# Homework Help: Probability that smoke detectors work

1. Jun 28, 2007

### TSN79

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

A smoke detector has a 92% chance of functioning properly.
If there are 3 of these, what are the chances at least one of them will go off during a fire?

2. Relevant equations

Standard binomial distribution I believe. X~B(n, p)

3. The attempt at a solution

P(X>=1) = P(X=3)+P(X=2)+P(X=1)
When placing this into the equation I end up with ~0.996.
Can someone verify?

2. Jun 28, 2007

### Brinx

I'm getting a different answer, you probably made a calculation error in there.

Consider: if you want to calculate the probability of at least one detector going off, it's easier to work 'the other way round'. You can calculate the probability that none of the smoke detectors will go off, and then apply that figure in a certain way to solve the problem. How exactly to do that is something I'll leave up to you to figure out for the moment! It's less work than what you did, in any case.

I believe you should arrive at an exact answer of 0.999488. Of course, take care in rounding off that figure to the correct number of decimals!

3. Jun 28, 2007

### HallsofIvy

What is the probability that one of them will not go off? What is the probability that none of them will go off? What is the probability that that will not happen?

4. Jun 28, 2007

### TSN79

Are you saying I used correct procedure but made some arithmatic error?

5. Jun 28, 2007

### Brinx

Yes, I think so. When I calculate those separate chances (P(X=1) + P(X=2) + P(X=3)) I still arrive at the answer I mentioned. Could you post your calculations here? We'll be able to point out where things go wrong.

6. Jun 28, 2007

### TSN79

Thx, I got the answer right now. Don't really know what I did wrong the first time...

7. Jun 28, 2007

### Curious3141

You're doing it in an inefficient way. Fine for 3 trials, but what if you're given 100 trials (say)? Will you manually add 100 terms?

Consider HallsofIvy's comment. That's the standard way to approach this sort of problem, saves a lot of work.