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.
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