Binomial probability question (1 Viewer)

sadpwner

Member
Joined
Feb 12, 2013
Messages
242
Gender
Male
HSC
N/A
A machine is powered by three similar batteries and will function provided that two of the batteries are in working order. The probability of any one battery failing during the first eight hours of operation of machine is 1/5, the probability of failure during the next eight ours is 2/5.
Find the chance that the machine will continue to function for eight hours and 16 hours.

For 16 hours, how do you do this and why don't you use the probability from 8 hours at all? The answer is 0.648
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
A machine is powered by three similar batteries and will function provided that two of the batteries are in working order. The probability of any one battery failing during the first eight hours of operation of machine is 1/5, the probability of failure during the next eight ours is 2/5.
Find the chance that the machine will continue to function for eight hours and 16 hours.

For 16 hours, how do you do this and why don't you use the probability from 8 hours at all? The answer is 0.648
One way to do it is to find the probability of the machine failing during the first 16 hours, and then subtracting this from 1.

To find that probability of failure, you can do it by cases. Failure occurs if either we have

• 0 batteries fail during the first 8 hours, and 2 OR 3 fail during the next 8 hours
• 1 battery fails during the first 8 hours and 1 OR 2 out of two that remain fail during the next 8 hours (so use n = 2 for the binomial probability calculation of the second 8-hour block for this).
• 2 or 3 batteries fail during the first 8 hours.

Calculate these and add them all up to get the probability of failure, then subtract from 1.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top