# Homework Help: Finding the lifespan of a telescope mirror (probability question)

1. Aug 30, 2011

### Apple&Orange

1. The problem statement, all variables and given/known data

A telescope contains 3 large mirrors. The time (in years) until a single mirror fails has been investigated and we know that the probability that a mirror is still fully functional after t years is e^(-(t/10)^5 )

a) All mirrors must be working to take the most detailed photographs. What is the probabilty that the telescope ca produce these types of pictures for at least 5 years?

b) The lowest resolution photographs can be taken as long as at least one mirror is working. What is the probability that these photographs can be taken for at least 7 years?

c) The most common photographs the telescope will be taking are of medium resolution. This is possible as long as at least two mirrors are working. What is the probability that this remains possible for at least 6 years?

2. Relevant equations

3. The attempt at a solution

a)

e^(-(4/10)^5 ) = 0.9898122503
0.9898122503 × 1/3 = 0.3299374168
P(X ≥ 5) = (1 - 0.3299374168) = 0.6700625832
Probability that all mirrors are functional for at least 5 years is 67%

b)
e^(-(6/10)^5 ) = 0.9251864446
P(X ≥ 7) = (1 – 0.9251864446) = 0.07481355535
Probability that at least one mirror is functional for at least 7 years is 7%

c)
e^(-(5/10)^5 ) = 0.9692332345
0.9692332345× 2/3 = 0.6461554897
P(X ≥ 6) = (1 – 0.6461554897) = 0.3538445103
Probability that at least 2 mirrors is functional for at least 6 years is 35%

I'm not too sure if this is the right working, could someone please verify this for me?

Last edited: Aug 30, 2011
2. Aug 30, 2011

### Ray Vickson

It's all wrong. Why do you suppose time must be measured exactly in integer numbers of years? Can't a mirror operate for 6.7375 years?

In part (a), why do you divide by 3? In part (b), why do you say P{X >= 7} = 1 - p? Doesn't your calculation just talk about one mirror, instead of 2 out of 3? In part (c), why do you compute something and then multiply it by 2/3?

Are you actually using well-established probability rules, or are you just making things up at random?

By the way: my answers are (a): 0.9105; (b): 0.9963; (c): 0.9840. And NO: I won't tell you how I got them, but I am willing to give hints here and there.

Hint 1: assume mirror failures are independent; that is, the failure (or not) of one mirror has no effect on failure probabilities of the others.

RGV

3. Aug 30, 2011

### Apple&Orange

Heres my logic

In question a), it asks for the probability of "at least for 5 years". To me, it says find the probability between 5 $\leq$ X $\leq \infty$.
Since I can't substitute infinity into t, so what I did was put in 4 instead. This way I can was hoping to find the probability of the mirror lasting less than 5 years.
Right now I realised a mistake in my already mistaken answer, but what I was hoping to do was 1-(probabilty of the mirror lasting less than 5 years), which would give me the probabilty of it lasting 5 or more years.
After that, I times it by 1/3, to calculate the probabilty of all three mirrors lasting since the given formula only applies to 1 mirror.

4. Aug 30, 2011

### Ray Vickson

The question says that exp(-(t/10)^5) is the probability the mirror lasts at least t years, so just putting in t = 5 gives you P{lifetime >= 5 yr}. There is no need to fool around with "infinity" or 4 or anything else. In (a), all three mirrors must last at least 5 years, and individual mirror failures (or not) are independent. What do you know about probabilities of independent events?

RGV

5. Aug 30, 2011

### Apple&Orange

Oh I see what you've done.

Independant Events are probabilities that either one occurs is not affected by the occurance of the other P(A $\cup$ B) =P(A)P(B)

So for question one; [e^-(5/10)^5]*[e^(5/10)^5]*[e^(5/10)^5] = 0.9105
question two; [1-P(All less than 7 years)] = [1 - 0.154705338^3] = 0.9962973224 :rofl:

Got it, thank you very much for your help =)