Lightbulb question

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mussi

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A room has two lamps that use bulbs of type A and B, respectively.
The lifetime, X, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:

for type-A Bulbs: fX(x) =
e^−x, if x ≥ 0,
0, otherwise;

for type-B Bulbs: fX(x) =
3e^−3x, if x ≥ 0,
0, otherwise.

Both lamps are lit at time zero. Whenever a bulb is burned out it is immediately
replaced by a new bulb.
(a) What is the expected value of the number of type-B bulb failures until time t?

The time is infinity? how is this possible?

(b) What is the PDF of the time until the first failure of either bulb type?

lamdatotal = lamda 1 + lamda2
4te^(-4t) ?

(c) Find the expected value and variance of the time until the third failure of a
type-B bulb.

???
(d) Suppose that a type-A bulb has just failed. How long do we expect to wait until a subsequent type-B bulb failure?

??
 
You seem to be totally confused. Given the question "(a) What is the expected value of the number of type-B bulb failures until time t?" you ask "The time is infinity? how is this possible?"
No, the time is NOT "infinity". I don't know where you got this idea. "t" in this problem is a variable which can be any positive real number. But not "infinity".

The next question is "(b) What is the PDF of the time until the first failure of either bulb type?"
and you have:
"lamdatotal = lamda 1 + lamda2
4te^(-4t) ?"
Okay, why are you adding the two lambdas? Part of your solution should be an explanation of why you are using specific formulas or operations. I'm not saying this is wrong- I am saying you should have a reason for adding the lambdas and that reason should tell you whether this is right or wrong.

For the last two you simply have question marks. Is this work not for a course you are taking? Have you not learned something about these? Do you know what "expected value" and "variance" mean?
 
Last edited:
mussi said:
A room has two lamps that use bulbs of type A and B, respectively.
The lifetime, X, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:

for type-A Bulbs: fX(x) =
e^−x, if x ≥ 0,
0, otherwise;

for type-B Bulbs: fX(x) =
3e^−3x, if x ≥ 0,
0, otherwise.
Click to expand...
Waiting to see your response to HallsofIvy - we need to see your work!

These are exponential distributions - can you find half-life? or Mean Time Between Failures (MTBF)?

What methods do you know? What have you tried?

 
Given the hints and my work

a) E[x] = lamda x t = 3t
# of arrivals given time period

b)
added/merged the possion process because it said either. Thus lamda type 1 = 1 lamda; type 2= 3.
thus:
4te^(-4t)


c)
E[time 3rd failure type B] = k/lamda = 3/3 = 1
V[time 3rd failure type B] = k/lamda^2 = 1/3
3 arrivals/rate

d)
Type A doesnt depend on type B so
E[Type B fail] = 1/3
 
mussi said:
A room has two lamps that use bulbs of type A and B, respectively.
The lifetime, X, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:

for type-A Bulbs: \(\displaystyle f_1(t) = \lambda_1\ e^{−\lambda_1 t}\), if x ≥ 0, 0 otherwise;

.........................where \(\displaystyle \lambda_1 = 1\)

for type-B Bulbs: \(\displaystyle f_2(t) = \lambda_2\ e^{−\lambda_2 t}\), if x ≥ 0, 0 otherwise;

.........................where \(\displaystyle \lambda_2 = 3\)

Both lamps are lit at time zero. Whenever a bulb is burned out it is immediately
replaced by a new bulb.
(a) What is the expected value of the number of type-B bulb failures until time t?

(b) What is the PDF of the time until the first failure of either bulb type?

(c) Find the expected value and variance of the time until the third failure of a
type-B bulb.

(d) Suppose that a type-A bulb has just failed. How long do we expect to wait until a subsequent type-B bulb failure?
Click to expand...
I have rewritten the two PDFs, hopefully appropriately! I think it is important to show lambda_1 explicitly, so that the units of exponents are dimensionless. Both lambda_1 and lambda_2 have units of 1/time. I have changed the dummy variable name from "x" to "t" to emphasize that these are time distributions. Note that these are exponential distributions of lifetimes - NOT Poisson distributions of numbers of events.
mussi said:
a) E[x] = lamda x t = 3t....OK
# of arrivals given time period

b)
added/merged the possion process because it said either. Thus lamda type 1 = 1 lamda; type 2= 3.
thus:
4te^(-4t) ....OK - but I would say "exponential" instead of "Poisson" when talking about times


c)
E[time 3rd failure type B] = k/lamda = 3/3 = 1....OK
V[time 3rd failure type B] = k/lamda^2 = 1/3....I assume you found a formula for this
3 arrivals/rate

d)
Type A doesnt depend on type B so
E[Type B fail] = 1/3....OK
Click to expand...
Very good. Just keep track of the difference between time between events, and probable number of events.
 
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