The probability is 0.6 that a single burglar alarm will detect a burglar. How many alarms should be used to be at least 90% certain that a burglar is detected?

What formula do I use to solve this ?

June must take a physical exam and a written exam. The probabilty of passing a written exam is 0.82. The probability that a candidate passes the written exam given that he or she has passed the physical exam is 0.58. What is the probability that she passes both exams?

How do I go about solving this one?

Okay so your first alarm catches 0.6 of a burglar. Now y our 2nd alarm catches 0.4*0.6 of a burglar (that is the chance of the first one failing times the chance of this one working). Then your third alarm catches 0.4*0.4*0.6 of a burglar and so on.

So you need the point where 0.6 + 0.4*0.6 + 0.4*0.4*0.6 + 0.4*0.4*0.4*0.6... is above 0.9.

I believe that this is 0.936 with 3 alarms.

The 2nd question is quite simple. I think I would give the question away if I tried to help you. Hopefully my above answer to the first question will help you see how to combine probabilities. How do you think you would solve it?

For the 1st problem I find it sometimes helpful to think about the inverse question - this can be especially useful when you are trying to find the probability where any one positive outcome means success. In this case Capuchin has shown how to calculate the probablilty that either the first alarm works, or if the first alarm fails that the second works, or if both the first and second fail that the third works, etc. It might be easier to think of the opposite question - that is, how many alarms do you need so that the probability of all the alarms working together and yet NOT catching the burglar is less than 0.1. This is an easier thing to calculate, as the probability that any one alarm will fail is 1- .6 = .4, and the probabilty that N alarms will all fail is 0.4 x 0.4 x 0.4 X... (N terms). So how many times do you need to multiply 0.4 by itself to generate a number that is less than 0.1? You get the same answer as Capuchin gave, but this approach might be a little easier to envision.

As for the 2nd problem - I could be wrong but I think this is a bit more complicted than Capuchin suggested. It looks to me like a conditional probability problem, where you would use the form P(W & P) = P(W|P) x P(P), where P(P&W) is the probability of passing both exams, P(W|P) is the probaility of passsing the written given that you pass the physical, and P(P) means passing the physical. However, the data you gave doesn't quite match the terms here ,which makes me wonder - are you sure you wrote the problem correctly?

As A Security System Installer not College Taught( Never got into probability) . For Maximum Security, 2 systems can provide Very effictive system.
One system might use 2 panels with supvervised loops protecting each other.( Has to have certain resistance. Jumping zone will violate zone. Another migh be a second system,
hid. Could be anywhere. Also Use Various methods getting signal out. Land line, Network communicator, RF, Satellite. Fiber. Metal detectors can't locate fiber. I also like to use small fuse at 2 sirens and 1 lrger fuse at panel. If someone rips siren off wall, will trip tamper, shorting of fuse will disable 1 siren. The probality formula should include Installer
capability, creativeness, and Negative points for Maximum Profit Companies(Bottom Line),Tend to use Cheap.
Another thing is second recessed magnetic contact. Close enough, so if someone places magnet near trying to bypass, it will trip second contact(24 hour zone)
Loud interior sirens can be annoying.
Burglars like Dark and Quiet(sometimes)