Question

The life times, Y in years of a certain brand of low-grade lightbulbs follow an exponential distribution with a mean of 0.5 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.

Answer 1

Answer:

Part a:

Assume that each test is independent.

You want P(Y1<1, Y2<1, Y3<1, Y4<1, Y5>1) = (P(Y<1))4 P(Y>1) by independence

Part b:

Here we have failures on 4,5,6, and 7 and success on 8. So 4 failures and a success. This is exactly the same as part a

Part c:

P(odd) = P(Y>1) + P(Y>1)(P(Y<1))2 + P(Y>1)(P(Y<1))4 . . .

= P(Y>1) Σ (P(Y<1))2n for n in the natural numbers.

P(even) = 1 - P(odd)