Find the probability of this problem...

A certain mechanical device has four independent parts to it... If any 1 part fails, the is no longer functional. Suppose the 4 year fail rate is as follows:

Part 1: .01
Part 2: .06
Part 3: .04
Part 4: .02

What is the probability the device will function for 4 years?

What I did was add all of the probabilities together...

.01+.06+.04+.02 = .13 Then subtract it from 1 and got .87.

However I got that problem wrong and the teacher wrote "Failure rates aren't mutually exclusive".

I see why that doesn't make sense now I am trying a different approach.

Probability of them not failing. .99 .94 .96 .98

and multiply them together .99*.94*.96*.98 and I got a total of .876 as the probability it will function all 4 years.

Does this work?

Thanks.

Alpha6 said:

A certain mechanical device has four independent parts to it... If any 1 part fails, the is no longer functional. Suppose the 4 year fail rate is as follows:

Part 1: .01
Part 2: .06
Part 3: .04
Part 4: .02


Probability of them not failing. .99 .94 .96 .98

and multiply them together .99*.94*.96*.98 and I got a total of .876 as the probability it will function all 4 years.

Does this work?

Thanks.

Click to expand...

Good idea - but it looks to me like you have the probability of working after ONE year. One more step needed to extend the warranty to 4 years . . .

DrPhil said:

Good idea - but it looks to me like you have the probability of working after ONE year. One more step needed to extend the warranty to 4 years . . .

Click to expand...

Sorry, so many components to this its so easy to miss one... lol

I'm thinking I'd raise .876 to the fourth power...

.876^4 = .5889 ??

Alpha6 said:

Sorry, so many components to this its so easy to miss one... lol

I'm thinking I'd raise .876 to the fourth power...

.876^4 = .5889 ??

Click to expand...

Yes - but I wouldn't keep so many decimal places. Actually, if you don't round off the 1-year probability, the 4th power is 0.5875. Just a simple 59% would be all the precision I would ask for.