1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Exponential Distribution Probability

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    The life times, Y in years of a certain brand of low-grade lightbulbs follow an exponential distribution with a mean of 0.6 years. A tester makes random observations of the life times of this particular brand of lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise.

    Part a) Find the probability to 3 decimal places that the first success occurs in the fifth observation.

    Part b) Find the probability to 3 decimal places of the second success occurring in the 8th observation given that the first success occurred in the 3rd observation.

    Part c) Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.

    2. Relevant equations
    Z= (x-mean)/stddev

    3. The attempt at a solution

    for part a)
    Assuming that each test is independent, I needP(Y1<1, Y2<1, Y3<1, Y4<1, Y5>1) = (P(Y<1))^4 P(Y>1) by independence.
    I found E(X) = 0.6 and Var(X) = 0.36 so STDEV = 0.6. Then I found the Z score to be 0.667, which corresponded to P(Y<1) = 0.7486.

    So I then did (0.7486)4 * (1 -0.7486) which gave me 0.0789, but this is incorrect. Do you see where I made a mistake?

    Am I doing it completely wrong?

    Any help is appreciated, thanks
  2. jcsd
  3. Mar 1, 2015 #2


    User Avatar
    Homework Helper

    You calculated normal distribution and not exponential.
  4. Mar 1, 2015 #3
    What formula am I meant to use? I only see F(x) = 1 - e^(lamda*t) but that doesn't look like the appropriate formula for this question
  5. Mar 1, 2015 #4


    User Avatar
    Homework Helper

    Check that formula for how lambda relates to the mean.
    I think normally you want exp(-t/mean) for exponential.
  6. Mar 1, 2015 #5
    I realize that lamda is just 1/mean but I am unsure what to use for t, can I just use t = 5? This seems like the probability that the first success is WITHIN the first 5 tries rather than on the 5th try like required
  7. Mar 1, 2015 #6


    User Avatar
    Homework Helper

    T is the time of the event. You are running the test for 1 year. Find the probability of failure before one year. Success is 1-p(fail).
    You can use the rest of your logic from above. P(first success on 5th) = p(fail)^4*p(success)
  8. Mar 1, 2015 #7

    got it, thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted