? And so on. with the short-lived bulbs?

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    $\begingroup$ five bulbs tested (presumably out of five) Think you should move that at the top of the answer, since it's not stated in the question, and (while sensible) it's not all that obvious an assumption to make. $\endgroup$
    – dxiv
    Aug 26, 2017 at 6:17
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    $\begingroup$ @dxiv: Now that your Comment has brought extra attention to this point, I think all is well. (I guess independence is another sensible unstated assumption.) $\endgroup$
    – BruceET
    Aug 26, 2017 at 6:22
  • $\begingroup$ Thanks @BruceET I get the picture that we gonna use Bayes theorem but still unsure that why we have assumed only 5 trials. can we use E as a event of 5 bulbs (gamma distribution n =5) surviving 300 hrs. or whether the question is incomplete (total bulbs not given) $\endgroup$ Aug 27, 2017 at 3:22
  • $\begingroup$ The sentence "After approximately 300 hours you find that 5 bulbs are alive." is unclear. I took it to mean that five bulbs are tested independently and all are alive after 300 hours. Put in another way the minimum lifetime of five bulbs exceeds 300. // The sum of the lifetimes, not the minimum would be gamma distributed. // Because we do not know how many bulbs are in the box, various other possible interpretations would make the problem difficult to solve. Also the word approximately seems strange. // My interpretation makes the question similar to others I have seen--and solvable. $\endgroup$
    – BruceET
    Aug 27, 2017 at 7:16
  • $\begingroup$ You still there? Next steps would be to find the probability that a 'short lived' bulb survives longer than 300 hours; then the probability that five such bulbs all survive longer than 300 hours to get $P(E|S) \approx 0.05.$ Then similarly to get $P(E|L) .$ Look at CDF formula for exponential dist'n. $\endgroup$
    – BruceET
    Aug 27, 2017 at 18:00

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