Question 1166037
this looks like it's a binomial distribution probability type of problem.


the formula for binomial distribution probability is:


p(x) = a ^ x * b ^ (n-x) * c(n,x)


c(n,x) is equal to n! / (x! * (n-x)!)


in your problem:
n is equal to 5.
x is equal to any integer from 0 to 5 inclusive.
a is equal to the probability that the life of a randomly selected battery is less than 130 hours.
b is equal to the probability  that the life of a randomly selected battery is greater than 130 hours.
the probability of b would be equal to 1 minus the probability of a.


using a normal distribution calculator such as the one found at <a href = "https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html" target = "_blank">https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html</a>, the probability that any randomly selected battery will have a life less than 130 hours is equal to 0.09121.
the probability of a randomly selected battery having a life greater than 130 hours is therefore equal to 1 - .09121 = .90879.


here's a display of the calculator results for the probability of a random selected battery having a life of less than 30.


<img src = "http://theo.x10hosting.com/2020/092802.jpg" >


the binomial distribution probability formula becomes:


p(x) = .09121 ^ x * .90879 ^ (5-x) * c(5,x)


the flashlight will not operate for more than 130 hours if 1 or more of its batteries dies in less than 130 hours.


this means that the flashlight will operate for more than 130 hours if 0 of its batteries dies in less than 130 hours.


what you are then looking for is x equal to 0 in the formula of p(x) = .09121 ^ x * .90879 ^ (5 - x) * c(5,x)


the formula becomes p(0) = .09121 ^ 0 * .90879 ^ (5 - 0) * c(5,0).


c(5,0) is equal to 5! / (0! * 5!) which is equal to 5! / 5! which is equal to 1.


the formula becomes p(0) = .09121 ^ 0 * .90879 ^ 5 which is equal to .6198943783.


that means that the probability that 0 batteries will fail in less than 130 hours is .6198943783.


if 0 batteries fail in less than 130 hours, then the flashlight will be operating for more than 130 hours.


i believe that .6198943783 is the answer they are looking for.


it is the probability that 0 of the flashlight batteries dies in less than 130 hours.


you can round that answer to whatever number of decimal points they require.