# Discrete Probability: Mean Number of Red Bulb Non-Failure Cycles

• Astudious
Yes.The probability that the bulb operates for n cycles without failure is P(No Fail)^n = (1-pR)nYes, but that is without stipulating what happens next. I.e. it's probability of at least n cycles without failure...So then E(N)=Σ(n*(1-pR)n)... but here you have used it as though it is the prob of exactly n cycles.Yes.OK, but my original question still stands. I can see two possible scenarios:(1) The unit keeps functioning until all its lights fail, in which case the distribution of the number of cycles until red fails will be given by

## Homework Statement

A traffic signal operates in the following cyclic regime: amber (A) light for 5 seconds,
then red (R) for 30 seconds, then amber again for 5 seconds, then green (G) for 40 seconds
(thus making a cycle ARAG), and then in the cyclic manner, i.e. ARAGARAG... .

Let us assume that the amber, green and red bulbs can fail every time they switch on
with independent probabilities pA, pG and pR, respectively.

What is the mean number of cycles of non-failure operation of the red bulb?

## Homework Equations

I suppose the E(X)=Σ(x*P(X=x)) over all valid x is probably relevant here.

## The Attempt at a Solution

The probability that the bulb operates for n cycles without failure is P(No Fail)^n = (1-pR)n. So then E(N)=Σ(n*(1-pR)n) where n is summed from 0 to infinity perhaps? But I wouldn't be able to reduce this (and from the solution, it isn't right anyway ...)

Astudious said:

## Homework Statement

A traffic signal operates in the following cyclic regime: amber (A) light for 5 seconds,
then red (R) for 30 seconds, then amber again for 5 seconds, then green (G) for 40 seconds
(thus making a cycle ARAG), and then in the cyclic manner, i.e. ARAGARAG... .

Let us assume that the amber, green and red bulbs can fail every time they switch on
with independent probabilities pA, pG and pR, respectively.

What is the mean number of cycles of non-failure operation of the red bulb?

## Homework Equations

I suppose the E(X)=Σ(x*P(X=x)) over all valid x is probably relevant here.

## The Attempt at a Solution

The probability that the bulb operates for n cycles without failure is P(No Fail)^n = (1-pR)n. So then E(N)=Σ(n*(1-pR)n) where n is summed from 0 to infinity perhaps? But I wouldn't be able to reduce this (and from the solution, it isn't right anyway ...)

Please clarify: what happens when a colored bulb fails? Does the unit continue to function--but using only the other, non-failed bulbs--or does it stop functioning altogether when any of its bulbs fails? If it does continue to operate, does it just "by-pass" the failed bulb, or does it still occupy the time of the failed bulb, but without light? For example, if the first Amber fails (but the unit keeps operating) does that collapse to a RAGRAG... situation with each cycle shortened by the missing 5 seconds, or does it remain at BRAGBRAG..., where now B = blank (lasting 5 sec)?

Ray Vickson said:
Please clarify: what happens when a colored bulb fails? Does the unit continue to function--but using only the other, non-failed bulbs--or does it stop functioning altogether when any of its bulbs fails? If it does continue to operate, does it just "by-pass" the failed bulb, or does it still occupy the time of the failed bulb, but without light? For example, if the first Amber fails (but the unit keeps operating) does that collapse to a RAGRAG... situation with each cycle shortened by the missing 5 seconds, or does it remain at BRAGBRAG..., where now B = blank (lasting 5 sec)?
I believe there is only one Amber bulb.
I read the question as referring to cycles of the red bulb, so at this stage what the other bulbs are doing is irrelevant. I assume this is one part of a multipart question. Astudious, can you confirm that?

haruspex said:
I believe there is only one Amber bulb.
I read the question as referring to cycles of the red bulb, so at this stage what the other bulbs are doing is irrelevant. I assume this is one part of a multipart question. Astudious, can you confirm that?

Yes.

Astudious said:
The probability that the bulb operates for n cycles without failure is P(No Fail)^n = (1-pR)n
Yes, but that is without stipulating what happens next. I.e. it's probability of at least n cycles without failure...
Astudious said:
So then E(N)=Σ(n*(1-pR)n)
... but here you have used it as though it is the prob of exactly n cycles.

Astudious said:
Yes.

OK, but my original question still stands. I can see two possible scenarios:
(1) The unit keeps functioning until all its lights fail, in which case the distribution of the number of cycles until red fails will be given by some probability distribution A.
(2) The unit functions only as long as none of the lights have failed, and in that case the distribution of the number of cycles until red fails (if ever) is some other probability distribution, B. (In that case, "red fails" = "red no longer shines" = "anything fails".)

My guess is that he/she means (1), but asking for clarification does no harm.