Question

Is it Exponential distribution? or Geometric distribution?
The life times, Y in years of low-grade lightbulbs follow an exponential distribution with a mean of 0.7 years. A tester makes random observations of the life times of this 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 that the first success occurs in the fifth observation.

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

Answer 1

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 2

I believe you use the exponential distribution to find the probability of success (say "p"). Then, you use the Geometric(p) distribution using the "p" you calculated from the exponential as your success probability. Also, in b), you use a negative binomial (which is really just a generalized geometric distribution).

Answer 3

So I can calculate part a with the "p" I calculated from exponential??

Answer 4

Yeah. Some more information:
Let's define X = # of failures before the 1st success.
Then, the pmf of X is a Geometric(p) distribution: \(f_X(x)=(1-p)^xp,\,\,\,\ x=0,1,2...\)

Now, what is \(p\), the probability of success??
p = probability of success
= probability of lifetime exceeding 1 year

Now, We know Y = life time of light-bulb, hence,
p = P(Y > 1)

Notice that the mean of the exponential is 0.7, and \(E(Y)=\frac{1}{\theta}\) if the pdf is parameterized as \[f_Y(y)=\theta e^{-\theta x}, \,\, x>0\], thus giving you \(\theta=1/0.7 = 10/7\)

So,\[p=P(Y>1)=\int_1^{\infty}\frac{10}{7} e^{-\frac{10}{7} x}\]

Answer 5

\[p=P(Y>1)=\int_1^{\infty}\frac{10}{7} e^{-\frac{10}{7} y}dy\]
Sorry the pdf of \(f_Y(y)\) should be in terms of "y", not "x" .

Answer 6

So that integral gives you p. Now back to the original problem, to find the prob of 1st success in the 5th observation... this means you have 4 failures and the 5th try is a success. So, X = 4 (we defined X as the # of failures). So, you simply find \(f_X(4)=(1-p)^4p\)

Answer 7

I ve got e^(-10/7) as p. Does it seem right?

Answer 8

Yes

Answer 9

I hope it's clear

Answer 10

Yeah Thanks. I will try part b and c too.

Answer 11

:)

Answer 12

I ve got wrong! :(
SO I said part a is 0.182(2dp) if you calculated fX(4)=(1−p)4p.

Answer 13

Isn't it P(x=5) instead of 4? Or it that just my calculating mistake?

Answer 14

What's 2dp?

Also, you can define the geometric distributions in two ways, whether X is the number of TRIALS (here X = 5), or X is the # of FAILURES before the 1st success (here X = 4).

If X = # trials, then \(f_X(x)=(1-p)^{x-1}p, \,\,\, x=1,2, 3, \ldots \)

If X = # failures, then \(f_X(x)=(1-p)^xp, \,\,\, x=0,1,2, \ldots \)